cp's OEIS Frontend

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).

%I A111235 #26 Sep 08 2022 08:45:20
%S A111235 1,1,1,1,2,3,7,23,167,3862,645115,2491437971,1607264007306619,
%T A111235 4004398577225334507664179,6436125704084114770053956998574742562466,
%U A111235 25772812612277833490303309040566300172816894832780792086674335463
%N 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).
%C A111235 a(5*n) is always even. Every other term of the sequence is odd.
%C A111235 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 A111235 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 A111235 Alois P. Heinz, <a href="/A111235/b111235.txt">Table of n, a(n) for n = 1..21</a>
%p A111235 a:= proc(n) a(n):= `if`(n<5, 1, a(n-1)*a(n-2) +a(n-3)*a(n-4)) end:
%p A111235 seq(a(n), n=1..16);  # _Alois P. Heinz_, Mar 30 2014
%t A111235 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 A111235 (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 A111235 Cf. A239967.
%K A111235 easy,nonn
%O A111235 1,5
%A A111235 _Leroy Quet_, Oct 28 2005
%E A111235 More terms from _Stefan Steinerberger_, Jan 21 2006
%E A111235 More terms from _Joshua Zucker_, May 04 2006