A111235 a(1)=a(2)=a(3)=a(4)=1. For n >= 5, a(n)= a(n-1)*a(n-2) + a(n-3)*a(n-4).

1, 1, 1, 1, 2, 3, 7, 23, 167, 3862, 645115, 2491437971, 1607264007306619, 4004398577225334507664179, 6436125704084114770053956998574742562466, 25772812612277833490303309040566300172816894832780792086674335463

Offset: 1

Author

Leroy Quet, Oct 28 2005

Keywords

• easy
• nonn

Comments

a(5*n) is always even. Every other term of the sequence is odd.

It is easy to see that a(n) >= A000301(n-3) for all n. From that we can deduce that a(n) >= 2^(Fibonacci(n-3)). Can anybody give a formula for the asymptotic behavior? - Stefan Steinerberger, Jan 21 2006

As n->infinity, log(a(n))/phi^n approaches t-(-1)^n*u/phi^(2*n), where phi=(1+sqrt(5))/2, t=0.0672009781433377128..., and u=0.766475715574332057.... - Jon E. Schoenfield, Sep 14 2013

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..21

Crossrefs

Cf. A239967.

Programs

Magma

I:=[1,1,1,1]; [n le 4 select I[n] else Self(n-1)*Self(n-2) +Self(n-3)*Self(n-4): n in [1..16]]; // Vincenzo Librandi, Mar 30 2014

Maple

a:= proc(n) a(n):= `if`(n<5, 1, a(n-1)*a(n-2) +a(n-3)*a(n-4)) end:
seq(a(n), n=1..16);  # Alois P. Heinz, Mar 30 2014

Mathematica

RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==1,a[n]==a[n-1]a[n-2]+a[n-3] a[n-4]}, a,{n,20}] (* Harvey P. Dale, Jun 06 2017 *)

Extensions

More terms from Stefan Steinerberger, Jan 21 2006

More terms from Joshua Zucker, May 04 2006