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

Author

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

Keywords

• nonn

Comments

The sequence is conjectured to go to positive infinity.

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

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
				

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