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).

1, 2, 9, 2081, 90004330561, 2081000000000008100779519733758574721

Offset: 1

Author

Peter Bala, Nov 29 2010

Keywords

• nonn
• easy
• base

Comments

The calculations for the first few values of the sequence are

... 2^3 = 8 so a(3) = 18/2 = 9

... 9^3 = 729 so a(4) = 18729/9 = 2081

... 2081^3 = 9011897441 so a(5)=187299011897441/2081 = 90004330561.

For similarly defined sequences see A181754 through A181756 and

A181864 through A181870.

Crossrefs

Cf. A006190 (F(n,3)), A181754, A181755, A181756, A181864, A181866, A181867, A181868, A181869, A181870

Programs

Maple

#A181865
M:=7:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1]^3,string):
s[2]:=cat(s[1],convert(a[2]^3,string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n]^3,string));
end do:
seq(a[n],n = 1..M);

Formula

DEFINITION

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

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

RECURRENCE RELATION

For n >= 2

(2)... a(n+2) = a(n+1)^2 + 10^F(n,3)*a(n),

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

F(n,3) = A006190(n).