A181870 - OEIS

%I #7 Mar 30 2012 18:40:13

%S 2,1,18,32401,1049824801000018,

%T 110213211279472739469283600032400000000000000032401

%N 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 The calculations for the first few values of the sequence are

%C ... 2^3 = 8 so a(3) = 18/1 = 18

%C ... 18^3 = 5832 so a(4) = 583218/18 = 32401

%C ... 32401^3 = 34015373377201 so a(5) = 34015373377201583218/32401 = 1049824801000018.

%C 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 DEFINITION

%F a(1) = 2, a(2) = 1, and for n >= 3,

%F (1)... a(n) = concatenate (a(n-1)^3,a(n-2)^3,...,a(1)^3)/a(n-1).

%F RECURRENCE RELATION

%F It appears that for n >= 2,

%F (2)... a(n+2) = 100^F(n-1,3)*a(n+1)^2 + a(n)

%F = 100^A006190(n-1)*a(n+1)^2 + a(n)

%F = 10^A052991(n-1)*a(n+1)^2 + a(n),

%F where F(n,3) is the n-th Fibonacci polynomial F(n,x) evaluated at x = 3.

%e 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

%p M:=6: a:=array(1..M):s:=array(1..M):

%p a[1]:=2: a[2]:=1:

%p s[1]:=convert(a[1]^3,string): s[2]:=cat(convert(a[2]^3,string),s[1]):

%p for n from 3 to M do

%p a[n] := parse(s[n-1])/a[n-1];

%p s[n]:= cat(convert(a[n]^3,string),s[n-1]);

%p end do:

%p seq(a[n],n = 1..M);

%Y Cf. A006190, A052991, A181754, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181869

%K nonn,easy,base

%O 1,1

%A _Peter Bala_, Dec 09 2010