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).

2, 1, 18, 32401, 1049824801000018, 110213211279472739469283600032400000000000000032401

Offset: 1

Author

Peter Bala, Dec 09 2010

Keywords

• nonn
• easy
• base

Comments

The calculations for the first few values of the sequence are

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

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

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

The value of a(7) is given in the Example section below. For similarly defined sequences see A181754 through A181756 and A181864 through A181869.

Examples

			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.
		

Crossrefs

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

Programs

Maple

#A181870
M:=6: a:=array(1..M):s:=array(1..M):
a[1]:=2: a[2]:=1:
s[1]:=convert(a[1]^3,string): s[2]:=cat(convert(a[2]^3,string),s[1]):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(convert(a[n]^3,string),s[n-1]);
end do:
seq(a[n],n = 1..M);

Formula

DEFINITION

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

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

RECURRENCE RELATION

It appears that for n >= 2,

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

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

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

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