A225918 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.

1, 9, 32, 98, 287, 828, 2377, 6812, 19510, 55866, 159958, 457987, 1311283, 3754381, 10749290, 30776629, 88117519, 252291984, 722344942, 2068168017, 5921435438, 16953843853, 48541071558, 138979434294, 397916291012, 1139286366040, 3261925819973, 9339320097349, 26739694491713

Offset: 1

Author

Clark Kimberling, May 21 2013

Keywords

• nonn

Comments

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:

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,... ?

1/(n+6) ... A225921: 1,14,50,... ?

1/(n+7) ... A225922: 1,16,48,... ?

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.

For sequences A225920-A225922, linear recurrence is unlikely to exist. - Max Alekseyev, Jan 27 2022

Examples

			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).
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) < ...
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... .
		

Crossrefs

Cf. A003462, A134931, A116952.

Cf. A225919, A225920, A225921, A225922.

Programs

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 *)

PARI

lista(nn) = {default(realprecision, 100); my(k=5, r=1, s); print1(1); for(n=2, nn, s=0; while((s+=1./k)Jinyuan Wang, Jun 14 2020

Formula

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

Extensions

a(12)-a(18) from Robert G. Wilson v, May 22 2013

a(19) from Jinyuan Wang, Jun 14 2020

Terms a(20) on from Max Alekseyev, Jan 27 2022