aa r X i v : . [ m a t h . N T ] M a r Updating An Upper Bound Of ErikWestzynthius
Gerhard R. PasemanMarch 28, 2014
Abstract
Inspired by a paper of Erik Westzynthius, we build on work ofHarlan Stevens and Hans-Joachim Kanold. Let k > n . In 1977, Stevensused Bonferroni inequalities to get an explicit upper bound on Jacob-sthal’s function g ( n ), which is related to the size of largest intervalof consecutive integers none of which are coprime to n . Letting u ( k )be the base 2 log of this bound, Stevens showed u ( k ) is O ((log k ) ),improving upon Kanold’s exponent O ( √ k ). We use elementary meth-ods similar to those of Stevens to get u ( k ) is O (log k (log log k )) in oneform and O ( σ − ( n ) log k ) in another form. We also show how thesebounds can be improved for small k . Erik Westzynthius provided a ground-breaking result in [1] on primegaps, showing that for any constant
D > p n so that p n +1 > p n + D log( p n ). In the same paper, he alsoprovided an upper bound for what was to be later called g ( P k ). Thisquantity measures how many consecutive integers we can find havinga ”small” ( ≤ p k ) prime factor.Ernst Jacobsthal in [2] defined and showed g ( n ) ≤ k k + 2 k − k ,where k is the number of distinct prime factors of n . Improved explicitbounds were later given by Kanold [3] (2 k for all k and 2 √ k for k ≥ e )and Stevens [4] (2 k e log k ), and some additional but less explicitresults given by Paul Erd˝os, Henryk Iwaniec, and others. e show several explicit bounds for g ( n ), some depending on thequantities σ − ( n ) and π − ( n ), including g ( n ) < P si =1 (cid:0) ki (cid:1) π − ( n ) − ( σ − ( n )) s +1 / ( s + 1)! , with s an odd integer bounded by (1 plus a constant times σ − ( n )),which gives g ( n ) < k B + C log log k for k > B and C , both < . σ − ( n ) and π − ( n ) in the next section, and also list someresults which apply when σ − ( n ) is small ( < / q ). The followingsection recalls work of Jacobsthal and Westzynthius, and shows howStevens’s bound can be tightened with a little effort.With elementary means, we also show the bound above that uses σ − ( n ) and π − ( n ), and follow that section with some supplementaryresults as well as suggestions for further research. The remainder ofthis article contains some history, a recommended reading list, and anAppendix as well as acknowledgments and a list of citations. We use ω ( n ) for the number of distinct (positive) prime factors of thepositive integer n , and declare k = ω ( n ). We also require k >
0, so n >
1. We list the prime factors q i of n in increasing order: q < q <. . . < q k . We recall the k th primorial P k as P k = Q k ≤ i p i . Definition : We define (here − is part of a label, not an exponent) σ − ( n ) = k X ≤ i /q i , and π − ( n ) = k Y ≤ i (1 − /q i ) . Note that nπ − ( n ) = φ ( n ), Euler’s function for counting positive in-tegers coprime to and less than n . Definition : After Jacobsthal, for n > g ( n ) to be thesmallest positive integer m such that for any integer a , the set of m consecutive integers { a + 1 , . . . , a + m } has an integer a + j , where1 ≤ j ≤ m , such that gcd( a + j, n ) = 1. If n >
1, define L ( n ) as thelargest integer l so that there is an interval of l consecutive integers { b + 1 , . . . , b + l } such that each b + j satisfies gcd( b + j, n ) > g ( n ) = L ( n ) + 1 for all n >
1. Also g ( n ) = max i ( c i +1 − c i ) where the c i denote all integers coprime to totatives of) n in increasing order. Consequently, g ( n ) dependsonly on the set of distinct prime factors of n ; we will use n squarefreeat times in this article.One has g ( p i ) = 2, and g ( n ) < n for n >
