OFFSET
1,5
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
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 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Leroy Quet, Oct 28 2005
EXTENSIONS
More terms from Stefan Steinerberger, Jan 21 2006
More terms from Joshua Zucker, May 04 2006
STATUS
approved