Primitive abundant and weird numbers with many prime factors
Gianluca Amato, Maximilian F. Hasler, Giuseppe Melfi, Maurizio Parton
aa r X i v : . [ m a t h . N T ] F e b PRIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS
GIANLUCA AMATO, MAXIMILIAN F. HASLER, GIUSEPPE MELFI, AND MAURIZIO PARTON
Abstract.
We give an algorithm to enumerate all primitive abundant numbers (briefly, PAN) with a fixed Ω(the number of prime factors counted with their multiplicity), and explicitly find all PAN up to Ω = 6, countall PAN and square-free PAN up to Ω = 7 and count all odd PAN and odd square-free PAN up to Ω = 8. Wefind primitive weird numbers (briefly, PWN) with up to 16 prime factors, improving the previous results of [1]where PWN with up to 6 prime factors were given. The largest PWN we find has 14712 digits: as far as weknow, this is the largest example existing, the previous one being 5328 digits long [14]. We find hundreds ofPWN with exactly one square odd prime factor: as far as we know, only five were known before. We find allPWN with at least one odd prime factor with multiplicity greater than one and Ω = 7 and prove that there arenone with Ω <
7. Regarding PWN with a cubic (or higher power) odd prime factor, we prove that there arenone with Ω ≤
7, and we did not find any with larger Ω. Finally, we find several PWN with 2 square odd primefactors, and one with 3 square odd prime factors. These are the first such examples. Introduction
Let n ∈ N be a natural number, and let σ ( n ) = P d | n d be the sum of its divisors. If σ ( n ) > n , then n is called abundant , whereas if σ ( n ) < n , then n is called deficient . Perfect numbers are those for which σ ( n ) = 2 n . If n is abundant and can be expressed as a sum of distinct proper divisors, then n is called semiperfect , or sometimesalso pseudoperfect . A weird number is a number which is abundant but not semiperfect.If n is abundant and it is not a multiple of a smaller non-deficient number than n is called a primitive abundantnumber , PAN in this paper. Similarly, a primitive weird number , PWN in this paper, is a weird number whichis not a multiple of any smaller weird number.In this paper, we study primitive abundant and weird numbers.Leonard Eugene Dickson, in two papers from 1913 on the American Journal of Mathematics [7, 8], proves thatthe sets of PAN having any given number ω of distinct prime factors is finite (for even PAN, one also needs tofix the exponent of 2). He then explicitly finds all odd PAN with ω ≤
4, and all even PAN with ω ≤
3, seealso [10, 13] for errata in Dickson’s tables. The number of odd PAN with ω = 5 was found in 2017, while for ω ≥ ω = 6 is given).Motivated by the above discussion, in this paper we focus on the set of PAN with a given number Ω of primefactors counted with their multiplicity . In this way we have been able to explicitly find all PAN with Ω up to 6,to count PAN with Ω ≤
7, and to count odd PAN with Ω ≤
8, see Table 2 and OEIS sequences A298157 andA287728. All these results are new, to the best of our knowledge.Weird numbers were defined in 1972 by Stan Benkoski [3], and appear to be rare: for instance, up to 10 wehave only 7 of them [17]. Despite this apparent rarity, which is the reason for the name, weird numbers areeasily proven to be infinite: if n is weird and p is a prime larger than σ ( n ), then np is weird (see for example[11, page 332]). But a much stronger property is true: Benkoski and Paul Erd˝os, in their joint 1974 paper [4],proved that the set of weird numbers has positive asymptotic density. Nevertheless, it is not yet known whetherPWN are infinite: Conjecture 1.1. [4, end of page 621]
There exist infinitely many PWN.
The search for new PWN is made more interesting by the fact that we still do not know whether they areinfinite. The list of the first PWN is regularly updated by Robert G. Wilson v, and at the time of writing(February 2018) the first 1081 PWN are known, see OEIS A002975.
Mathematics Subject Classification.
Primary 11A25; Secondary 11-04, 11N25, 11Y55.
Key words and phrases. deficient numbers, abundant numbers, primitive abundant numbers, weird numbers, primitive weirdnumbers, sum-of-divisor function.The third and the fourth author were supported by the GNSAGA group of INdAM and by the MIUR under the PRIN Project“Variet`a reali e complesse: geometria, topologia e analisi armonica”.
Looking for the largest possible PWN is also very interesting. One approach is to consider patterns in the primefactorization of PWN, see [1]. At the time of writing (February 2018), only 12 PWN with 6 distinct primefactors [1] and just one PWN with 7 distinct prime factors [5] are known.In this paper we dramatically improve these figures. We find hundreds of PWN with more than 6 distinct primefactors. In particular, we find PWN with up to 16 distinct prime factors, see Tables 3 and 4. The largest PWNwe have found has 16 distinct prime factors and 14712 digits. As far as we know, this is the largest PWNknown, the previous one being 5328 digits long [14].Another strange behavior in the prime decomposition of PWN is the fact that only five PWN with non square-free odd prime factors were known, see OEIS sequence A273815, and no PWN with an odd prime factor ofmultiplicity strictly greater than two is known. We explain this fact with Theorem 4.7:
Theorem (PWN with non square-free odd part and Ω ≤ . There are no PWN m with a quadratic or higherpower odd prime factor and Ω( m ) < . There are no PWN m with 2 quadratic odd prime factor and Ω( m ) = 7 .There are no PWN m with a cubic or higher power odd prime factor and Ω( m ) = 7 . These results are new, to the best of our knowledge.Finally, in this paper we find hundreds of new PWN with a square odd prime factor, see Table 5 for a selectionof them. We find several new PWN with 2 square odd prime factors, and one with 3 square odd prime factors,see Table 6. These are the first examples of this kind.In the following, we describe with some details the methods of the paper.In Section 2 we start with a careful analysis of the set A Ω of PAN m with a fixed number Ω = Ω( m ) of primefactors counted with their multiplicity. These sets are finite, a corollary of Dickson’s theorems [7, 8], see alsoTheorem 3.2. The main result in this section is Theorem 2.8 about the structure of PAN: they are of the form mp e , where p is a prime larger than the largest prime factor of m , and m is a deficient number satisfying certainconditions involving the center c ( m ) = σ ( m ) / (2 m − σ ( m )) of m , see Definition 2.6. Note that some results inthis section are easy consequences of the definitions, and some of them are well-known: however, we leave themin the paper because we will use them extensively in Sections 3 and 4.In Section 3 we face the problem of explicitly computing A Ω , or some statistics on it, for specific values of Ω.Here we distinguish the square-free case from the general case, since the former appears to be notably simplerthan the latter.Every square-free PAN is then given by p · · · p k − p k for certain primes p < · · · < p k , and p · · · p i is recursivelybuilt from p · · · p i − by imposing p · · · p i deficient. This gives an explicit construction for A Ω in the square-freecase, see Algorithm 1. However, since the condition for p · · · p i to be deficient is open (see Proposition 2.5),we need a termination condition. This is done by exploiting Theorem 3.3, stating that a deficient sequence ofprimes m = ¯ p e · · · ¯ p e r r can be completed to a PAN mpq for suitable primes p, q . Applying this machinery weexplicitly find all square-free PAN with Ω ≤
