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A077473 Greedy powers of (5/9): Sum_{n>=1} (5/9)^a(n) = 1.

%I A077473 #12 May 22 2025 21:36:45
%S A077473 1,2,4,6,8,11,13,18,21,24,27,28,30,32,35,37,40,43,45,50,51,59,62,64,
%T A077473 73,76,79,82,83,86,87,93,96,99,100,103,106,108,110,112,113,117,118,
%U A077473 121,123,126,127,131,137,139,140,143,145,146,148,154,155,157,163,165,166
%N A077473 Greedy powers of (5/9): Sum_{n>=1} (5/9)^a(n) = 1.
%C A077473 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 A077473 A heuristic argument suggests that the limit of a(n)/n is m - Sum_{n>=m} log(1 + x^n)/log(x) = 2.8326013771..., where x=5/9 and m=floor(log(1-x)/log(x))=1. - _Paul D. Hanna_, Nov 16 2002
%F A077473 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=5/9 and frac(y) = y - floor(y).
%F A077473 a(n) seems to be asymptotic to c*n with c around 2.8... - _Benoit Cloitre_
%e A077473 a(3)=4 since (5/9) +(5/9)^2 +(5/9)^4 < 1 and (5/9) +(5/9)^2 +(5/9)^3 > 1.
%t A077473 s = 0; a = {}; Do[ If[s + (5/9)^n < 1, s = s + (5/9)^n; a = Append[a, n]], {n, 1, 173}]; a
%t A077473 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[5/9], 20]
%Y A077473 Cf. A077468, A077469, A077470, A077471, A077472, A077474, A077475.
%K A077473 easy,nonn
%O A077473 1,2
%A A077473 _Paul D. Hanna_, Nov 06 2002
%E A077473 Edited and extended by _Robert G. Wilson v_, Nov 08 2002. Also extended by _Benoit Cloitre_, Nov 06 2002