A077475 Greedy powers of (8/13): Sum_{n>=1} (8/13)^a(n) = 1.

1, 2, 11, 14, 25, 28, 30, 37, 39, 41, 43, 46, 48, 51, 54, 57, 60, 64, 66, 71, 76, 78, 80, 82, 84, 90, 95, 101, 103, 106, 110, 113, 115, 117, 127, 133, 135, 140, 146, 152, 157, 160, 162, 165, 167, 170, 173, 179, 181, 185, 189, 196, 200, 203, 206, 209, 212, 215, 220

Offset

1,2

Comments

The n-th greedy power of x, when 0.5 < x < 1, is the smallest integer exponent a(n) that does not cause the power series sum_{k=1..n} x^a(k) to exceed unity.

A heuristic argument suggests that the limit of a(n)/n is m - sum_{n=m..inf} log(1 + x^n)/log(x) = 3.8170308430..., where x=8/13 and m=floor(log(1-x)/log(x))=1. - Paul D. Hanna, Nov 16 2002

By the time you reach sum_{n=1..59} (8/13)^a(n), the difference between that sum and 1 is only 1.6*10^-47.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Formula

a(n)=sum_{k=1..n}floor(g_k) where g_1=1, g_{n+1}=log_x(x^frac(g_n) - x) (n>0) at x=(8/13) and frac(y) = y - floor(y).

a(n) seems to be asymptotic to c*n with c around 3.7... - Benoit Cloitre

Example

a(3)=11 since (8/13) +(8/13)^2 +(8/13)^11 < 1 and (8/13)+(8/13)^2+(8/13)^10 >1.

Maple

V:= Vector(100):
V[1]:= 1: T:= 1 - 8/13:
for n from 2 to 100 do
V[n]:= -floor(log[13/8](T));
T:= T - (8/13)^V[n];
od:
convert(V, list); # Robert Israel, Aug 11 2020

Mathematica

s = 0; a = {}; Do[ If[s + (8/13)^n < 1, s = s + (8/13)^n; a = Append[a, n]], {n, 1, 250}]; a
heuristiclimit[x_] := (m=Floor[Log[x, 1-x]])+1/24+Log[x, Product[1+x^n, {n, 1, m-1}]/DedekindEta[I Log[x]/-Pi]*DedekindEta[ -I Log[x]/2/Pi]]; N[heuristiclimit[8/13], 20]

Crossrefs

Cf. A077468, A077469, A077470, A077471, A077472, A077473, A077474.

Sequence in context: A031192 A283930 A034039 * A371071 A374446 A168498

Adjacent sequences: A077472 A077473 A077474 * A077476 A077477 A077478

Keyword

easy,nonn

Author

Paul D. Hanna, Nov 06 2002

Extensions

Edited and extended by Robert G. Wilson v, Nov 08 2002. Also extended by Benoit Cloitre, Nov 06 2002

Status

approved