6, count the square-free PAN with Ω ≤ ≤
8, see Table 1 and OEIS sequences A295369 and A287590.Adapting the techniques to the non square-free case essentially means allowing consecutive primes in m = p , . . . , p k to be equal, and being more careful in identifying which sequences of primes give origin to PAN. Asalready said, we explicitly find all PAN with Ω ≤
6, count PAN with Ω ≤
7, and count odd PAN with Ω ≤ k prime factors for large k , it is not computationallyfeasible to explicitly find all PAN and then check for weirdness. Thus, when computing the deficient seed m = p · · · p k − , we choose an amplitude a and limit the choice for p i to the first a primes larger than c ( p · · · p i − ),and the choice for p k to the last a primes smaller than c ( p · · · p k − ). In order to be able to deal with the hugenumbers involved, we represent them in a form we have called index sequence , see Definition 4.4. Finally, inRemarks 4.5, 4.6, and in Section 5, we explain what we noticed from experiments, as a possibly useful futurereference. The new findings in this section are: PWN with more than 6 distinct prime factors, PWN with upto 16 distinct prime factors, see Tables 3 and 4, PWN with a square odd prime factor, see Table 5, PWN with2 and 3 square odd prime factors, see Table 6, and Theorem 4.7 on patterns for PWN.Note that the problem of finding a PWN with a cubic or higher power odd prime factor is still open. This, andother open questions, are listed in Section 5.All the software we have developed and results of our experiments are available on-line at the GitHub repository https://github.com/amato-gianluca/weirds . RIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS 3 Deficient, perfect and abundant numbers
In line with [12], we will refer to ∆( n ) := σ ( n ) − n as the abundance of n , and to d ( n ) := 2 n − σ ( n ) = − ∆( n )as the deficiency of n . It is sometimes convenient to use the notation σ ℓ ( n ) := P d | n d ℓ for the sum of the ℓ -thpowers of divisors, so that σ is the number of divisors including 1 and the number itself, σ := σ their sum,and σ − ( n ) = σ ( n ) /n is the abundancy of n . One can characterize deficient, perfect and abundant numbersrespectively by σ − ( n ) < σ − ( n ) = 2, σ − ( n ) > n = p e · · · p e k k with p < · · · < p k primes, then for each p i we can choose an exponent from 0 to e i to build adivisor of n . Therefore, the function σ ℓ is multiplicative on prime factors, that is, we have:(1) σ ℓ ( n ) = k Y i =1 (1 + p ℓi + · · · + p ℓe i i ) = k Y i =1 p ( e i +1) li − p ℓi − k Y i =1 σ ℓ ( p e i i )Moreover, since σ ℓ ( p e ) ≤ σ ℓ ( p ) e , then σ ℓ is sub-multiplicative, i.e., σ ℓ ( mn ) ≤ σ ℓ ( m ) σ ℓ ( n ).If a number is non-deficient (i.e., either perfect or abundant) and all of its proper divisors are deficient, then itis called primitive non-deficient . A primitive abundant number PAN is a primitive non-deficient number whichis also abundant . Proposition 2.1. If m is non-deficient and n ∈ N , n > , then mn is abundant.Proof. If d and d ′ are distinct divisors of m , then dn and d ′ n are distinct divisors of mn . If P d | m d ≥ m , then P d | mn d > P d | m dn ≥ mn . (cid:3) Corollary 2.2.
All perfect numbers are primitive non-deficient.Proof. If m is perfect and n is a proper divisor of m , n should be deficient. Otherwise, by Proposition 2.1, m would be abundant. (cid:3) The following corollary states that, whenever we want to check if m is primitive, we need to look only at asubset of its divisors. Corollary 2.3. If m is abundant and m/p is deficient for all primes p | m , then m is primitive abundant.Proof. If m is not primitive abundant, then there exists some non-deficient number m ′ | m , with m ′ < m .Moreover, there exists a prime p | m such that m ′ | m/p , and since m/p is deficient, by Proposition 2.1 thiscontradicts the fact that m ′ is non-deficient. (cid:3) Proposition 2.4.
Let m = p e · · · p e k k with p < · · · < p k . Choose a position i ≤ k and a prime p such that ( m, p ) = 1 . Let e m be the result of substituting p e i i with p e i in the decomposition of m , i.e., e m = mp e i /p e i i . Then • if m is abundant or perfect and p < p i , then e m is abundant; • if m is deficient or perfect and p > p i then e m is deficient.Proof. Note that replacing p i with p in m results in replacing p i with p in all divisors of m . This means thatwhenever p i appears in a summand of σ − ( m ), it is replaced with p . Thus, σ − ( m ) is decreasing in the p i ’s.Therefore, if m is abundant or perfect and p < p i , we have σ − ( e m ) > σ − ( m ) ≥
2, hence ˜ m is abundant.Similarly for the second case. (cid:3) Note that, if m = p e · · · p e k k is primitive abundant and we replace p e i i with p e i for some p < p i , we are not surethat the number we obtain is primitive abundant (although we know it is abundant). For example, 3 · · · · · ·
103 is not, since 2 · · · · ·
13 is primitive abundant but 2 · · ·
13 is not, since 2 · ·
11 isalready abundant. Some authors define a PAN to be an abundant number with no abundant proper divisors. The two definitions differ on multiplesof perfect numbers. For example, 30 = 2 · · · G. AMATO, M. F. HASLER, G. MELFI, AND M. PARTON
Adding a new coprime factor p e to a deficient number. The following proposition considers theproblem of starting with a deficient number m and adding a new prime factor p e with ( p, m ) = 1. We want tostudy under which conditions mp e is perfect, (primitive) abundant or deficient. The reason we are interestedin this problem is that, in Section 3, we will build PAN by adjoining one prime factor at a time to a startingdeficient number. Proposition 2.5. If m is deficient, e ∈ N and p is a prime such that ( m, p ) = 1 , • if p e /σ ( p e − ) < σ ( m ) /d ( m ) then mp e is abundant; • if p e /σ ( p e − ) = σ ( m ) /d ( m ) then mp e is perfect; • if p e /σ ( p e − ) > σ ( m ) /d ( m ) then mp e is deficient.Proof. With simple algebraic manipulations one can show that d ( mp e ) = d ( m ) p e − σ ( m ) σ ( p e − ). The thesisimmediately follows by imposing d ( mp e ) less, equal or greater than zero. (cid:3) Since the term σ ( m ) /d ( m ) will have a major role in the following, we introduce a more succinct notation. Seealso [7, Formula (10)]. Definition 2.6 (Center of a deficient number) . Given m ∈ N , we call center of m the value c ( m ) := σ ( m ) /d ( m ).Let m be deficient and p a prime such that ( m, p ) = 1 and p < c ( m ). By Proposition 2.5, it turns out that mp is abundant. However, it is not guaranteed to be primitive abundant. Consider for example m = 16, with c ( m ) = 31. If we take p = 7, we have that 16 · · m = 2 · ·
31 = 806. Then5 < c ( m ) <
6. If we take p = 3, then mp is abundant but not primitive abundant, since 2 · ·
13 is abundant.
