A225921 - OEIS

%I #24 Jan 27 2022 20:54:38

%S 1,14,50,150,427,1195,3324,9226,25587,70942,196672,545212,1511411,

%T 4189842,11614806,32197786,89256522,247430866,685911016,1901435842,

%U 5271031028,14611993445,40506373648,112289011899,311279955644,862909105217,2392097886032,6631210937220,18382591594862

%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+6) and a(1) = 1.

%C Conjecture:  a(n) is linearly recurrent.  See A225918 for details.

%C The sequence does not satisfy any linear recurrence of order below 50, which suggests it's unlikely to exist. - _Max Alekseyev_, Jan 27 2022

%F For n>=3, a(n) = ceiling( (a(n-1)+6.5)^2 / (a(n-2)+6.5) - 6.5 ) 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) = 14 because 1/8 + ... + 1/19 < 1 < 1/8 + ... + 1/(14+6), so that a(3) = 50 because 1/21 + ... + 1/55 < 1/8 + ... + 1/20 < 1/21 + ... + 1/(50+6).

%e Successive values of a(n) yield a chain: 1 < 1/(1+7) + ... + 1/(14+6) < 1/(14+7) + ... + 1/(50+6) < 1/(50+7) + ... + 1/(150+6) < ...

%e Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.7721..., it appears that lim_{n->infinity} b(n) = log(R) = 1.0196... .

%t nn = 11; f[n_] := 1/(n + 6); a[1] = 1; g[n_] := g[n] = Sum[f[k], {k, 1, n}]; s = 0; a[2] = NestWhile[# + 1 &, 2, ! (s += f[#]) >= a[1] &];

%t 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 Cf. A225918.

%K nonn

%O 1,2

%A _Clark Kimberling_, May 21 2013

%E a(13)-a(16) from _Robert G. Wilson v_, May 22 2013

%E a(17)-a(18) from _Robert G. Wilson v_, Jun 13 2013

%E a(19) from _Jinyuan Wang_, Jun 14 2020

%E Terms a(20) on from _Max Alekseyev_, Jan 27 2022