A181866 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the fourth powers of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

1, 2, 58, 200195112, 580000000008023436288643185139644928

Offset: 1

Author

Peter Bala, Nov 29 2010

Keywords

• nonn
• easy
• base

Comments

The calculations for the first few values of the sequence are

... 2^4 = 16 so a(3) = 116/2 = 58

... 58^4 = 11316496 so a(4) = 11611316496/58 = 200195112.

The value of a(6) is calculated in the Example section below.

For similarly defined sequences see A181754 through A181756 and

A181864 through A181870.

Examples

			The recurrence relation (2) above gives
a(6) = a(5)^3+10^144*a(4)
= 200 19511 20000 00000 00000 00000 00000 00000 00000 00195 11200
00080 97251 90261 07158 64917 75226 28886 69453 43420 83613 55167
37330 42401 44438 01550 47183 94579 01959 53586 66752.
a(6) has 153 digits.
		

Crossrefs

Cf. A000045, A001076 (F(n,4)), A181754, A181755, A181756, A181864, A181865,

A181867, A181868, A181869, A181870

Programs

Maple

#A181866
M:=5:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1]^4,string):
s[2]:=cat(s[1],convert(a[2]^4,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]^4,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)^4,a(2)^4,...,a(n-1)^4)/a(n-1).

RECURRENCE RELATION

For n >= 2

(2)... a(n+2) = a(n+1)^3 + (100^F(n,4))*a(n)

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

where F(n,4) is the Fibonacci polynomial F(n,x) evaluated at x = 4

and where F(n) denotes the n-th Fibonacci number A000045(n).

F(n,4) = A001076(n).