cp's OEIS Frontend

A181865 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the cubes of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

%I A181865 #5 Mar 30 2012 18:40:13
%S A181865 1,2,9,2081,90004330561,2081000000000008100779519733758574721
%N A181865 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the cubes of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).
%C A181865 The calculations for the first few values of the sequence are
%C A181865 ... 2^3 = 8 so a(3) = 18/2 = 9
%C A181865 ... 9^3 = 729 so a(4) = 18729/9 = 2081
%C A181865 ... 2081^3 = 9011897441 so a(5)=187299011897441/2081 = 90004330561.
%C A181865 For similarly defined sequences see A181754 through A181756 and
%C A181865 A181864 through A181870.
%F A181865 DEFINITION
%F A181865 a(1) = 1, a(2) = 2, and for n >= 3
%F A181865 (1)... a(n) = concatenate(a(1)^3,a(2)^3,...,a(n-1)^3)/a(n-1).
%F A181865 RECURRENCE RELATION
%F A181865 For n >= 2
%F A181865 (2)... a(n+2) = a(n+1)^2 + 10^F(n,3)*a(n),
%F A181865 where F(n,3) is the Fibonacci polynomial F(n,x) evaluated at x = 3.
%F A181865 F(n,3) = A006190(n).
%p A181865 #A181865
%p A181865 M:=7:
%p A181865 a:=array(1..M):s:=array(1..M):
%p A181865 a[1]:=1:a[2]:=2:
%p A181865 s[1]:=convert(a[1]^3,string):
%p A181865 s[2]:=cat(s[1],convert(a[2]^3,string)):
%p A181865 for n from 3 to M do
%p A181865 a[n] := parse(s[n-1])/a[n-1];
%p A181865 s[n]:= cat(s[n-1],convert(a[n]^3,string));
%p A181865 end do:
%p A181865 seq(a[n],n = 1..M);
%Y A181865 Cf. A006190 (F(n,3)), A181754, A181755, A181756, A181864, A181866, A181867, A181868, A181869, A181870
%K A181865 nonn,easy,base
%O A181865 1,2
%A A181865 _Peter Bala_, Nov 29 2010