2. If n has ”large” primefactors ( q i > k for all i ), then g ( n ) = L ( n ) + 1 is ”small”. Specifically, Proposition (Jacobsthal): If q > k , then L ( n ) = k .Proof Sketch: Any integer interval of length k + 1 has at most onemultiple of q i , giving at most k integers in that interval having a primefactor in common with n . So at least one integer in that interval isnot a multiple of any q i . So L ( n ) < k + 1.Conversely, for any permutation τ of the k indices, the ChineseRemainder Theorem gives an integer b τ such that b τ + τ ( i ) = 0 mod q i for 1 ≤ i ≤ k , so L ( n ) ≥ k . End of Proof Sketch.We could have chosen τ to be the identity permutation, but wewant to point out that this proof gives k ! many intervals in (0 , n )which achieve the L ( n ) bound for this kind of n . This proof alsoshows g ( n ) > k for any n > Proposition (Kanold- P .): Let 0 < r < n such that r + σ − ( n ) <
1. Then L ( n ) < k/r , and g ( n ) ≤ ⌈ k/r ⌉ .Proof Sketch: For an interval containing L consecutive integers,at most ⌈ L/q i ⌉ ≤ L − /q i of them are multiples of q i . Since σ − ( n ) < − r , the count of numbers not coprime to n is at most( L − σ − ( n ) + k < ( L − − r ) + k . Now whenever L ≥ k/r ,( L − − r ) + k ≤ L − r , so this count is less than L . One gets g ( n ) ≤ ⌈ k/r ⌉ , giving L ( n ) < k/r . End of Proof Sketch.This gives a weaker bound than Jacobsthal’s proposition, but itapplies to more cases, even when q is about 2 √ k and r > /q . Itsuggests the following Variation : Let k > σ − ( n ) < q . Then L ( n ) < q q − k − − σ − ( n ))2 q , so g ( n ) < kq q q − . Proof Sketch: The estimate above for multiples of q i , 1 < i ≤ k ,among the L numbers is refined by subtracting that portion which arealso multiples of q ; we underestimate it by L/q i q − < ⌊⌈ L/q i ⌉ /q ⌋ .Thus ( L − σ − ( n ) + k − P k (2 k − − σ − ( n )) . The assumption (1 + 1 /q − σ − ( n )) > / q yields that if L/ q > (2 k − − σ − ( n ))( q / ( q − L is greater than our overcount;as above, we get L ( n ) < q (2 k − − σ − ( n ))( q / ( q − g ( n ) < kq ( q / ( q − End of Proof Sketch.We now present one bound of general character, and one whichasymptotically improves upon Kanold’s smaller bound. We use the
Fact : Let b, d and f be integers with gcd( f, d ) = 1. There is aninteger z so that for all integers t, gcd( b + tf, d ) = gcd( zb + t, d ).Pick z so that zf = 1 mod d . gcd( b + tf, d ) = gcd( z ( b + tf ) , d ) =gcd( zb + t, d ), giving the Fact above and the application below thatcoprimality in sequences of consecutive integers behaves the same wayin certain arithmetic progressions.Assume k > , n squarefree and d | n with d neither 1 nor n . Set f = n/d . 1 + tf is coprime to f , and the Fact above shows that 1 + tf always has at least one out of g ( d ) consecutive members coprime to d . Any interval of ( g ( d ) f )-many integers thus has one coprime to n ofthe form 1 + tf . Considering c + tf for all totatives c of f , we get the Observation : g ( f d ) ≤ g ( d ) f − f + g ( f ) . Proof Sketch: When c = 1, there is at least one number of the form c + tf coprime to n = f d in the interval (1 + a, a + g ( d ) f ), where a = t f and we assume that if 1 + tf is coprime to n , then it is outsidethis interval. There are at least ( φ ( f ) − g ( d ) f − f , then the numbers must be in the subintervals(1 + a, a + f ) and (1 + a + ( g ( d ) − f, a + g ( d ) f ). If the largestin the first interval is c + a , then the smallest in the second interval isat most c ′ + a + ( g ( d ) − f , where c ′ is the next largest totative to f after c . But c ′ − c ≤ g ( f ), giving the bound. End of Proof Sketch.We use weaker forms of the Observation and Variation to improveon Kanold’s smaller bound.Let n be squarefree. Consider the tail of σ − ( n ), that is, find q l smallest such that P kl ≤ i /q i < / q l . Let d = Q kl ≤ i q i . From theVariation we have g ( d ) < k − l + 1)( q l / ( q l − k is large, q l is small enough that n/d is smaller than 