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A225919 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+4) and a(1) = 1.

%I A225919 #22 Jan 27 2022 20:42:04
%S A225919 1,11,40,124,367,1070,3104,8989,26016,75280,217815,630210,1823388,
%T A225919 5275597,15263836,44162700,127775471,369691398,1069624136,3094731965,
%U A225919 8953954568,25906380024,74954649447,216865477466,627454543012,1815407450477,5252498763364,15196997925652
%N A225919 a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+4) and a(1) = 1.
%C A225919 Conjecture: a(n) is linearly recurrent. See A225918 for details.
%F A225919 For n>=3, a(n) = ceiling( (a(n-1)+4.5)^2 / (a(n-2)+4.5) - 4.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - _Max Alekseyev_, Jan 27 2022
%e A225919 a(1) = 1 by decree; a(2) = 11 because 1/6 + ... + 1/14 < 1 < 1/6 + ... + 1/(11+4), so that a(3) = 40 because 1/16 + ... + 1/43 < 1/6 + ... + 1/15 < 1/16 + ... + 1/(40+4).
%e A225919 Successive values of a(n) yield a chain: 1 < 1/(1+5) + ... + 1/(11+4) < 1/(11+5) + ... + 1/(40+4) < 1/(40+5) + ... + 1/(124+4) < ...
%e A225919 Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.8931..., it appears that lim_{n->infinity} b(n) = log(R) = 1.0623... .
%t A225919 nn = 11; f[n_] := 1/(n + 4); 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]]
%Y A225919 Cf. A225918.
%K A225919 nonn
%O A225919 1,2
%A A225919 _Clark Kimberling_, May 21 2013
%E A225919 a(11)-a(16) from _Robert G. Wilson v_, May 22 2013
%E A225919 a(17)-a(18) from _Robert G. Wilson v_, Jun 13 2013
%E A225919 a(19) from _Jinyuan Wang_, Jun 14 2020
%E A225919 Terms a(20) on from _Max Alekseyev_, Jan 27 2022