A Linear Division-Based Recursion with Number Theoretic Applications
aa r X i v : . [ m a t h . N T ] J a n A Linear Division-Based Recursion with NumberTheoretic Applications
Jonathan L. MerzelJanuary 27, 2021
Abstract
A simple remark on infinite series is presented. This applies to aparticular recursion scenario, which in turn has applications related toa classical theorem on Euler’s phi-function and to recent work by RonBrown on natural density of square-free numbers.
In a recent paper [1], Ron Brown has computed the natural density of the set ofsquare-free numbers divisible by a but relatively prime to b , where a and b arerelatively prime square-free integers. Here we note a simple remark on infiniteseries, one of whose consequences generalizes a key argument in that work. Wethen derive a consequence of a well-known result on the Euler ϕ -function. The” m = p ” case of that consequence follows from En-Naoui[2] who anticipatessome of our arguments.. Remark 1
Let ∞ P i =1 a i be an absolutely convergent series of complex numbers,and for i ≥ , f i : N ∪ { } → C with lim N →∞ f i ( N ) = D (independent of i ) andthe f i uniformly bounded. Then lim N →∞ ∞ P i =1 a i f i ( N ) = D ∞ X i =1 a i . Proof.
This is a special case of the Lebesgue Dominated Convergence Theorem(using the counting measure and applied to the sequence { a i f i ( n ) } ∞ n =1 ). Topreserve the elementary character of the arguments here, we give an ”Introduc-tory Analysis” proof.Let ε > B for which | f i ( N ) − D | < B for all i and N . Choose k ∈ N with ∞ P i = k +1 | a i | < ε B , and chooseM such that for all N ≥ M and 1 ≤ i ≤ k, | f i ( N ) − D | < ε/ (1 + 2 k P j =1 | a j | ) . N ≥ M (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X i =1 a i f i ( N ) − D ∞ X i =1 a i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X i =1 a i ( f i ( N ) − D ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ k X i =1 | a i | | ( f i ( N ) − D ) | + ∞ X i = k +1 | a i | | ( f i ( N ) − D ) | < k X i =1 | a i | · ε/ (1 + 2 k X i =1 | a i | ) + ε B · B < ε
For all applications of the remark above, we first derive the following conse-quence involving a ”linear division-based” recursion.
Lemma 2
Let,
F, G : N ∪ { } → C , < m ∈ N , α, β, D ∈ C satisfy theconditions (1) lim N →∞ F ( N ) /N = D , (2) | β | < m , (3) G ( N ) = αF ( ⌊ N/m ⌋ ) + βG ( ⌊ N/m ⌋ ) , and (4) F (0) = G (0) = 0 . Then lim N →∞ G ( N ) /N = Dαm − β . Proof.
Recursively expand (using condition (3) and ⌊⌊ a/b ⌋ /c ⌋ = ⌊ a/ ( bc ) ⌋ forpositive integers a, b, c ) we have for N > G ( N ) /N = αm · F ( ⌊ N/m ⌋ ) N/m + αβm F ( (cid:4) N/m (cid:5) ) N/m + · · · + αβ j − m j F ( (cid:4) N/m j (cid:5) ) N/m j + αβ j − m j G ( (cid:4) N/m j (cid:5) ) N/m j (*) . By properties (1), (2) and (4), this implies we have G ( N ) /N = ∞ X i =1 αβ i − m i F ( (cid:4) N/m i (cid:5) ) N/m i After all, for any fixed N this is actually a finite sum by (4) and the finalterm in display (*) above is 0 for large j . Now by Lemma 1, taking a i = αβ i − m i and f i ( N ) = F ( ⌊ N/m i ⌋ ) N/m i , it follows that lim N →∞ G ( N ) /N = D ∞ X i =1 αβ i − m i = Dαm − β .We can derive some simple applications.Application 1. Let m be an integer greater than 1. Call an integer n oddlydivisible by m if the largest nonnegative integer t with m t | n is odd. Similarly de-fine evenly divisible. (Note that by this definition, a number not divisible by m isevenly divisible by m .) Set F ( n ) = n and G ( n ) = |{ i ∈ N : 1 ≤ i ≤ n , i oddly divisible by m }| .Since there is a 1-1 correspondence between { i ∈ N : 1 ≤ i ≤ n , i oddly divisible by m } { i ∈ N : 1 ≤ i ≤ ⌊ n/m ⌋ and i is evenly divisible by m } , we quickly see that G ( n ) = F ( ⌊ n/m ⌋ ) − G ( ⌊ n/m ⌋ ). Now apply the Lemma with D = α = − β = 1to get lim N →∞ G ( N ) /N = m +1 . So the natural density of numbers oddly divisibleby m is m +1 . (This is also easily arrived at by an inclusion-exclusion argument.)Application 2. In Brown[1] the natural density of the set of square-freenumbers divisible by primes p , · · · , p k is shown to be 6 /π k Q i =1 1 p k +1 . (In fact, hemore generally computes the density of the set of such numbers also not divisibleby a further set of primes and reduces that problem to this one.) Using that thenatural density of the set of square-free numbers is 6 /π , the cited result followsdirectly from [?] Lemma 3, which states that, for a square-free integer t and aprime p not dividing t , if the natural density of the set of square-free numbersdivisible by t is D , then the natural density of the set of square-free numbersdivisible by tp is D/ ( p + 1). To do this (converting to our notation), letting C be the set of square-free numbers, F ( x ) = |{ r ∈ C : t | r, r ≤ x }| and G ( x ) = |{ r ∈ C : pt | r, r ≤ x }| Brown quickly establishes that F ( x/p ) = G ( x/p ) + G ( x ) . Noting that we can replace arguments here with their greatest integers, and thatall hypotheses are in place, we can apply Lemma 2 with α = 1 , β = − , m = p to arrive at lim N →∞ G ( N ) /N = Dp +1 . ϕ -function It is well-known that lim N →∞ N X n =1 ϕ ( n ) n ! /N = 6 /π . (See for example [3].)From this we can derive the following proposition, where we sum only overmultiples of an integer m : Proposition 3
Let m be a positive integer, and let p , · · · , p k the distinct primedivisors of m . Then lim N →∞ X m | n ≤ N ϕ ( n ) n /N = 6 π m k Y j =1 p j p j Some numerical evidence: N = 1000 , m = 5 . Here P | n ≤ ϕ ( n ) n ≈ . π · ≈ . N = 100000 , m = 200 . Here P | n ≤ ϕ ( n ) n ≈ . π · · ≈ . = 10000000 , m = 12348 . Here P | n ≤ ϕ ( n ) n ≈ . π · · · ≈ . Proof.