2 √ k . Improvement : For k sufficiently large, n/d < k . , giving g ( n ) < k . / k − l +1) q l q l − . As k gets large, log g ( n ) < k ( ǫ +1 /e ) eventu-ally for any fixed ǫ > roof Idea: Let m satisfy p m = q l − . We defer to the Appendixshowing that p m < k . when k ≥ e . . As n/d ≤ P m , and for anexplicit positive c < . P m < e p m (1+ c ) < p m / ; the Observationimplies the weaker form g ( n ) < ( n/d ) g ( d ) < k . / g ( d ).About log g ( n ), we use that P kl ≤ j /q j ≤ P k − l +11 ≤ i /p m + i , andthen truncate to the first u ≤ k − l + 1 terms while still having P u ≤ i /p m + i > − / p m . We use an approximation of Mertensfor this last sum to get log(log( p m + u ) / log p m ) > − O (1 / log p m ).Picking a convenient positive ǫ and ǫ , as m goes large, p m + u > p e − ǫ m eventually, and log of the bound is majorized by p m (1 + c ′ ) for somepositive small c ′ and is eventually dominated by k /e + ǫ . The Appendixprovides more detail. End of Proof Idea.We take credit for this form of presentation, but are influenced byKanold’s paper; his proof for 2 √ k has many of the ideas above, andwe wonder if perhaps he did discover it.We end this section with lower bounds: Westzynthius shows aneasily demonstrated lower bound 2 p k − ≤ g ( P k ) for k >
1, and thenshows a better lower bound (for k sufficiently large after choosing ǫ >
0) of e γ (1 − ǫ ) p k log log p k log log log p k . Both Westzynthius and Jacobsthal use a simple sieve argument toestablish an upper bound on g ( n ), using n squarefree. Since Stevensuses part of this, we show the argument here.Recap: We use inclusion-exclusion to count integers in the openinterval L = ( a, a + x ) that are coprime to squarefree n , for realnumbers a and x > I counts the totatives of n in L , and for t > t | n, I t counts multiples of t in L . Then I = P t | n ( − ω ( t ) I t . As inthe previous section we replace the count I t , this time by x/t + E ( t ). E ( t ) is an error term which depends on both t and a actually, butalways | E ( t ) | ≤ , and will be removed below. I = X t | n ( x/t + E ( t ))( − ω ( t ) ≥ x X t | n ( − ω ( t ) /t − X t | n . Now rewrite the sum of ( − ω ( t ) /t as a product Q (1 − /q i ), and notethe second sum is 2 k . We get I ≥ xπ − ( n ) − k . ow this relation above is essentially independent of a . If we pick x large enough, then I > g ( n ) will be at most x . So choose x = 2 k /π − ( n ) + ǫ . End of Recap.The above shows g ( n ) ≤ k /π − ( n ). One would like to improve onthis since Kanold has an asymptotically better result.In our view Stevens has two main ideas in his 1977 paper. Thefirst is to observe that the sum produced from inclusion-exclusion canbe truncated a la Bonferroni to a smaller sum to give fewer terms toapproximate. The second is that the denominator T s can be writtenas π − ( n ) − T ′ s , and that T ′ s can be easier to handle. We adapt hisproof slightly, and then we tighten up the estimates he uses to improvethe exponent.Adaptation: We use x, I , t, and I t where Stevens used Q, L, B and N ( . . . ) . We use integral x , following Stevens. We assume k > t and uses a result of Landau for the first idea.His display (3) in our notation says: for any odd value of s , I ≥ s X ≤ i ( − i X t | n,ω ( t )= i I t . Using the estimate | I t − x/t | ≤ t = 1 when I = x ),we write what Stevens has in his (4) and (5) as I ≥ x s X ≤ i ( − s X t | n,ω ( t )= i t − s X ≤ i ki ! = def xT s − SB. (We’ve written SB for P s ≤ i (cid:0) ki (cid:1) , and T s for P s ≤ i ( − s P t | n,ω ( t )= i t .)Defining T ′ s by the relation π − ( n ) − T ′ s = T s , Stevens notes (using anapproximation of k s for SB instead of SB directly) that if s is chosenso that π − ( n ) > T ′ s (so that T s > x > SB/ ( π − ( n ) − T ′ s ),then I > g ( n ) ≤ x . He and we now look for a suitable s . T ′ s can be written as a sum over i of sums of terms t with ω ( t ) = i ,just like T s , but with s < i ≤ k . Before doing this, Stevens observes:for k > h ( k ), a putative upper bound for σ − ( n )) r ! X t | n,ω ( t )= r /t < ( σ − ( n )) r < h ( k ) r . Stevens uses log k for h ( k ); later we will use log log p k + 1 /
