A079403 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.
3, 6, 13, 27, 54, 109, 219, 439, 879, 1759, 3518, 7037, 14075, 28151, 56303, 112606, 225212, 450424, 900848, 1801696, 3603393, 7206787, 14413574, 28827148, 57654296, 115308593, 230617186, 461234373
Offset: 2
Examples
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.
Links
- J. Zhao, Partial sums of multiple zeta value series II: finiteness of p-divisible sets, arXiv:math/0303043 [math.NT], 2003-2010. See (23) p. 11.
Crossrefs
Cf. A079404.
Programs
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Maple
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: -
Mathematica
nmax = 15; dhs = Array[HarmonicNumber[# - 1]/# &, 2^nmax] // Accumulate; Print["dhs finished"]; f[s_] := IntegerExponent[s // Denominator, 2]; 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]]]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 2, nmax}] (* Jean-François Alcover, Jan 22 2018 *)
Formula
From Benoit Cloitre, Jan 24 2003: (Start)
a(n+1) - 2*a(n) = (a(n+1) mod 2);
a(n) = floor(c*2^n) where c = 1.718232... = 3/2 + Sum_{k>=2} (a(k+1) - 2*a(k))/2^k. (End)
Extensions
a(23)-a(29) from Sean A. Irvine, Aug 12 2025
Comments