OFFSET
2,1
COMMENTS
The n-th partial sum of double harmonic series is defined to be Sum_{1 <= k < l <= n} 1/(kl).
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.
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)
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
more,nonn
AUTHOR
Jianqiang Zhao (jqz(AT)math.upenn.edu), Jan 06 2003
STATUS
approved