Proposition 2.7.
The center enjoys the following properties: (1) c ( m ) = 2 md ( m ) − σ − ( m ) − , for any deficient m ∈ N ; (2) if n > and mn is deficient, then c ( mn ) > c ( m ) ; (3) for any prime p , c ( p e ) is increasing in e and lim e → + ∞ c ( p e ) = pp − ; (4) if m is deficient and p, q are primes coprime with m , q > p > c ( m ) , then c ( mq ) < c ( mp ) .Proof. For (1), we have c ( m ) = σ ( m )2 m − σ ( m ) = σ ( m ) − m + 2 m m − σ ( m ) = 2 md ( m ) − c ( m ) = 12 m − σ ( m ) σ ( m ) = 12 σ − ( m ) − σ − ( m ) <
2, we have that c ( m ) is increasing with σ − ( m ).Since σ − ( mn ) > σ − ( m ), under the hypothesis of this proposition, we have c ( mn ) > c ( m ), proving (2). From σ − ( p e ) = 1 − (cid:18) p (cid:19) e +1 − p , lim e → + ∞ σ − ( p e ) = 11 − p = pp − c ( p e ) is increasing in e andlim e → + ∞ c ( p e ) = 12( p − p − pp − d ( mp ) and d ( mq ) are positive, by Propo-sition 2.5, and conclude by an algebraic manipulation of c ( mq ) < c ( mp ). (cid:3) We want to give appropriate conditions ensuring that mp is primitive abundant. We know from Corollary 2.3and Proposition 2.5, that a necessary condition for mp e to be primitive abundant is p e σ ( p e − ) > c ( m/q ) for eachprime q | m . Since our aim is to implement a program to enumerate PAN (see Section 3), we would like toreduce the number of tests we need to perform each time. The following will be useful. RIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS 5
Theorem 2.8 (Structure Theorem for PAN) . Let m be a deficient number, e ∈ N and p a prime such that ( m, p ) = 1 . Then mp e is primitive abundant iff p e /σ ( p e − ) < c ( m ) , p e /σ ( p e − ) > σ ( m ) d ( m ) + 2 mσ ( q α ) − for each q α || m , and either e = 1 or p e − /σ ( p e − ) > c ( m ) .Proof. When e >
1, by Corollary 2.3 and Proposition 2.5, we have that mp e is primitive abundant iff p e /σ ( p e − )
1) 2 m − d ( m ) d ( m )( σ ( q α ) −
1) + 2 m < σ ( q α ) − (cid:3) Remark . Due to the approximations in the previous proof, it is evident that the condition p ≥ σ ( q α ) − m = 8 and p = 7. Although 7 < σ (8) −
1, it turns out that 8 · Corollary 2.11. If p < · · · < p k are primes such that m = p · · · p k is deficient, p > p k is a prime such that mp is abundant, then mp is primitive abundant.Proof. Since p > p k then ( m, p ) = 1. Moreover, for each p i , we have p i || m , hence p ≥ σ ( p i ) − p i . Thethesis follows from Corollary 2.9. (cid:3) Adding any prime factor to a deficient number.
We now consider the case when we start with adeficient number m and add a prime factor p with p α || m . We want to study under which conditions mp isperfect, (primitive) abundant or deficient.First of all, consider that Proposition 2.5 does not hold when ( m, p ) = 1. For example, for m = 10 = 2 · c ( m ) = 9 but 2 · is deficient. We may change Proposition 2.5 in the following way: Proposition 2.12. If m is deficient and p is a prime such that p α || m , then • if pσ ( p α ) < c ( m ) then mp is abundant; G. AMATO, M. F. HASLER, G. MELFI, AND M. PARTON • if pσ ( p α ) = c ( m ) then mp is perfect; • if pσ ( p α ) > c ( m ) then mp is deficient.Proof. We have that d ( mp ) = 2 mp − σ ( mp ) = 2 mp − σ ( m ) p α +2 − p α +1 − = 2 mp − σ ( m )( p + p − p α +1 − ) = d ( m ) p − σ ( m ) p − p α +1 − = d ( m ) p − σ ( m ) /σ ( p α ). (cid:3) Example 2.13. If m = 2 ·
5, we have c ( m ) = 9 but 5 · σ (5) = 30 hence m · m = 2 · · · · < c ( m ) < · σ (61) = 3782 < c ( m ), we have that m ·
61 is abundant.We may also adapt Theorem 2.8 to the case when p is not coprime with m as follows: Theorem 2.14. If m is deficient and p is a prime such that p α || m , we have that mp is primitive abundantiff pσ ( p α ) < c ( m ) and pσ ( p α ) > σ ( m ) d ( m ) + 2 mσ ( q β ) − for each q β || m with q = p .Proof. By Corollary 2.3 and Proposition 2.12, it is immediate that mp is primitive abundant iff pσ ( p α ) < c ( m )and pσ ( p α ) > c ( m/q ) for each prime q | m with q = p . In the proof of Theorem 2.8 we have shown that c ( m/q ) = σ ( m ) d ( m ) + 2 mσ ( q β ) − (cid:3) We may repeat the same considerations we have made for Theorem 2.8, regarding the fact we might onlyconsider the largest q β || m . Moreover, Equation 3 still holds.3. Enumerating primitive abundant numbers
Theorems 2.8 and 2.14 allow us to devise an algorithm for enumerating PAN or, more generally, primitivenon-deficient numbers. We will enumerate PAN on the basis of their factorization. For this reason, when m = p e · · · p e k k , we will always assume p < · · · < p k . Moreover, we will denote with ω ( m ) := k the number ofdistinct prime factors in m and with Ω( m ) := e + · · · + e k the number of prime factors in m counted with theirmultiplicity.Note that, if we fix the number of prime factors counted with multiplicity, then enumeration terminates, thanksto the following results. Lemma 3.1.
