cp's OEIS Frontend

A181870 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the cubes of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).

%I A181870 #7 Mar 30 2012 18:40:13
%S A181870 2,1,18,32401,1049824801000018,
%T A181870 110213211279472739469283600032400000000000000032401
%N A181870 a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the cubes of the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).
%C A181870 The calculations for the first few values of the sequence are
%C A181870 ... 2^3 = 8 so a(3) = 18/1 = 18
%C A181870 ... 18^3 = 5832 so a(4) = 583218/18 = 32401
%C A181870 ... 32401^3 = 34015373377201 so a(5) = 34015373377201583218/32401 = 1049824801000018.
%C A181870 The value of a(7) is given in the Example section below. For similarly defined sequences see A181754 through A181756 and A181864 through A181869.
%F A181870 DEFINITION
%F A181870 a(1) = 2, a(2) = 1, and for n >= 3,
%F A181870 (1)... a(n) = concatenate (a(n-1)^3,a(n-2)^3,...,a(1)^3)/a(n-1).
%F A181870 RECURRENCE RELATION
%F A181870 It appears that for n >= 2,
%F A181870 (2)... a(n+2) = 100^F(n-1,3)*a(n+1)^2 + a(n)
%F A181870 = 100^A006190(n-1)*a(n+1)^2 + a(n)
%F A181870 = 10^A052991(n-1)*a(n+1)^2 + a(n),
%F A181870 where F(n,3) is the n-th Fibonacci polynomial F(n,x) evaluated at x = 3.
%e A181870 a(7) = 12 14695 19405 33697 08585 73749 64887 08343 06977 30753 71161 37983 82152 46308 85158 49299 58480 00000 00001 04982 48010 00000 00000 00000 00000 00000 00000 00000 00000 00000 00001 04982 48010 00018 has 167 digits.
%p A181870 #A181870
%p A181870 M:=6: a:=array(1..M):s:=array(1..M):
%p A181870 a[1]:=2: a[2]:=1:
%p A181870 s[1]:=convert(a[1]^3,string): s[2]:=cat(convert(a[2]^3,string),s[1]):
%p A181870 for n from 3 to M do
%p A181870 a[n] := parse(s[n-1])/a[n-1];
%p A181870 s[n]:= cat(convert(a[n]^3,string),s[n-1]);
%p A181870 end do:
%p A181870 seq(a[n],n = 1..M);
%Y A181870 Cf. A006190, A052991, A181754, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181869
%K A181870 nonn,easy,base
%O A181870 1,1
%A A181870 _Peter Bala_, Dec 09 2010