2. Then T ′ s = k X s
4, where we willuse β/ log k . He then has π − ( n ) − T ′ s > /k − k/ s +1 > /k − k − e log 2 > , since s + 1 is an even integer greater than 2 e log k . So I > g ( n ) ≤ x ) if s is an odd integer greater than 2 e log k − x > k s +1 > k s / (1 /k − k/ s +1 ) > SB/ ( π − ( n ) − T ′ s ). Stevens replaces s + 1 with 2 e log k + 2 to ensure the bound holds for all k . End ofAdaptation.We repeat the above, using log log p k + 1 / h ( k ) and β/ log p k for π − ( n ). (The Appendix discusses the validity of these choices.)Again asking for odd s with s + 1 ≥ eh ( k ) , then T ′ s ≤ e h ( k ) / s +1 , and π − ( n ) − T ′ s > β/ log p k − (log p k ) e / / s +1 > β/ log p k − (log p k ) e / / e (log log p k +1 / = β/ log p k − (log p k )( e/ e ) / / (log p k ) e log 2 > (1 / log p k )[ β − (log p k ) − e log 2) / > . In the last line, we use that ( e/ e ) / < /
3, that we can pick 1 / . >β > /
3, that log p k > k >
4, and that e log 2 > π − ( n ) − T ′ s > / p k by choosing β > / p k ) e log 2 − > k >
4. Using suchan estimate we have whenever s + 1 is even and ≥ e (log log p k + 1 / π − ( n ) − T ′ s > / p k leading to Theorem: g ( n ) ≤ (4 log p k ) X ≤ i ≤ s ki ! < (4 log p k ) k e (1 / p k ) . We could tweak the choice of s slightly to get a smaller exponent, aswell as use a better approximation for the sum of binomial coefficients.In the next section, we will find a bound which depends directly on σ − ( n ) which not only does both, but gives a tighter bound in general. σ − ( n ) and π − ( n ) We modify Stevens’s argument with a better upper bound for thenumerator and express T ′ s as an alternating and eventually decreasingsum, allowing us a smaller s .Note that T ′ s is an alternating sum and that one has σ − ( n ) X t | n,ω ( t )= j /t > ( j + 1) X t | n,ω ( t )= j +1 /t, so that when s > σ − ( n ), one can bound T ′ s by σ − ( n ) s +1 / ( s + 1)! . Now instead of Taylor’s theorem and h ( k ), we use (see Appendix) e < (1 /π − ( n )) /σ − ( n ) ≤ r ≥ σ − ( n )gives π − ( n ) − T ′⌈ r ⌉− > π − ( n ) − σ − ( n ) ⌈ r ⌉ / ( ⌈ r ⌉ )! > e < / , so e (1 /π − ( n )) / σ − ( n ) < e / < σ − ( n ) < π − ( n ) / σ − ( n ) σ − ( n ) /e ≤ π − ( n ) /r ( r/e )so σ − ( n ) ⌈ r ⌉ < π − ( n )( ⌈ r ⌉ /e ) ⌈ r ⌉ < π − ( n )( ⌈ r ⌉ )! . We now claim
Theorem:
Let s be the smallest odd integer with s +1 ≥ σ − ( n ) . For k > , P ≤ i ≤ s (cid:0) ki (cid:1) π − ( n ) − σ − ( n ) s +1 / ( s + 1)! > g ( n ) . We collapse the summands in the numerator slightly, increasingthe total by 1, and as ( s + 1)! > p π ( s + 1)(( s + 1) /e ) s +1 , one seesthe denominator is larger than ( p π ( s + 1) − σ − ( n ) s +1 / ( s + 1)!, sowe can write a weaker upper bound as a corollary :( s + 1)! P ≤ j g ( n ) . This may seem intimidating, but when we take into account thatfor k > σ − ( n ) is at most 1 / k (1 + log 2 / log k ) it is thenseen that the dominant term in the sum is (cid:0) k +1 s (cid:1) and the expressionis O ((( k + 1) /σ − ( n )) k + ǫ ) when σ − ( n ) ≥
1. (The portionthat is s +1( √ π ( s +1) − σ − ( n ) is less than 1 for large enough σ − ( n ); for σ − ( n ) near or smaller than 1 we have the more elementary bounds.)This argument only needs K such that ( s + 1) > Kσ − ( n ) andalso that e/K < π − ( n ) /Kσ − ( n ) . This holds for K > .
89, and when − ( n ) >
1, one can lower K from 4 to 3 .
81. However, even for largevalues of σ − ( n ), the argument still expects ( K/e ) K > e , which means K can’t be shown smaller than 3 .
