0,4

The recurrence can also be written a(n)+1=(a(n-1)+1)*(a(n-2)+1) or log_p(a(n)+1)=log_p(a(n-1)+1)+log_p(a(n-2)+1), respectively. Setting a(1)=p-1 for any natural p>1, it follows that log_p(a(n)+1)=Fib(n). Hence any other sequence p^Fib(n)-1 could also serve as a valid solution to that recurrence, only depending on the value of the term a(1). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007

Written in binary, a(n) contains Fib(n) 1's (Fib(n)=A000045(n)). Thus the sequence converted to base-2 is A007088(a(n))=0,1,1,11,111,11111,11111111,... - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007

The solution to the recurrence a(0) = 0; a(1) = 1; a(n) = a(n-1)*a(n-2) + a(n-1) + a(n-2).

a(n)=A000301(n+1)-1. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 26 2007

a(n)=a(n-2)*2^ceiling(log_2(a(n-1)))+a(n-1) for n>1. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007

a[0, k_] = 0; a[1, k_] = 1; a[n_, k_] := (k - 1)*a[n - 1, k]*a[n - 2, k] + a[n - 1, k] + a[n - 2, k]; Table[ a[n, 2], {n, 0, 14} ]

Cf. A000045 and A000301.

Cf. A000045, A061107.

See A131293 for a base-10 analog with Fib(n) 1's.

Sequence in context: A073917 A030521 A105767 this_sequence A074047 A121810 A081475

Adjacent sequences: A063893 A063894 A063895 this_sequence A063897 A063898 A063899

nonn

Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 29 2001