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

1,2

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.

Links

Table of n, a(n) for n=1..5.

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).

Example

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.

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);

Crossrefs

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

A181867, A181868, A181869, A181870

Sequence in context: A290306 A113635 A136095 * A106897 A244269 A195325

Adjacent sequences: A181863 A181864 A181865 * A181867 A181868 A181869

Keyword

nonn,easy,base

Author

Peter Bala, Nov 29 2010

Status

approved