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A087910 Exponent of the greatest power of 2 dividing the numerator of 2^1/1 + 2^2/2 + 2^3/3 + ... + 2^n/n.

%I A087910 #25 Feb 22 2020 10:35:35
%S A087910 1,2,2,5,8,5,5,13,9,10,10,12,12,12,12,22,17,18,18,21,22,21,21,27,25,
%T A087910 26,26,27,27,27,27,40,33,34,34,37,39,37,37,48,41,42,42,44,44,44,44,54,
%U A087910 49,50,50,53,54,53,53,58,57,59,62,58,58,58
%N A087910 Exponent of the greatest power of 2 dividing the numerator of 2^1/1 + 2^2/2 + 2^3/3 + ... + 2^n/n.
%C A087910 Problem 9 of the 2002 Sydney University Mathematical Society Problems competition asked for a proof that a(n) tends to infinity with n. While this is immediate from the theory of the 2-adic logarithm, elementary proofs are available.
%C A087910 a(n) tends to infinity with n implies that log(-1) = 0 in the 2-adic field, by setting x = 2 in -log(1-x) = Sum_{k>=1} x^k/k. - _Jianing Song_, Aug 05 2019
%D A087910 A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see p. 278.
%H A087910 Robert Israel, <a href="/A087910/b087910.txt">Table of n, a(n) for n = 1..5000</a>
%H A087910 Sydney University Mathematical Society <a href="http://www.maths.usyd.edu.au/u/SUMS/">Problems Competitions</a>, see Problem 9 of the 2002 competition.
%F A087910 a(n) = A007814(A108866(n)). - _Michel Marcus_, Feb 22 2020
%F A087910 Sum_{k=1..n} 2^k/k = (2^n/n)*Sum_{k=0..n-1} 1/binomial(n-1,k), so a(n) >= n - v(n,2) - max_{k=0..n-1} v(binomial(n-1,k),2) = n - A007814(n) - A119387(n) = n - floor(log_2(n)), where v(n,2) is the 2-adic valuation of n. It seems that the equality holds if and only if n = 2^m - 1 for some m. - _Jianing Song_, Feb 22 2020
%e A087910 a(5) = 8 as 2^1/1 + 2^2/2 + 2^3/3 + 2^4/4 + 2^5/5 = 256/15 whose numerator is divisible by 2^8 but not by 2^9.
%p A087910 S:= 0:
%p A087910 for n from 1 to 100 do
%p A087910   S:= S + 2^n/n;
%p A087910   a[n]:= padic:-ordp(numer(S),2);
%p A087910 od:
%p A087910 seq(a[n],n=1..100); # _Robert Israel_, Jun 09 2015
%t A087910 s[n_] := -2^(n + 1) LerchPhi[2, 1, n + 1] - I Pi;
%t A087910 a[n_] := IntegerExponent[Numerator[Simplify[s[n]]], 2];
%t A087910 Array[a, 62] (* _Peter Luschny_, Feb 22 2020 *)
%o A087910 (PARI) a(n) = valuation(sum(k=1,n,2^k/k), 2) \\ _Jianing Song_, Feb 22 2020
%Y A087910 Cf. A108866, A007814, A119387, A000523.
%K A087910 easy,nonn
%O A087910 1,2
%A A087910 _Robin Chapman_, Oct 17 2003