Given a number m and k ≥ , there are only finitely many PAN of the form mn with ( m, n ) = 1 and Ω( n ) = k .Proof. By induction on k . For k = 0 the result is trivial, either m is primitive abundant and n = 1 or it is not.If k >
1, we distinguish whether m is deficient or not. If m is not deficient, then mn is never primitive abundantand the lemma holds. If m is deficient, consider an n such that mn is primitive abundant and Ω( n ) = k . Then n has the form p e · · · p e ℓ ℓ with p < · · · < p ℓ and P ℓi =1 e i = k . Since mn is abundant, σ − ( n ) > /σ − ( m ).However, the abundancy of n is bounded by σ − ( n ) = σ − ( p e ) · · · σ − ( p e ℓ ℓ ) ≤ σ − ( p e ) · · · σ ℓ ( p e ℓ ) ≤ (1 /p + 1) k Therefore, (1 /p + 1) k > /σ − ( m ), i.e., 1 /p > k p /σ ( m ) −
1. Since m is deficient, σ − ( m ) <
2. Hence, theright hand side of this inequality is positive and p is bounded from the above. Given one of the finitely many p satisfying this condition and e ∈ { , . . . , k } , by inductive hypothesis there are only finitely many n ′ coprimewith mp e , with Ω( n ′ ) = k − e and such that mp e n is abundant. Varying p , these cover all possible values of n in the statement of this lemma. (cid:3) Theorem 3.2.
For any k > , there are only finitely many PAN n with Ω( n ) = k .Proof. This follows immediately from the previous lemma for m = 1. (cid:3) We remark that Theorem 3.2 is a Corollary of [7, 8] about finiteness of PAN with a fixed number of odd primefactors (counted without multiplicity) and a fixed power of 2.
RIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS 7
Square-free PAN.
We consider the special case of enumerating square-free PAN (SFPAN in the rest ofthe paper) with k prime factors. Note that the more general case of primitive square-free non-deficient numbersis not interesting, since it is well-known that there is only one square-free perfect number which is 6.The algorithm is a recursive procedure which takes a deficient number m = ¯ p · · · ¯ p r with ¯ p < · · · < ¯ p r and r < k as input. Initially m = 1. If r < k −
1, for each prime p > c ( m ) we consider the number e m = mp , whichis deficient by Proposition 2.5, and recursively call the procedure. If r = k −
1, then we consider all primes p contained in the possibly empty open interval (¯ p r , c ( m )). By Corollary 2.11, each number of the form mp is aPAN.The algorithm needs a stopping condition in the case r < k −
1, since we cannot actually test all the countablyinfinite primes p > c ( m ). We decide to try primes in increasing order, stopping as soon as we find a p such thatthere are no PAN starting with mp . The complete description may be found in Algorithm 1. The algorithmis easily checked to be correct. Completeness, i.e., the fact that the algorithm finds all SFPAN of the chosenform, will be discussed later. Algorithm 1:
Enumerating SFPAN with k prime factors. Function sfpan( k: nat, m: nat ) isInput: k is a natural number; m = ¯ p · · · ¯ p r is a square-free deficient number, with ¯ p < · · · < ¯ p r Output: all primitive abundant numbers of the form m · p · · · p k , with ¯ p r < p < · · · < p k Result: the number of square-free primitive abundant numbers of the form above count ← if k = 1 then foreach p prime s.t. ¯ p r < p < c ( m ) do Print( mp ) ; count ← count + 1 end return count else foreach p prime s.t. p > max(¯ p r , c ( m )) do innerCount ← sfpan( k-1, mp ) ; count ← count + innerCount ; if innerCount = 0 then return count end end endend When we only want to count PAN, steps 3–6 of the algorithm may be replaced by a prime counting function.Using an implementation in SageMath of the algorithm and the prime counting function provided by KimWalisch’s primecount library, we managed to count the number of SFPAN from 1 up to 7 distinct prime factorsand odd SFPAN from 1 up to 8 distinct prime factors. The result is shown in Table 1 and form the OEISsequences A295369 and A287590.We have also computed a list of SFPAN with up to 6 distinct prime factors, which is available on GitHub at https://github.com/amato-gianluca/weirds . ω Table 1.
Number of SFPAN and odd SFPAN with given number of distinct prime factors.
G. AMATO, M. F. HASLER, G. MELFI, AND M. PARTON
Completeness of the enumeration algorithm.
