A083697 a(n) = 2^(2^n - 1) * Fibonacci(2^n).

1, 2, 24, 2688, 32342016, 4677882957791232, 97861912906883207538212742365184, 42829440312913272520181533609472356498655100482256687829780267008

Offset: 0

Author

Mario Catalani (mario.catalani(AT)unito.it), May 22 2003

Keywords

• easy
• nonn

Comments

A083696(n)/a(n) converges to sqrt(5).

Similar to A081460: a(n) is the denominator of the same mapping f(r)=(1/2)(r+5/r) but with initial value r=1.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10

Crossrefs

Cf. A000045, A058635, A058891, A081460, A083696.

Programs

Magma

[2^(2^n -1)*Fibonacci(2^n): n in [0..8]]; // G. C. Greubel, Jan 14 2022

Mathematica

Table[Sum[Product[2^n -k, {k,0,2*r}]k^r/(2*r+1)!, {r,0,2^n -1}], {n,0,8}]
Table[2^(2^n -1)*Fibonacci[2^n], {n,0,8}] (* G. C. Greubel, Jan 14 2022 *)

Sage

[2^(2^n -1)*lucas_number1(2^n, 1, -1) for n in (0..8)] # G. C. Greubel, Jan 14 2022

Formula

a(n) = 2*a(n-1)*A083696(n-1).

a(n) = A058635(n) * A058891(n).

a(n) = 2^(2^n - 1) * A000045(2^n).

a(n) = Sum_{r=0..(2^n -1)} (5^r/(2*r+1)!)*Product_{k=0..2*r} (2^n - k).

Extensions

The next term is too large to include.

Better description from Ralf Stephan, Aug 29 2004