A069954 a(n) = binomial(2^(n+1), 2^n)/2 = binomial(2^(n+1) - 1, 2^n) = binomial(2^(n+1) - 1, 2^n-1).

1, 3, 35, 6435, 300540195, 916312070471295267, 11975573020964041433067793888190275875, 2884329411724603169044874178931143443870105850987581016304218283632259375395

Offset

0,2

Comments

Terms are always odd. a(1) = A061548(2), a(2) = A061548(3), a(3) = A061548(5), a(4) = A061548(9), a(5) = A061548(17), ... Hence it seems that a(n) = A061548(A000051(n)).

C(2*k, k)/2 = C(2*k-1, k) = C(2*k-1, k-1) is odd if and only if k = 2^n. - Michael Somos, Mar 12 2014

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10

Formula

From Harry Richman, May 18 2023: (Start)

a(n) = A001790(2^n).

a(n) = 1/2 * A000984(2^n).

a(n) = 1/2 * (2^n + 1) * A000108(2^n).

log log a(n) ~ (log 2) * (n + 1) + log log 2 + O(n / 2^n). (End)

Example

C(2,1)/2 = C(1,0) = C(1,1) = 1. C(4,2)/2 = C(3,1) = C(3,2) = 3. C(8,4)/2 = C(7,3) = C(7,4) = 35. - Michael Somos, Mar 12 2014

Mathematica

Table[Binomial[2^(n+1) -1, 2^n -1], {n, 0, 10}] (* Vincenzo Librandi, Mar 14 2014 *)

Prog

(Magma) [Binomial(2^(n+1)-1, 2^n-1): n in [0..10]]; // Vincenzo Librandi, Mar 14 2014
(SageMath) [binomial(2^(n+1) -1, 2^n) for n in (0..9)] # G. C. Greubel, Aug 16 2022

Crossrefs

Cf. A000051, A000108, A000984, A001790, A061548.

Sequence in context: A132941 A068002 A132557 * A134098 A132513 A034174

Adjacent sequences: A069951 A069952 A069953 * A069955 A069956 A069957

Keyword

easy,nonn

Author

Benoit Cloitre, Apr 27 2002

Extensions

a(0) = 1 added by Michael Somos, Mar 12 2014

Status

approved