The critical point of this algorithm is the stoppingcondition. Are we sure we do not loose any PAN? In order to ensure completeness of the search procedure, weneed to prove that, if there is no SFPAN n such that ω ( n ) = k and whose factorization starts with p · · · p r ,then there is no SFPAN m with ω ( m ) = k and whose factorization starts with p · · · p r − · p for any p > p r .We actually prove the contrapositive, i.e., that if p · · · p r − · p · p r +1 · · · p k is primitive abundant and p r −
8. Let p be the smallest prime largerthan c ( m ). Then p > c ( m ) ≥ ¯ p r , and mp is deficient by Proposition 2.5. We need to find a prime q > p suchthat mpq is abundant. This requires q < c ( mp ). We have c ( mp ) = σ ( m )( p + 1)2 mp − σ ( m )( p + 1) = σ ( m )( p + 1) d ( m ) p − σ ( m ) = p + 1 pc ( m ) − x ≥ x and 3 x/ p , using x = c ( m ) in Nagura’s Theorem, we have p < c ( m ) / c ( mp ) > p + 1) = 2 p + 2Again by Nagura’s Theorem (or even weaker results), there is a prime q in the interval ( p, p + 2). Thus, q < p + 2 < c ( mp ), and this concludes the case c ( m ) ≥ c ( m ) <
8, which implies ¯ p r < p r = 7, then c ( m ) ≥
7. Let us take p = 11 and q = 13. Then mp is deficient and c ( mp ) = 12 / (11 /c ( m ) − ≥ / (11 / −
1) = 21, hence mpq is abundant.If ¯ p r = 5, then r = 1, because c (5 e r ) < / (5 −
2) = 5 / e r by (3) in Proposition 2.7. If both 2 and 3 areother factors of m , then m is abundant by Proposition 2.1, because 2 · · m is eitherof the form 2 e e or 3 e e for e , e ≥
1. Since c (2 ·
5) = 9 >
8, we only consider the case m = 3 e e , by (2)in Proposition 2.7. Since c (3 · ) >
8, the only cases remaining are: m = 3 · m = 3 · m = 3 · .However, c (3 ·
5) = 4 < c (3 · ) = 62 / <
5. Therefore, the only m satisfying the hypothesis of thetheorem is 3 ·
5, for which we may take p = 7 and q = 11.If ¯ p r = 3, then r = 1: if 2 also appears as a prime factor in m , then m cannot be deficient since 2 · m = 3 e for some e . However, c (3 e ) < / (3 −
2) = 3, hence m does not satisfy thehypothesis of the theorem.If ¯ p r = 2, then m = 2 e r and σ ( m ) = c ( m ) = 2 e r +1 −
1. If e r ≥ c ( m ) ≥ m = 2, take p = 5 and q = 7; if m = 4, take p = 11 and q = 13. (cid:3) Remark . In the hypothesis of the previous theorem, when m is not square-free, it might not be possible toobtain p, q such that mpq is primitive abundant. Consider m = 3 ·
5, so that 8 < c ( m ) <
9. If we determine p as the smallest prime p > c ( m ) and q as the largest q < c ( mp ) as in Proposition 2.5, we get p = 11 , q = 53 and m · ·
53 which is abundant but not primitive abundant, since 3 · · ·
53 is abundant, too. If we replace 53with smaller primes q the abundance increases, because in general ∆( mq ) − ∆( mq ′ ) = ∆( m )( q − q ′ ) whenever q, q ′ are coprime with m , hence m · · q is abundant and the number we obtain cannot be primitive abundant byProposition 2.4. By increasing p and computing the corresponding largest possible q < c ( mp ), we get m · · m · ·
19, but none of them is primitive abundant. We have [ c ( m · p ≥
19 weget c ( m · p ) ≥ c ( m · >
17 by (4) in Proposition 2.7, and we have no primes q > p making mpq abundant.Even relaxing the condition ¯ p r < p < q into ¯ p r ≤ p ≤ q , we do not get any PAN of the form mpq . Actually, m · is deficient, hence no number of the form mpq is abundant when p = q ≥
11, by Proposition 2.4. Ifwe take p = 5, we have 13 < c ( m · <
14. Hence, m · ·
13 is abundant, but not primitive abundant, since3 · ·
13 is abundant, too. Finally, m · is not abundant. RIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS 9
Corollary 3.5. If m = ¯ p e · · · ¯ p e r r is deficient and there exists a prime p > ¯ p r such that mp is abundant, thenfor each s > there are primes p < · · · < p s such that p > ¯ p r , mp · · · p s is abundant and mp · · · p i isdeficient for each i < s . Moreover, if m is square-free, then mp · · · p s is primitive abundant.Proof. For s = 1 the result follows by choosing p = p . For s >
1, it follows by repeatedly applying Theorem 3.3.Note that since mp is abundant and p > ¯ p r , then c ( m ) > p > ¯ p r by Proposition 2.5, hence the hypothesis ofTheorem 3.3 hold and they are preserved by repeated applications. The result for m square-free follows fromCorollary 2.11. (cid:3) If m is not square-free, the fact that p , . . . , p s may be chosen in such a way that p , . . . p s is primitive abundantis not always true: take for instance m = 3 · m · p , q such that 5 ≤ p ≤ q and mpq is primitive abundant.The following theorem proves that the algorithm enumerating SFPAN is complete. Theorem 3.6.
Let m = p · · · p k be an abundant number, with p < · · · < p k . Let j < k and p j − < ˜ p j < p j such that p · · · p j − ˜ p j is deficient. Then, there are primes ˜ p j +1 < · · · < ˜ p k such that ˜ p j < ˜ p j +1 , e m = p · · · p j − ˜ p j · · · ˜ p k is primitive abundant and p · · · p j − ˜ p j · · · ˜ p i is deficient for every i < k .Proof. Let r be the first index such that p · · · p j − ˜ p j p j +1 · · · p r is abundant. Then r > j by hypothesis, and r ≤ k because m ˜ p j /p j is abundant by Proposition 2.4. Then we just apply Corollary 3.5 in order to add k − r + 1prime factors to p · · · p j − ˜ p j p j +1 · · · p r − . (cid:3) Possibly non square-free PAN.
An extension of the algorithm to find (non necessarily square-free)PAN n with a fixed Ω( n ) may be devised by allowing consecutive primes to be equal.In other words, we see a number m as the product of primes ¯ p · · · ¯ p r with ¯ p ≤ · · · ≤ ¯ p r . When called with r < k −
1, the recursive procedure tries to extend m to a deficient number e m = mp using either p = ¯ p r or p > c ( m ) as for the square-free case. When r = k −
1, the procedure tries to obtain an abundant number mp by choosing either p = ¯ p r or p < c ( m ). In both cases, when p = ¯ p r , Proposition 2.12 is used to decide whether mp is abundant or deficient.In the square-free case, when r = k −
1, it is enough to choose p > ¯ p r in order to ensure that mp is not onlyabundant, but also primitive abundant. In the non square-free case this is not enough: we need to use a differentlower bound for p , which can be computed using Theorem 2.14.Another difference with respect to the square-free case is the stopping condition. The reason lies in the extensionof Theorem 3.6 to possibly non square-free number. Theorem 3.7.
