OFFSET
0,1
COMMENTS
Naturally a subsequence of A260004.
Using f(2,n,2) = f(1, f(2,n,1), f(2,n,1)+2) = 2^(f(2,n,1)+2)*(f(2,n,1)+2)-f(2,n,1)-4 and f(2,n,1) = f(1, n, n+1) = 2^(n+1)*(n+2)-(n+3) we have:
a(n)=f(2,n,2)
=f(1, 2^(n+1)*(n+2)-(n+3), 2^(n+1)*(n+2)-(n+3)+2)
=2^(2^(n+1)*(n+2)-(n+3)+2)*(2^(n+1)*(n+2)-(n+3)+2)-2^(n+1)*(n+2)+(n+3)-4
=2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1).
LINKS
Natan Arie' Consigli, Table of n, a(n) for n = 0..6
Wikipedia, Sudan function (see 3rd line of "Values of F2(x, y)" table).
FORMULA
a(n) = 2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1).
EXAMPLE
a(1) = f(2,1,2) = f(1,f(2,1,1),f(2,1,1)+2) = f(1,8,10) = 2^10(8+2)-10-2 = 10228.
MATHEMATICA
Table[2^(2^(n + 1) (n + 2) - (n + 1)) (2^(n + 1) (n + 2) - (n + 1)) - 2^(n + 1) (n + 2) + (n - 1), {n, 0, 5}] (* Vincenzo Librandi, Jul 27 2015
PROG
(Magma) [2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1):n in [0..5]]; // Vincenzo Librandi, Jul 27 2015
(PARI) a(n) = 2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1);
vector(10, n, a(n-1)) \\ Altug Alkan, Oct 01 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Natan Arie Consigli, Jul 23 2015
STATUS
approved