59 with this method.As a rough comparison, Jacobsthal’s bound is larger than Kanold’sbound for all k . Stevens’s bound is smaller than 2 k for k > k > k > e , and the exponent s ≤ σ − ( n ) above is smaller than that of Stevens for k > The recap is our interpretation of Westzynthius’s upper bound argu-ment published in 1931, generalized to arbitrary n with ω ( n ) = k instead of P k , which Westzynthius did not publish as far as we know.In a footnote Westzynthius did hint at sieving with just odd numbers,and we considered extending that argument with thinner sets. Thisled us to asking the question [6] on MathOverflow in 2010.Correspondence on MathOverflow led us eventually to ThomasHagedorn’s paper [5] and Jacobsthal’s function. Jacobsthal in [2] usesa slightly different argument, and (with the notation of this article)instead of using 2 k /π − ( n ) he bounds π − ( n ) by 1 / ( k + 1) and gives abound of ( k +1)(2 k −
1) on L ( n ). Hagedorn’s paper quoted the boundsof Kanold and Stevens, and after studying those papers we adaptedStevens’s argument and posted the results on MathOverflow in 2011,as well as producing a private manuscript with small circulation.Since then we have accumulated and posted other accessible re-sults, and arranged some of them for this article. The Observationrepresents a small improvement on Kanold’s result which involves g ( f d ) ≤ g ( d ) f + 1 − φ ( f ) instead of g ( f d ) ≤ g ( d ) f + g ( f ) − f ; thetwo are the same for f prime. The Improvement is intended to shownot just improved asymptotic results but also that Kanold’s boundholds for k smaller than e . Indeed the name is earned once k ≥ e .We admit the work is in showing the bounds hold for small k , whichmakes the Improvement not as elementary or accessible as we hope.Hagedorn also mentions work of Erd˝os, Iwaniec, and others. Erd˝osshows for any given positive real ǫ that | − g ( n ) π − ( n ) /k | > ǫ only for n in a set of zero density. Erd˝os also comments that Brun’s methodcan yield a constant c such that g ( n ) is O ( k c ), but we have not founda version of this that is both explicit and accessible. Iwaniec shows he existence of a constant C independent of n such that there are atleast k many totatives of n in an interval of size C ( k log k ) /π − ( n ),which implies g ( n ) is O (( k log k ) ); again we do not know what C is. We intended this article to give simpler, more accessible, and moreexplicit proofs of upper bounds on Jacobsthal’s function. We areoptimistic about improving upon the results of Erd˝os and Iwaniec.In particular, we think there is more to the Observation: we hopeto achieve a subquadratic in k upper bound using this direction bynoting how large intervals of numbers with factor common to n aredistributed in (0 , n ). At present, the difference in (base 2) exponentsbetween O (log k log log k ) and O (log k ) is substantial.Except for the bound depending on σ − ( n ), all of these bounds arealso bounds on Jacobsthal’s C ( k ), given by C ( k ) = max k = ω ( n ) L ( n ).It was shown recently [8] that the conjecture C ( k ) + 1 = g ( P k ) holdsfor 1 ≤ k ≤
23 and fails at k = 24.Note that the bound involving σ − ( n ) can represent a substantialimprovement even if n cannot be factored; for those n which do nothave small factors, σ − ( n ) can be substantially smaller as can k , evenfor numbers near 10 . Of course, when k or a partial factorizationof n are better known, better bounds on g ( n ) become available.We are interested in tweaking the Variation to handle more square-free n by sieving out small prime factors. Our beginning efforts havenot yielded much improvement on bounds obtained by the Obser-vation. It seems better estimates on the number of totatives in aninterval of arbitrary length are needed to carry out an argument likethat in the Variation.Some questions of interest:1) Pick a small odd prime p and odd n with q > p . We know g ( pn ) /g ( n ) < p : can we get anything sharper? In particular, what arethose integers n such that g (3 n ) > g ( n )? Such that g (5 n ) > g ( n )?2) Let a ( n ) be the smallest positive integer such that gcd( n, a ( n ) + i ) > < i < g ( n ). One can show a ( n ) < n/
2; how much canthis be improved? If b ( n ) is the number of such longest intervals ofnontotatives of n in (1 , n ), can we hope for a ( n ) b ( n ) < n ?3) How close are two such intervals? If one hopes for a subquadratic(in k ) bound on g ( n ), this will be an important bit of information. ven in the case q > k , it should be related to how primes are dis-tributed, which suggests that some interesting perspective is needed.4) Not much asymptotic improvement should be expected fromthese arguments in the case that σ − ( n ) is large, say σ − ( n ) > T = (1 /π − ( n )) /σ − ( n ) is expected to decrease as σ − ( n ) increases. Howdoes T behave with σ − ( n ), and can one use this in bounding g ( n )?5) Even the simple estimates with small σ − ( n ) have some slop,primarily in overestimating multiples of q i with L/q i . Often this re-sults in an estimate about twice as large as needed. Can something besaid about this ”noise” vector L/q i − I q i and what approaches avoidthe error introduced by this?We recommend the bibliography and also the following reading list,which provides additional information related to Jacobsthal’s functionand applications.Ernst Jacobsthal, ¨Uber Sequenzen ganzer Zahlen, von denen keinezu n teilerfremd ist. I-III. Det Kongelige Norske Videnskabers SelskabsForhandlinger
Bd 33 1960, Nr. 24, Trondheim I Kommisjon Hos F.Bruns Bokhandel 1961, pp. 117-124,125-131,132-139. (Also see IVand V published in a later edition.)Hans-Joachim Kanold, Neuere Untersuchungen ¨uber die Jacobsthal-Funktion g ( n ). Monatshefte Math.