Let m = p · · · p k be a PAN, with p ≤ · · · ≤ p k . Let j < k and p j − < ˜ p j < p j such that p · · · p j − ˜ p j is deficient. Then, there are primes ˜ p j +1 ≤ · · · ≤ ˜ p k such that ˜ p j ≤ ˜ p j +1 , e m = p · · · p j − ˜ p j · · · ˜ p k is abundant and p · · · p j − ˜ p j · · · ˜ p i is deficient for every i < k .Proof. Since σ − is sub-multiplicative, if we replace in m the prime p j with ˜ p j , the resulting number m ˜ p j /p j is abundant. Actually 2 < σ − ( m ) = σ − ( mp j /p j ) ≤ σ − ( m/p j ) σ − ( p j ) ≤ σ − ( m/p j ) σ − (˜ p j ) = σ − ( m ˜ p j /p j ).Let r be the first index (which by hypothesis is strictly larger than j ) such that p . . . p j − ˜ p j p j +1 · · · p r isabundant. Then we just apply Corollary 3.5 in order to add k − r + 1 prime factors to p . . . p j − ˜ p j p j +1 · · · p r − . (cid:3) We cannot guarantee that e m is primitive abundant. For example, although 3 · · ·
31 is primitive abundant,there is no p ≥
11 such that m = 3 · · · p is primitive abundant.Since Theorem 3.7 does not ensure that e m is primitive abundant, the procedure should return a boolean sayingwhether an abundant number (not necessarily a primitive abundant number) has been found, and stop whenthe recursive call returns false.The complete description may be found in Algorithm 2.Using an implementation in SageMath of the algorithmwe managed to count the number of PAN with from 1 to 7 prime factors (counted with their multiplicity) andodd SFPAN with from 1 to 8 prime factors. The results are shown in Table 2 and form the OEIS sequencesA298157 and A287728. We have also computed a list of PAN with up to 6 prime factors, which is available onGitHub at https://github.com/amato-gianluca/weirds . Ω
Table 2.
Number of PAN and odd PAN with given number of prime factors counted with multiplicity.4.
Weird numbers
In a previous paper [1], we developed search algorithms which allowed us to find primitive weird numbers(PWN) with up to 6 different prime factors. However, we were not able to proceed further, because of thecomputational complexity involved. It was clear that a different approach was needed, which was suggested tous by the following known result.
Proposition 4.1.
A number is primitive weird iff it is weird and primitive abundant.Proof. If n is primitive weird, by definition it is weird and abundant. We prove that, for any m | n , m isdeficient. Assume n = mk . For the sake of contradiction, assume m is non-deficient. Since m cannot be weirdby hypothesis, there is a subset S of divisors of m such that m = P d ∈ S d . If d is a divisor of m , dk is a divisorof n . Hence n = mk = P d ∈ S dk is not weird, contradicting our hypothesis.On the other side, let n be weird and primitive abundant. If m | n then m is deficient, hence it cannot be weird.Therefore, n is primitive weird. (cid:3) Given that PWN are only a particular case of PAN, we use the algorithms for enumerating PAN shown inthe previous section, and add a straightforward check for weirdness, transforming them into algorithms forenumerating PWN.Checking for weirdness can actually be made more efficient using the following well-known fact.
Proposition 4.2.
An abundant number n is weird iff ∆( n ) cannot be expressed as a sum of distinct properdivisors of n .Proof. For a proof one can see, for instance, [14, Lemma 2]. (cid:3)
The square-free case.
We consider again Algorithm 1 for the square-free case. Since we are interestedin finding PWN with several prime factors, and since it is not computationally feasible to enumerate all PANin such cases, we provide as an additional input to the algorithm an amplitude value a . At each step of theprocedure, when iterating over primes larger than c ( m ) (or smaller then c ( m ) in the case r = k − a primes.Another generalization consists in starting the search procedure from a possibly non square-free deficient number m . This means that, in the Algorithm 1, each ¯ p i may be a power of a prime number, although new primesadded by the procedure are always square-free. However, when r = k −
1, we only consider primes p whichare larger than σ ( q α ) for each q α || m . In such a way, by Corollary 2.9, the abundant numbers found by thesearch procedure turns out to be primitive abundant. When m is a power of 2, c ( m ) = σ ( m ) and there are noadditional constraints on the choice of the last prime. Remark . In determining whether a number is weird, the sufficient conditions in Theorem 3.1 of our previouspaper [1] could be employed. However, experimental evaluation has shown that most of the weird numbersgenerated with our approach fail to satisfy these conditions. Therefore, a direct proof of weirdness usingProposition 4.2 is employed.The weird numbers generated by this procedure tend to be huge. At each step, since we choose p close to c ( m ),we minimize the deficiency of e m = mp . However, when recursively calling the search procedure on e m , since d ( e m ) is small, c ( e m ) is quite large. This is repeated step after step, leading to very large prime factors. Forexample, all the PWN we have generated with ω = 12 are larger than 10 . Since dealing with these hugenumbers is cumbersome, we represent them in a form we have called index sequence , that turned out to be veryuseful. RIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS 11
Algorithm 2:
Enumerating primitive non-deficient numbers with k prime factors, counted with their multi-plicity. Function pndn( k: nat, m: nat = 1 ) isInput: k is a natural number; m = ¯ p e · · · ¯ p e r r is a deficient number with ¯ p < · · · < ¯ p r Output: all primitive non-deficient numbers of the form m · p · · · p k , with ¯ p r ≤ p ≤ · · · ≤ p k . Result: a pair ( count , found ) where count is the number of primitive non-deficient number of the formabove, and found is a boolean which is true when a (possibly non-primitive) non-deficient numberof the form above has been found. begin count ← found ← false; if j = 1 then if there is a prime p s.t. ¯ p r < p ≤ c ( m ) then found ← true; lowerbound ← max { c ( m/p ) | p is a divisor of m } ; foreach p prime s.t. max(¯ p r , lowerbound ) < p ≤ c ( m ) do Print( mp ) ; count ← count + 1 end end if ¯ p r · σ (¯ p e r r ) ≤ c ( m ) then found ← true; lowerbound ← max { c ( m/p ) | p < ¯ p r is a divisor of m } ; if ¯ p r · σ (¯ p e r r ) > lowerbound then Print ( m ¯ p r ); count ← count +1 end end return ( count , found ) else if m ¯ p r is deficient then ( innerCount , innerFound ) ← pndn( k − , m ¯ p r ) ; count ← count + innerCount ; found ← found or innerFound end foreach p prime s.t. p > max(¯ p r , c ( m )) do ( innerCount , innerFound ) ← pndn( k − , mp ) ; count ← count + innerCount ; found ← found or innerFound ; if innerFound = false then return ( count , found ) end end end endendDefinition 4.4 (Index sequence) . Given a number m = p e · · · p e k k , we define ι ( m ), the index sequence associatedto m , as the sequence [( ι , e ) , . . . , ( ι k , e k )] with ι , . . . , ι k ∈ Z such that: • if ι i = 0, then p i = c ( w i − ); • if ι i >
0, then p i is the ι i -th prime larger than c ( w i − ); • if ι i <
0, then p i is the | ι i | -th prime smaller than c ( w i − ),where w i := Π ij =1 p e j j . To ease notation, we write each pair ( ι, e ) as ι e or just ι if e = 1. ω factored weird number ∆ ( w ) index sequence · · , , − · ·
19 8 [1 , , − · ·
127 16 [1 , , − · ·
71 16 [1 , , − · ·
61 56 [1 , , − · ·
43 16 [1 , , − · ·
31 16 [1 , , − · · ·
53 4 [1 , , , − · · ·
31 4 [1 , , , − · · ·
251 8 [1 , , , − · · ·
241 88 [1 , , , − · · ·
67 8 [1 , , , − · · ·
439 8 [1 , , , − · · · , , , − · · · , , , − · · · , , , − · · ·
787 16 [1 , , , − · · ·
139 16 [1 , , , − · · · ·
647 20 [1 , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · , , , , − · · · · · , , , , , − · · · · · , , , , , − · · · · · , , , , , − · · · · · , , , , , − · · · · · , , , , , − · · · · · , , , , , − · · · · · , , , , , − · · · · · , , , , , − · · · · · · , , , , , , − · · · · · · , , , , , , − Table 3.
