A067667 a(n) = (2^n)!/2^(2^n-1).

1, 1, 3, 315, 638512875, 122529844256906551386796875, 13757108753595648665519665029568345104465749222289382342659100341796875

Offset: 0

Author

Benoit Cloitre, Feb 04 2002

Keywords

• nonn

Comments

a(n) is also the number of knockout tournament seedings with 2^n teams. - Alexander Karpov, Aug 09 2015

From Zhujun Zhang, Jun 17 2019: (Start)

a(n) is also the number of heap-ordered binomial trees of order n (i.e., binomial heaps with 2^n nodes), see the Mark R. Brown reference.

a(n) is also the largest odd divisor of (2^n)!. (End)

Links

• Jianing Song, Table of n, a(n) for n = 0..8
• Mark R. Brown, Implementation and analysis of binomial queue algorithms, SIAM Journal on Computing, 1978, 7(3):298-319.
• Alexander Karpov, A theory of knockout tournament seedings, Heidelberg University, AWI Discussion Paper Series, No. 600.
• Zhujun Zhang, A Note on Counting Binomial Heaps, ResearchGate, June 2019.

Crossrefs

Cf. A049606, A069954.

Programs

Magma

[Factorial(2^n)/2^(2^n-1): n in [1..6]]; // Vincenzo Librandi, Aug 10 2015

Mathematica

Table[(2^n)! / 2^(2^n - 1), {n, 6}] (* Vincenzo Librandi, Aug 10 2015 *)

PARI

a(n) = (2^n)!/2^(2^n-1) \\ Jianing Song, Jul 15 2021

Formula

From Alexander Karpov, Aug 09 2015: (Start)

a(n) = (2^n)!/2^(2^n-1).

a(n) = (2^n-1)!!*a(n-1).

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

(End)

a(n) = A049606(2^n). - Zhujun Zhang, Jun 16 2019

a(n) = Product_{odd k < 2^n} k^(n - floor(log_2(k))). - Harry Richman, May 18 2023

Extensions

a(0) prepended by Jianing Song, Jul 15 2021