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%I #10 Nov 26 2015 00:20:46
%S 0,1,6,1,15,6,42,16,24,30,44,6,1365,42,150,256,3855,72,74898,30,336,
%T 1364,44620,240,900,2730,262656,336,39672,1650,195225786,65536,
%U 1198956,131070,92190,216,616318176,524286,2123940,61680,26815350376,43344
%N a(0) = 0; for n>0, a(n) = period length of the decimal expansion of the number Sum_{i>=1} 2^(-n*i). Also period length of the fractions 1/b(n), where b(n) = 2*b(n-1) + 1, with b(1)=1.
%C In base 2 consider the numbers 0.1111111..., 0.01010101...., 0.001001001..., 0.000100010001.... where the period [0 k times, 1], where k=0,1,2,3,.... Then convert to base 10. The sequence gives the length of each period.
%C The period length of the fraction 1/A000225(n) = 1/(2^n-1) for n>0. - _Robert G. Wilson v_, Mar 30 2008
%t f[n_] := Length[RealDigits[Sum[2^(-n*k), {k, Infinity}]][[1, 1]]]; Array[f, 36] (* _Robert G. Wilson v_, Mar 30 2008 *)
%K easy,nonn,base
%O 0,3
%A _Paolo P. Lava_ & _Giorgio Balzarotti_, Mar 19 2008
%E More terms from _Robert G. Wilson v_, Mar 30 2008
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