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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).
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%I #26 Sep 08 2022 08:45:20

%S 1,1,1,1,2,3,7,23,167,3862,645115,2491437971,1607264007306619,

%T 4004398577225334507664179,6436125704084114770053956998574742562466,

%U 25772812612277833490303309040566300172816894832780792086674335463

%N 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).

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

%C 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

%C 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

%H Alois P. Heinz, <a href="/A111235/b111235.txt">Table of n, a(n) for n = 1..21</a>

%p a:= proc(n) a(n):= `if`(n<5, 1, a(n-1)*a(n-2) +a(n-3)*a(n-4)) end:

%p seq(a(n), n=1..16); # _Alois P. Heinz_, Mar 30 2014

%t 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 *)

%o (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

%Y Cf. A239967.

%K easy,nonn

%O 1,5

%A _Leroy Quet_, Oct 28 2005

%E More terms from _Stefan Steinerberger_, Jan 21 2006

%E More terms from _Joshua Zucker_, May 04 2006