A077472 Greedy powers of (5/8): Sum_{n>=1} (5/8)^a(n) = 1.
1, 3, 5, 8, 10, 13, 15, 23, 26, 30, 33, 36, 38, 46, 48, 51, 53, 57, 61, 64, 66, 69, 72, 76, 78, 84, 88, 93, 95, 104, 106, 110, 115, 117, 121, 126, 129, 131, 136, 138, 143, 148, 150, 152, 157, 160, 164, 169, 172, 175, 179, 181, 185, 187, 191, 196, 198, 201, 203
Offset: 1
Examples
a(3)=5 since (5/8) +(5/8)^3 +(5/8)^5 < 1 and (5/8) +(5/8)^3 +(5/8)^4 > 1.
Programs
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Mathematica
s = 0; a = {}; Do[ If[s + (5/8)^n < 1, s = s + (5/8)^n; a = Append[a, n]], {n, 1, 210}]; 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[5/8], 20]
Formula
a(n) = Sum_{k=1..n} floor(g(k)) where g(1)=1, g(n+1) = log_x(x^frac(g(n)) - x) at x=(5/8) and frac(y) = y - floor(y).
a(n) seems to be asymptotic to c*n with c around 3.4... - Benoit Cloitre
Extensions
Edited and extended by Robert G. Wilson v, Nov 08 2002. Also extended by Benoit Cloitre, Nov 06 2002
Comments