A280294 a(n) = a(n-1) + 2^n * a(n-2) with a(0) = 1 and a(1) = 1.

1, 1, 5, 13, 93, 509, 6461, 71613, 1725629, 38391485, 1805435581, 80431196861, 7475495336637, 666367860021949, 123144883455482557, 21958686920654707389, 8092381769059159562941, 2886261393833112966453949, 2124255587862077437434059453

Offset: 0

Author

Seiichi Manyama, Dec 31 2016

Keywords

• nonn

Comments

The Rogers-Ramanujan continued fraction is defined by R(q) = q^(1/5)/(1+q/(1+q^2/(1+q^3/(1+ ... )))). The limit of a(n)/A015459(n+2) is 2^(-1/5) * R(2).

Examples

			1/1 = a(0)/A015459(2).
1/(1+2/1) = 1/3 = a(1)/A015459(3).
1/(1+2/(1+2^2/1)) = 5/7 = a(2)/A015459(4).
1/(1+2/(1+2^2/(1+2^3/1))) = 13/31 = a(3)/A015459(5).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..114
• Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction

Crossrefs

Cf. A015463, A015459, A128915.

Cf. similar sequences with the recurrence a(n-1) + q^n * a(n-2) for n>1, a(0)=1 and a(1)=1: this sequence (q=2), A279543 (q=3), A280340 (q=10).

Programs

Mathematica

nxt[{n_,a_,b_}]:={n+1,b,b+2^(n+1)*a}; NestList[nxt,{1,1,1},20][[All,2]] (* Harvey P. Dale, Jul 17 2020 *)

Python

def a():
    a, b, p = 1, 0, 1
    while True:
        p, a, b = p + p, b, b + p * a
        yield b
A280294 = a()
print([next(A280294) for  in range(19)]) # _Peter Luschny, Dec 05 2017