A260005 a(n) = f(2,n,2), where f is the Sudan function defined in A260002.

19, 10228, 15569256417, 5742397643169488579854258, 36681813266165915713665394441869800619098139628586701684547

Offset

0,1

Comments

Naturally a subsequence of A260004.

See A260002-A260003 for the evaluation of the Sudan function.

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

Cf. A048493 (f(2,n,1)), A260002, A260003, A260004, A260006.

Sequence in context: A265964 A156973 A270972 * A276739 A093400 A265169

Adjacent sequences: A260002 A260003 A260004 * A260006 A260007 A260008

Keyword

nonn

Author

Natan Arie Consigli, Jul 23 2015

Status

approved