
COMMENTS

Suppose that f(n) is a sequence of positive real numbers for which the series f(1) + f(2) + ... diverges. Put a(1) = f(1) and a(n) = least k such that f(a(n1)+1) + ... + f(k) > f(a(n2)+1) + ... + f(a(n)1) for n > 1. Conjecture: a(n) is linearly recurrent for the choices of f(n) shown here:
f(n) ...... a(n)................ recurrence coefficients
1/n ....... A003462: 1,4,13,.... (4,3)
1/(n+1) ... A134931: 1,6,21,.... (4,3)
1/(n+2) ... A116952: 1,8,29,.... (4,3)
1/(n+3) ... A225918: 1,9,32,.... (3,0,1,0,1)
1/(n+4) ... A225919: 1,11,40,... (4,4,3,2)
1/(n+5) ... A225920: 1,13,48,... (3,1,4,1)
1/(n+6) ... A225921: 1,14,50,... ?
1/(n+7) ... A225922: 1,16,48,... (2,3,1,3)
Assuming linear recurrence, it appears that lim a(n+1)/a(n) is the greatest root, R, of the characteristic polynomial of the recurrence, and that lim( 1/(a(n1)+1) + ... + 1/a(n) ) = log R.


MATHEMATICA

nn = 11; f[n_] := 1/(n + 3); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &]; s = 0; a[3] = NestWhile[# + 1 &, a[2] + 1, ! (s += f[#]) >= g[a[2]]  f[1] &]; Do[s = 0; a[z] = NestWhile[# + 1 &, a[z  1] + 1, ! (s += f[#]) >= g[a[z  1]]  g[a[z  2]] &], {z, 4, nn}]; m = Map[a, Range[nn]] (* Peter J. C. Moses, May 13 2013 *)
