A079404 Let G(n) be the set of numbers between 2^(n-1) and 2^n-1, inclusive. There is a unique number m(n) in G(n) so that the denominator of the m(n)-th partial sum of the double harmonic series is divisible by smaller 2-power than that of others in G(n). This power is defined to be a(n).
0, 1, 1, 3, 4, 3, 3, 5, 7, 9, 10, 9, 10, 12, 14, 13, 13, 15, 17, 19, 19
Offset: 2
Keywords
Examples
a(3)=1 because G(3)={4,5,6,7} and among Sum_{1 <= k < l <= 4} 1/(kl) = 35/24, Sum_{1 <= k < l <= 5} 1/(kl) = 15/8, Sum_{1 <= k < l <= 6} 1/(kl) = 203/90, Sum_{1 <= k < l <= 7} 1/(kl) = 469/180, 90 has the smallest 2-power factor among the denominators.
References
- Partial sums of multiple zeta value series II: finiteness of p-divisible sets.
Links
- J. Zhao, Partial sums of multiple zeta value series II: finiteness of p-divisible sets, arXiv:math/0303043 [math.NT], 2003-2010. See (24) p. 11.
Crossrefs
Cf. A079403.
Programs
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Maple
sequ := proc(T) local b,counter,A,n,t,psum,innersum; psum := 0; innersum := 0; A := array(1..T-1); for t to T-1 do for n from 2^(t) to 2^(t+1)-1 do innersum := innersum+1/(n-1); psum := psum+innersum/n; if 2^(2*t)*psum mod 2^(2*t+1)=0 then print(`The conjecture that 2 never divides the numerators of partial sums of double harmonic series is wrong.`); else b := 0; counter := 2*t; while b=0 do b := 2^counter*psum mod 2; counter := counter-1; od; if counter
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Mathematica
nmax = 15; dhs = Array[HarmonicNumber[# - 1 ]/# &, 2^nmax] // Accumulate; Print["dhs finished"]; f[s_] := IntegerExponent[s // Denominator, 2]; a[n_] := Table[{f[dhs[[k]] ], k}, {k, 2^(n - 1), 2^n - 1}] // Sort // First // First; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 2, nmax}] (* Jean-François Alcover, Jan 22 2018 *)
Extensions
Typo in data corrected by Jean-François Alcover, Jan 22 2018
Comments