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a(n) = ((2^(3*(2^n))) - 1).
2

%I #20 Jan 14 2024 08:06:59

%S 7,63,4095,16777215,281474976710655,79228162514264337593543950335,

%T 6277101735386680763835789423207666416102355444464034512895,

%U 39402006196394479212279040100143613805079739270465446667948293404245721771497210611414266254884915640806627990306815

%N a(n) = ((2^(3*(2^n))) - 1).

%C Also "Denominators for Fibonacci Binary Verticals viewed as Periodic Binary Fractions": The cycle of bit-n of Fibonacci numbers in binary is (3*(2^n)). Looking from top to bottom they can be viewed as non-finite periodic binary fractions, with each fraction computed as the n-th element of A036286 divided by the n-th element of A036287.

%H Antti Karttunen, <a href="/A036284/a036284.c.txt">C program related to this sequence</a>

%t 2^(3 2^Range[0,10])-1 (* _Harvey P. Dale_, Jun 19 2011 *)

%o (PARI) A036287(n) = ((2^(3*(2^n))) - 1); \\ _Antti Karttunen_, Jan 14 2024

%Y Cf. A036286, A000045.

%K nonn,frac

%O 0,1

%A _Antti Karttunen_