Some PWN found by our search algorithm. The first column is the number of primefactors. For each ω , entries are in lexicographic order with respect to the index sequence.For example, the number m = 2 · · · · · , , , , , − c (1) = 1, 13 is the 2nd prime larger than c (2 ) = 7,17 is the 1st prime larger than c (2 ·
13) = 16 . ¯3, and so on. All index sequences generated by our searchprocedure have positive indices for all but the last position. All the indices have an absolute value smaller thanthe amplitude parameter a . Remark . Having to deal with huge numbers is a limitation of our approach: increasing the value of k has abig impact on performance because not only is the search space increased by a factor a (the amplitude of thesearch space) but the numbers we deal with also become much larger. Experimentally we see that, when p i isnear c ( p e · · · p e i − i − ), then each prime is roughly double the size of the preceding one, in terms of the numberof digits. Therefore, there is an exponential increase in the size of factors, which impacts all operations onthese numbers, but particularly the procedure for determining the (pseudo-)prime immediately preceding orfollowing a given number n . This procedure essentially works by repeatedly calling a (pseudo-)primality testwith consecutive odd numbers until a new (pseudo-)prime is found. Since in the average the gap betweenprimes is log n and the Baillie–PSW primality test [2, 16] used by SageMath takes time proportional to log n ,the computational complexity of determining the next prime is roughly log n , i.e., 4 k . This makes it extremelyhard to run our algorithms with values of Ω >
16, even with a small value for the amplitude.
RIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS 13 ω index sequence ω index sequence , , , , , , −
1] 11 [1 , , , , , , , , , , − , , , , , , −
6] 11 [1 , , , , , , , , , , − , , , , , , −
5] 11 [1 , , , , , , , , , , − , , , , , , −
6] 11 [1 , , , , , , , , , , − , , , , , , −
3] 11 [1 , , , , , , , , , , − , , , , , , −
1] 11 [1 , , , , , , , , , , − , , , , , , −
5] 11 [1 , , , , , , , , , , − , , , , , , −
2] 11 [1 , , , , , , , , , , − , , , , , , −
6] 11 [1 , , , , , , , , , , − , , , , , , −
2] 11 [1 , , , , , , , , , , − , , , , , , −
3] 11 [1 , , , , , , , , , , − , , , , , , −
4] 11 [1 , , , , , , , , , , − , , , , , , , −
2] 11 [1 , , , , , , , , , , − , , , , , , , −
2] 11 [1 , , , , , , , , , , − , , , , , , , −
3] 12 [1 , , , , , , , , , , , − , , , , , , , , −
3] 12 [1 , , , , , , , , , , , − , , , , , , , , −
3] 12 [1 , , , , , , , , , , , − , , , , , , , , −
3] 12 [1 , , , , , , , , , , , − , , , , , , , , −
1] 12 [1 , , , , , , , , , , , − , , , , , , , , , −
3] 12 [1 , , , , , , , , , , , − , , , , , , , , , −
2] 12 [1 , , , , , , , , , , , − , , , , , , , , , −
3] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
1] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
1] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
3] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
1] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
3] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
1] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
3] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
3] 13 [1 , , , , , , , , , , , − , , , , , , , , , −
1] 13 [1 , , , , , , , , , , , , − , , , , , , , , , −
2] 14 [1 , , , , , , , , , , , , − , , , , , , , , , −
3] 15 [1 , , , , , , , , , , , , , , − , , , , , , , , , −
1] 16 [1 , , , , , , , , , , , , , , , − Table 4.
Other PWN found with our search algorithm. Since these numbers are large, onlythe index sequence is shown. As an example, the first two entries are 54 and 37 digits long,while the last three entries are 3608, 7392 and 14712 digits long respectively. For each ω , entriesare in lexicographic order.On the other side, it seems that the abundant numbers m generated in this way are very likely going to beweird. This is, at least in part, due to the fact that ∆( m ) is low if compared to m and its prime factors. A lowabundance is unlikely to be expressible as sum of divisors of m , see Proposition 4.2.In line with the previous remark, many PWN are easily found starting from a power of two for m and a smallamplitude for a . Tables 3 and 4 contain some of the PWN we have found starting from the following parameters: • m = 2, a = 8, k ∈ { , . . . , } ; • m = 4, a = 3, k ∈ { , . . . , } ; • m = 8, a = 6, k ∈ { , . . . , } .Table 3 contains, for each PWN, both its factorization and its index sequence. Table 4 only contains indexsequences since the constituent primes would not fit on the page. In particular, we mention the following results: Ω factored weird number index sequence · · · · † [1 , , , , − · · · · , , , , − · · · · · † [1 , , , , , − · · · · · , , , , , − · · · · · † [1 , , , , , − · · · · , , , , − · · · · , , , , − · · · · · , , , , , − · · · · · , , , , , − · · † [1 , , − · · · · , , , , − · · · · , , , , − · · · · · · P [1 , , , , , , − · · · · · · P [1 , , , , , , − · · · · · P [1 , , , , , − · · · · · · P [1 , , , , , , − · · † [1 , , − · · · · , , , , − · · , , − · · , , − Table 5.