84, 1977, pp. 109-124.Paul Erd˝os, On the integers relatively prime to n and on a number-theoretic function considered by Jacobsthal. Math. Scand.
11 (1962)pp. 163-170.R.C. Vaughan, On the order of magnitude of Jacobsthal’s function.
Proc. Edinburgh Math. Soc.
20 (1976-77) pp. 329-331.Henryk Iwaniec, On the problem Of Jacobsthal.
DemonstratioMathematica v XI no. 1 1978 pp.225-231.MathOverflow Questions (Number refers to URL, so 37679 ex-pands to http://mathoverflow.net/questions/37679)88323 Analogues of Jacobsthal’s Function.56099 Lower bound of the number of relatively primes (each other) inan interval.68351 Least Prime Factor in a sequence of 2n consecutive integers.Shallit, J., State Complexity and Jacobsthal’s Function.
CIAA ’00Revised Papers from the 5th International Conference on Implemen-tation and Application of Automata Acknowledgments
We acknowledge Will Jagy, Aaron Meyerowitz, and Thomas Hagedornfor their support and assistance. We appreciate the stimulating en-vironment provided by the MathOverflow forum and its community,and are thankful for its providing a repository of these results. Weespecially appreciate Will and Aaron for their MathOverflow contri-butions, and are also grateful for discussions with users Larry Freeman(who asked a version of question 2) above) and asterios gantzounis.
We resolve some details on assertions made in earlier sections: that(1 /π − ( n )) /σ − ( n ) ∈ ( e, σ − ( n ) from above by 1 / p k and π − ( n ) from below by 1 / p k , and on showing theImprovement holds for k > e . .In getting a bound depending on σ − ( n ), we used the assumptionthat e < (1 /π − ( n )) /σ − ( n ) ≤ n >
1. We proved this alongwith related results in a private manuscript [7]. The proof was basedon observing that ( − log π − ( n )) /σ − ( n ) was a mediant sum of valuesof the form p i log( p i / ( p i − ab and cd give a mediant of a + cb + d ), sothat ( q k / ( q k − q k < (1 /π − ( n )) /σ − ( n ) < ( q / ( q − q . Also, if σ − ( n ) ≥
1, then (1 /π − ( n )) /σ − ( n ) < .
6, so one can improve theconstant C in g ( n ) ≤ Ak B + C log log k to a value less than 3 .
81. If we didnot care about the advantage given by using σ − ( n ), we could use ageneral bound of g ( n ) < k .
81 log log k for k >
2, which can be verifiedby hand for small values of k and which would be weaker (and thusvalid) than the Variation when σ − ( n ) ≤ n = P k , Mertens determines σ − ( n ) and π − ( n ) with errorby σ − ( n ) = log log p k + B + E ( k ) and π − ( n ) = e − ( γ + δ ( k )) / log p k where B is a constant with value near 0 . E ( k ) and δ ( k )are error terms in O (1 / log p k ). In using Stevens’s argument withtighter bounds, we used 1 / p k as an upper bound for σ − ( n );calculations show that σ − ( n ) < . p k for 7 < p k < , andMertens estimates of the error (or tighter estimates given by Rosserand Schoenfeld) show this holds for k >
4. If we were concerned onlywith k >
50, we could replace 0 .
