%I #21 Jan 22 2018 11:06:47

%S 3,6,13,27,54,109,219,439,879,1759,3518,7037,14075,28151,56303,112606,

%T 225212,450424,900848,1801696,3603393

%N Let G(t) be the set of numbers between 2^(t-1) and 2^t-1, inclusive. There is a unique number a(t) in G(t) so that the denominator of the a(t)-th partial sum of the double harmonic series is divisible by smaller 2-powers than its neighbors.

%C The n-th partial sum of double harmonic series is defined to be Sum_{1 <= k < l <= n} 1/(kl).

%H J. Zhao, <a href="https://arxiv.org/abs/math/0303043">Partial sums of multiple zeta value series II: finiteness of p-divisible sets</a>, arXiv:math/0303043 [math.NT], 2003-2010. See (23) p. 11.

%F From _Benoit Cloitre_, Jan 24 2003: (Start)

%F a(n+1) - 2*a(n) = (a(n+1) mod 2);

%F a(n) = floor(c*2^n) where c = 1.718232... = 3/2 + Sum_{k>=2} (a(k+1) - 2*a(k))/2^k. (End)

%e a(3)=6 because Sum_{1 <= k < l <= 6} 1/(kl) = 203/90, 4 does not divide 90, while 4 divides the denominators of both Sum_{1 <= k < l <= 5} 1/(kl) = 15/8 and Sum_{1 <= k < l <= 7} 1/(kl) = 469/180.

%p sequ := proc(T) local A,i,n,t,psum,innersum; psum := 0; innersum := 0; A := {}; for t to T-1 do for n from 2^t to 2^(t+1)-1 do innersum := innersum+2^T/(n-1) mod 2^(2*T); psum := psum+2^T*innersum/n mod 2^(2*T); if psum mod 2^(2*T-t+1)=0 then A := A union {n}; end if; od; od; RETURN(A); end:

%t nmax = 15; dhs = Array[HarmonicNumber[# - 1]/# &, 2^nmax] // Accumulate; Print["dhs finished"];

%t f[s_] := IntegerExponent[s // Denominator, 2];

%t a[2] = 3; a[n_] := a[n] = For[k = 2*a[n - 1], k <= 2^n - 1, k++, fk = f[dhs[[k]]]; If[f[dhs[[k-1]]] > fk && f[dhs[[k+1]]] > fk, Return[k]]];

%t Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 2, nmax}] (* _Jean-François Alcover_, Jan 22 2018 *)

%Y Cf. A079404.

%K more,nonn

%O 2,1

%A Jianqiang Zhao (jqz(AT)math.upenn.edu), Jan 06 2003