Some of the PWN with square odd prime factors that we have found. PWN alreadylisted in A273815 are marked with † . For Ω = 7, this is the complete list of all the PWN withat least one odd prime factor with exponent greater than one. P i denotes a prime with i digits.Entries are in lexicographic order of index sequences. • We have found PWN with up to 16 distinct prime factors. Previously, PWN with 6 distinct primefactors were shown in [1], and only one with 7 distinct prime factors was known [5], while no PWN wasknown with 8 or more distinct prime factors. • The PWN with 16 distinct prime factors has 14712 digits. This is, to the best of our knowledge, thelargest PWN known, the previously largest having 5328 digits [14].Note that, for the sake of efficiency, the search algorithm uses pseudo-primes. However, all the factors for theweird numbers in Tables 3, 4, 5 and 6 have been validated to be real primes, even using additional softwaresuch as Primo (a primality proving program based on the Elliptic Curve Primality Proving algorithm).
Remark . Explaining the fact we find so many PWN only on the basis of their abundance is not satisfactory.In particular, by looking at the tables, it is evident that the initial value m = 4 is the best choice for determiningPWN, at least for low values of the amplitude parameter: with a value of just 3, we could find PWN with k distinct prime factors for all k between 3 and 16. The results for m = 2 and m = 8 were less satisfactory, evenusing much larger values for the parameter a . We will investigate this behavior in a forthcoming paper.4.2. PWN with square factors.
Another weirdness in the realm of weirds is the rarity of PWN with oddprime factors of multiplicity greater than one. To the best of our knowledge, up to now there were only fiveknown PWN with a square odd prime factor, listed in the OEIS sequence A273815, and no PWN with an oddprime factor of multiplicity strictly greater than two is known.Using an extension of Algorithm 2 we have found hundreds of new PWN with at least one odd prime factor ofmultiplicity greater than one. A selection of them may be found in Table 5. We find that there are no suchPWN for Ω <
7, and the list for Ω = 7 is complete. From Ω = 8 onwards, our list is only partial. None of thePWN we have found has odd prime factors with exponent greater than two.On the other side we have found many PWN which have two odd prime factors with exponent greater thanone, which were not known up to now. One of them is:2 · · · · · · · · , , , , , , , , − RIMITIVE ABUNDANT AND WEIRD NUMBERS WITH MANY PRIME FACTORS 15
Ω factored weird number index sequence w · · P · P [1 , , , , , , , , − w · · P · P [1 , , , , , , , , − w ′ · · P · P [1 , , , , , , , , − w ′ · · P · P · P [1 , , , , , , , , , − w ′ · · P · P · P · P [1 , , , , , , , , , , − w ′ · · P · P · P · P [1 , , , , , , , , , , − Table 6.
Some of the PWN with 2 and 3 square odd prime factors that we have found. Here, w = 2 · · · · · w ′ = 2 · · · · · P i denotesa prime with i digits. Entries are in lexicographic order of index sequences.Other PWN with 2 square odd prime factors are given in Table 6. Actually, the last of them has 3 squareodd prime factors, so it is likely that there are weird numbers with any number of square odd prime factors,provided Ω is big enough.All of the above can be summed up in the following theorem: Theorem 4.7 (PWN with non square-free odd part and Ω ≤ . There are no PWN m with a quadratic orhigher power odd prime factor and Ω( m ) < . There are no PWN m with 2 quadratic odd prime factor and Ω( m ) = 7 . There are no PWN m with a cubic or higher power odd prime factor and Ω( m ) = 7 . Open problems
By examining Tables 5 and 6, together with other weird numbers found by our search procedure and whichmay be found on-line, we observe some facts which can be useful for further experiments.First of all, there are some prefixes in the factorization which occur in many PWN. One of this recurring prefixis 2 · · · · <
12, butwe think they are quite rare. The same thing may be said about PWN with 3 square odd prime factors, whichonly appear with Ω = 15. Unfortunately, with Ω >
15 the numbers become huge (thousands of digits) and thismakes experiments much more difficult.
Open Question . For each n ∈ N , find a PWN with exactly n square odd prime factors and the least Ω = Ω n .From the previous section and Theorem 4.7 we obtain Ω = 7, 8 ≤ Ω ≤
12, 8 ≤ Ω ≤
15, and in general if n ≥ n ≥ Open Question . Find a PWN with a cubic or higher power odd prime factor.From the experiments, odd square prime factors seems more common at the right end of the factorization,although in our search results they never appear in the last position.
Open Question . Find a PWN which has its largest prime factor squared or to a higher power.On OEIS A002975 it was asked if the following fact is true: a weird number is primitive iff divided by its largestprime factor it is not weird. The following would be a counterexample.
Open Question . Find a weird number w which is not primitive and such that w/ (largest prime factor) isnot weird.The following problem appears as an editor’s comment in [3]. Erd˝os offered 25$ for its solution. Open Question . Is σ ( m ) /m bounded when m ranges through the set of (not necessarily primitive) weirdnumbers?Finally, the following would settle a long-standing problem. Open Question . Find an odd weird number, or prove that all weird numbers are even.
The above problem was raised by Erd˝os, that offered 10$ for an example of an odd weird number, and 25$ fora proof that none can exist [3]. Wenjie Fang and Uwe Beckert proved, using parallel tree search, that there areno odd weird numbers up to 10 , and no odd weird numbers up to 10 with abundance not exceeding 10 [9,Section 4.2]. Acknowledgements.
We would like to thank Vincenzo Acciaro, Adam Atkinson, Rosa Gini, Francesca Scozzari,Agnese Telloni for their valuable discussions with us. The second and third authors wish to thank the first andfourth authors for invitation and hospitality at Chieti-Pescara University, where significant parts of this workhave been done.
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Universit`a di Chieti-Pescara, Dipartimento di Economia Aziendale, viale della Pineta 4, I-65129 Pescara, Italy
E-mail address : [email protected] Universit´e des Antilles, D.S.I., B.P. 7209, Campus de Schoelcher, F-97275 Schoelcher cedex, Martinique (F.W.I.)
E-mail address : [email protected]
University of Applied Sciences of Western Switzerland, HEG Arc, Espace de l’Europe, 21, CH-2000 Neuchˆatel
E-mail address : [email protected] Universit`a di Chieti-Pescara, Dipartimento di Economia, viale della Pineta 4, I-65129 Pescara, Italy
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