%I #36 Jan 27 2022 20:54:07
%S 1,9,32,98,287,828,2377,6812,19510,55866,159958,457987,1311283,
%T 3754381,10749290,30776629,88117519,252291984,722344942,2068168017,
%U 5921435438,16953843853,48541071558,138979434294,397916291012,1139286366040,3261925819973,9339320097349,26739694491713
%N 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+3) and a(1) = 1.
%C Suppose that f(n) is a sequence of positive real numbers for which the series f(1) + f(2) + ... diverges. Put a(1) = 1 and a(n) = least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1. Conjecture: a(n) is linearly recurrent for the choices of f(n) shown here:
%C f(n) ...... a(n)................ recurrence coefficients
%C 1/n ....... A003462: 1,4,13,.... (4,-3)
%C 1/(n+1) ... A134931: 1,6,21,.... (4,-3)
%C 1/(n+2) ... A116952: 1,8,29,.... (4,-3)
%C 1/(n+3) ... A225918: 1,9,32,.... (3,0,-1,0,-1)
%C 1/(n+4) ... A225919: 1,11,40,... (4,-4,3,-2)
%C 1/(n+5) ... A225920: 1,13,48,... ?
%C 1/(n+6) ... A225921: 1,14,50,... ?
%C 1/(n+7) ... A225922: 1,16,48,... ?
%C Assuming linear recurrence, it appears that lim_{n->infinity} a(n+1)/a(n) is the greatest root, R, of the characteristic polynomial of the recurrence, and that lim_{n->infinity} (1/(a(n-1)+1) + ... + 1/a(n)) = log R.
%C For sequences A225920-A225922, linear recurrence is unlikely to exist. - _Max Alekseyev_, Jan 27 2022
%F For n>=3, a(n) = ceiling( (a(n-1)+3.5)^2 / (a(n-2)+3.5) - 3.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - _Max Alekseyev_, Jan 27 2022
%e a(1) = 1 by decree; a(2) = 9 because 1/5 + ... + 1/11 < 1 < 1/5 + ... + 1/(9+3), so that a(3) = 32 because 1/13 + ... + 1/34 < 1/5 + ... + 1/12 < 1/13 + ... + 1/(32+3).
%e Successive values of a(n) yield a chain: 1 < 1/(1+4) + ... + 1/(9+3) < 1/(9+4) + ... + 1/(32+3) < 1/(32+4) + ... + 1/(98+3) < ...
%e Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.8631..., it appears that lim_{n->infinity} b(n) = log R = 1.0519... .
%t 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 *)
%o (PARI) lista(nn) = {default(realprecision, 100); my(k=5, r=1, s); print1(1); for(n=2, nn, s=0; while((s+=1./k)<r, k++); r=s; print1(", ", (k++)-4)); } \\ _Jinyuan Wang_, Jun 14 2020
%Y Cf. A003462, A134931, A116952.
%Y Cf. A225919, A225920, A225921, A225922.
%K nonn
%O 1,2
%A _Clark Kimberling_, May 21 2013
%E a(12)-a(18) from _Robert G. Wilson v_, May 22 2013
%E a(19) from _Jinyuan Wang_, Jun 14 2020
%E Terms a(20) on from _Max Alekseyev_, Jan 27 2022
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