A181866 - OEIS

%I #5 Mar 30 2012 18:40:13

%S 1,2,58,200195112,580000000008023436288643185139644928

%N 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).

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

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

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

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

%C For similarly defined sequences see A181754 through A181756 and

%C A181864 through A181870.

%F DEFINITION

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

%F (1)... a(n) = concatenate(a(1)^4,a(2)^4,...,a(n-1)^4)/a(n-1).

%F RECURRENCE RELATION

%F For n >= 2

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

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

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

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

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

%e The recurrence relation (2) above gives

%e a(6) = a(5)^3+10^144*a(4)

%e = 200 19511 20000 00000 00000 00000 00000 00000 00000 00195 11200

%e 00080 97251 90261 07158 64917 75226 28886 69453 43420 83613 55167

%e 37330 42401 44438 01550 47183 94579 01959 53586 66752.

%e a(6) has 153 digits.

%p #A181866

%p M:=5:

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

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

%p s[1]:=convert(a[1]^4,string):

%p s[2]:=cat(s[1],convert(a[2]^4,string)):

%p for n from 3 to M do

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

%p s[n]:= cat(s[n-1],convert(a[n]^4,string));

%p end do:

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

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

%Y A181867, A181868, A181869, A181870

%K nonn,easy,base

%O 1,2

%A _Peter Bala_, Nov 29 2010