41 by 0 .
28, closer to the value of B .We could also use 3 / k as an upper bound for k >
10, andreplace 3 / k sufficiently large. imilarly π − ( n ) log( p k ) approaches e − γ which is near 0 . π − ( n ) > / p k for 1 < p k < ;theory then gives the weak inequality for all k . We could use 1 / p k for k > σ − ( n ) in thesection following Stevens’s argument.The rest of this Appendix contains the verification of the claim thatthe choice of q l for n with k > e . satisfies q l < k . , and remarksexpanding on the proof idea of the Improvement.We first work with sums of the form P ≤ j ≤ u /p m + j which arewithin 1 / p m to 1, as they represent an extremal form with respectto the estimate. P ≤ j ≤ /p i < .
9, so when p m = 5 we already have u > p m if we want a sum close enough to 1. As one increases m by 1,one has to remove 1 /p m +1 and ”replace” it by more than 2 p m termsof size smaller than 1 / p m , so we already have m/u < /m log m forsuch sums.As m grows, log p m + u tends to e log p m and ( m + u ) log( m + u )approaches p em , and thus u/p em approaches 1, yielding the asymptotic(in k ) result of log( g ( n )) < k /e + ǫ . Because of oscillations aroundzero of the quantity ( P ≤ j ≤ u /p j − B − log log p u ), a proof of q l − ≤ p m < k . seems more challenging; we will use results of Rosser andSchoenfeld [9] to show this bound for special sums of the above formfor m ≥ m down to 20, andthen show how this implies the general result when k > e . .We start with getting log p m + u in terms close to log u . Lemma : log p m + u < log u + (log log u + mu )(1 + u ) + mu (log u ) when u > m .Sketch of Proof: Theorem 6 of [9] yields for k > p k < log k + log(log k + log log k ). We start with the more complicatedsubterm: log(log( m + u ) + log log( m + u )) = log log u + LL , where LL = log(log( m + u ) / log u + log log( m + u ) / log u ). LL = log(1 + [log(1 + m/u ) + log log( m + u )] / log u ) < [ m/u + log log( m + u )] / log u = [ m/u + log(log( u ) + log(1 + m/u ))] / log u = [ m/u + log log u + log(1 + log(1 + m/u ) / log u )] / log u< [ m/u + log log u + m/u log u ] / log u = ( m/u + log log u ) / log u + m/u (log u ) ow we use Theorem 6 to getlog p m + u < log( m + u ) + log(log( m + u ) + log log( m + u )) < log u + m/u + log log u + LL< log u + ( m/u + log log u )(1 + 1 / log u ) + m/u (log u ) . End of Sketch.Next, we want to get a good lower bound on log(log p m + u / log p m )when we have our special sum close enough to 1. Lemma : Suppose P ≤ j ≤ u /p m + j > − / p m and p m + u > p m + u / log p m ) > − / p m − . − / p m + u ) .Sketch of Proof: From [9] Theorem 5 we derive (where p ≤ x meansthe primes greater than 1 and at most x ) | X p ≤ x /p − log log x − B |≤ / x ) for all x ≥ X p ≤ x /p − log log x − B > ≤ x ≤ . Combining the results over the two ranges gives − log log p m − B + 1 / )) > = − X p ≤ p m /p for all m > X p m
e > m ≥
184 and 1 / log u < /
15. Then log(log p m + u / log p m ) > − / − . − / > .
99 and e . > . p m + u > .
691 log p m . oward a contradiction, assume that x =
20 log p m > log u . Thenthe Lemma concerning log p m + u gives1 . x = 2 .
691 log p m < log p m + u < x + (log x + m/u )(1 + 1 / log u ) + m/u (log u ) < x + (log x + 1 / /
15) + 1 / < x + (16 log x ) /
15 + 1 / . x > (16 log x ) /
15 + .
