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A077475
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Greedy powers of (8/13): Sum_{n>=1} (8/13)^a(n) = 1.
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19
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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a(3)=11 since (8/13) +(8/13)^2 +(8/13)^11 < 1 and (8/13)+(8/13)^2+(8/13)^10 >1.
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MAPLE
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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:
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MATHEMATICA
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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]
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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