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Greedy powers of (8/13): Sum_{n>=1} (8/13)^a(n) = 1.
19

%I #15 Aug 11 2020 19:40:47

%S 1,2,11,14,25,28,30,37,39,41,43,46,48,51,54,57,60,64,66,71,76,78,80,

%T 82,84,90,95,101,103,106,110,113,115,117,127,133,135,140,146,152,157,

%U 160,162,165,167,170,173,179,181,185,189,196,200,203,206,209,212,215,220

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

%C 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.

%C 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

%C 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.

%H Robert Israel, <a href="/A077475/b077475.txt">Table of n, a(n) for n = 1..10000</a>

%F 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).

%F a(n) seems to be asymptotic to c*n with c around 3.7... - _Benoit Cloitre_

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

%p V:= Vector(100):

%p V[1]:= 1: T:= 1 - 8/13:

%p for n from 2 to 100 do

%p V[n]:= -floor(log[13/8](T));

%p T:= T - (8/13)^V[n];

%p od:

%p convert(V,list); # _Robert Israel_, Aug 11 2020

%t s = 0; a = {}; Do[ If[s + (8/13)^n < 1, s = s + (8/13)^n; a = Append[a, n]], {n, 1, 250}]; a

%t 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]

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

%K easy,nonn

%O 1,2

%A _Paul D. Hanna_, Nov 06 2002

%E Edited and extended by _Robert G. Wilson v_, Nov 08 2002. Also extended by _Benoit Cloitre_, Nov 06 2002