01 for x >
14, which means forlog p m >
7, and we have a contradiction. Thus u ≥ p / m .For primes p m = 71 through p m = 857 ( m = 20 to m = 148),we verified through computation that if P u ≤ j /p m + i > − / p m ,then log u/ log p m > /
9. In particular for u as low as 13250 < e . and n a ratio of certain primorials the Improvement holds. Also, P ≤ j /p j > − / p , thus P u ≤ j /p m + j > − / p m im-plies u > p / m for p m running up to 1102 < / . So theimplication holds for all p m ≥ n with k ≥ e . and otherwise arbitrary. Find l small-est so that P kl ≤ i /q i < / q l . Let p m = q l − . Then 1 − / q l − ≤ P kl ≤ i /q i ≤ P k − l +11 ≤ j /p m + j . Either p m ≥
71 and so ( k − l + 1) . ≥ u . ≥ p m , or else q l − < < ( e . ) / ≤ k . . Shortly after version 1 of this article was posted, we found an upperbound based on estimates of Euler’s totient which does better thanthe Improvement.Aaron Meyerowitz asked about these estimates in question 88777on MathOverflow. The key observation is that the number of totativesto n in the interval [0 , x ) differs from xφ ( n ) by a periodic function E n ( x ) which has a maximum value at most 2 k − . Using this, we cangeneralize the Variation using a similar argument to show g ( n ) ≤ ( k − m + 1)(2 m + π − ( d )) π − ( d )(1 − σ − ( n/d )) , where one chooses m < k so that d = Q ≤ i ≤ m q i divides n and alsosatisfies t = 1 − σ − ( n/d ) >
0, and finally so that the right handside above leads to an optimal bound. (Hint: the Proof sketch of he Variation introduces an error term of size at most 2 for countingmultiples of q i which are not multiples of q ; now use an error of2 m + π − ( d ) for an interval [ y, x ] of length L when counting multiplesof q i that are coprime to d .)When such an optimal m > t > / (3 + q m +1 ), and the analysis from the Appendix can be adaptedto show m e < k . As 1 /π − ( d ) is bounded by 2 log q m , this will beat ourbound k Kσ − ( n ) when 2 m is substantially smaller than k Kσ − ( n ) − / .Computing both bounds for n = P k for k < show this new boundto be superior: we expect to show that it holds when k < and σ − ( n ) >
1, which would imply that K above can be taken near 3 . k > √ k actuallyholds for k > e . There are also improvements to be made on theerror term (2 m + π − ( d )); with such improvements we expect to showin a followup article a subquadratic in k upper bound for k < .We also found a statement of the Observation in a 1975 work ofKanold’s. (We thank Prof. Dr. Heiko Harborth for making this part ofthe literature available to us.) We are still looking for a published proofof the Observation as well as an appearance in the literature of theProposition that g ( n ) ≤ ⌈ k/ (1 − σ − ( n )) ⌉ for those n with σ − ( n ) < k . σ − ( n ) has notappeared in the literature.We have planned a series of forthcoming articles, tentatively titled’Adventures in finding bounds on Jacobsthal’s function.’ In additionto fleshing out some of the questions asked in an earlier section, we willconsider the computational complexity of g ( n ) and approximationsto g ( n ), various lower bounds coming from elementary (and not soelementary) considerations, applications using both conjectured andactual bounds, and generalizations in geometric and algebraic realms. References [1] Erik Westzynthius, ¨Uber die Verteilung der Zahlen, die zu den n ersten Primzahlen teilerfremd sind. Societas Scientiarum FennicaCommentationes Physico-Mathematicae V. 25, Helsingfors 1931,pp. 1-37.[2] Ernst Jacobsthal, ¨Uber Sequenzen ganzer Zahlen, von denen keinezu n teilerfremd ist. I. Det Kongelige Norske Videnskabers Selskabs orhandlinger Bd 33 1960, Nr. 24, Trondheim I Kommisjon HosF. Bruns Bokhandel 1961, pp. 117-124.[3] Hans-Joachim Kanold, ¨Uber eine zahlentheoretische Funktion vonJacobsthal.
Mathematische Annalen
170 1967, pp.314-326.[4] Harlan Stevens, On Jacobsthal’s g ( n )-function. MathematischeAnnalen
226 1977, pp. 95-97.[5] Thomas R. Hagedorn, Computation of Jacobsthal’s function h ( n )for n < Mathematics of Computation
78 2009, pp. 1073-1087[6] MathOverflow question Erik Westzynthius’s cool upper bound ar-gument: update? http://mathoverflow.net/questions/37679/[7] Gerhard Paseman, The Waltraud and Richard R. Paseman Theo-rem, private manuscript, March 2011[8] L. Hajdu and N. Saradha, Disproof of a conjecture of Jacobsthal.
Mathematics of Computation
V 81 n. 280 October 2012, pp. 2461-2471.[9] J. Barkley Rosser, Lowell Schoenfeld, Approximate formulas forsome functions of prime numbers.
Illinois Journal of Mathematics6 1962, pp 64-94.