A062878 a(n) is the position of A050614(n) in A062877.

1, 3, 6, 15, 24, 60, 102, 255, 384, 960, 1632, 4080, 6168, 15420, 26214, 65535, 98304, 245760, 417792, 1044480, 1579008, 3947520, 6710784, 16776960, 25166208, 62915520, 106956384, 267390960, 404232216, 1010580540, 1717986918, 4294967295, 6442450944

Offset: 0

Author

Antti Karttunen, Jun 26 2001

Keywords

• nonn

Comments

In binary this sequence looks like 1, 11, 110, 1111, 11000, 111100, 1100110, 11111111, 110000000, 1111000000, 11001100000, 111111110000, 1100000011000, 11110000111100, 110011001100110, ...

Sequence A282387 may be the same, but I cannot prove nor disprove this beyond a(22). - Robert Price, Feb 13 2017

Agrees with A282387 for at least 1000 terms. - Sean A. Irvine, Apr 14 2023

Programs

Mathematica

A050614 = Table[k = Floor[Log[2, n + 1]]; Product[j = 2^(i + 1); l = Fibonacci[j + 1] + Fibonacci[j - 1]; If[BitAnd[2^i, n] == 0, b = 0, b = 1]; l^b, {i, 0, k}], {n, 0, 200}]; A062877 = Union[Total /@ Subsets[Fibonacci[Range[1, 46, 2]]]]; Flatten[Table[Position[ A062877, A050614[[i]] ] - 1, {i, 1, 25}]] (* Robert Price, Feb 13 2017 *)

Formula

a(2^n-1) = 2^(2^n) - 1. - Philippe Deléham, Apr 05 2007

a(n) = Sum_{k=0..n} A127872(n,k)*2^k. - Philippe Deléham, Oct 09 2007

Extensions

a(15)-a(22) from Robert Price, Feb 13 2017

More terms from Sean A. Irvine, Apr 14 2023