A069954 -id:A069954 - cp's OEIS frontend

A037293 a(n) = Sum_{i=0..2^(n-1)} binomial(2^(n-1), i)^2.

1, 2, 6, 70, 12870, 601080390, 1832624140942590534, 23951146041928082866135587776380551750, 5768658823449206338089748357862286887740211701975162032608436567264518750790

Offset: 0

John Tromp, Dec 11 1999

For n >= 1, a(n) is the number of n-variable balanced Boolean functions. A Boolean function is balanced if it takes the values 0 and 1 an equal number of times. - Aniruddha Biswas, Nov 05 2024

Seiichi Manyama, Table of n, a(n) for n = 0..11
Aniruddha Biswas and Palash Sarkar, Counting unate and balanced monotone Boolean functions, arXiv:2304.14069 [math.CO], 2023.
Aniruddha Biswas and Palash Sarkar, Counting Unate and Monotone Boolean Functions Under Restrictions of Balancedness and Non-Degeneracy, J. Int. Seq. (2025) Vol. 28, Art. No. 25.3.4. See p. 2.
Niklas G. Johansson, Efficient Simulation of the Deutsch-Jozsa Algorithm, Master's Project, Department of Electrical Engineering & Department of Physics, Chemistry and Biology, Linkoping University, April, 2015. See Eq. (3.15).

Cf. A000079, A000984, A001405, A069954, A000721.

a:= n-> (t-> binomial(t, iquo(t, 2)))(2^n):
seq(a(n), n=0..8);  # Alois P. Heinz, Jan 14 2017
#
a:=n->sum(binomial(2^(n-1),i)^2,i=0..2^(n-1)); seq(a(n),n=0..8);
a:=n->piecewise(n=0,1,binomial(2^n,2^(n-1))); seq(a(n),n=0..8); # Martin Renner, Jan 16 2017

Flatten[{1,Table[Binomial[2^n, 2^(n-1)], {n, 1, 8}]}] (* Vaclav Kotesovec, Nov 13 2014 *)

a(n) = sum(i=0, 2^(n-1), binomial(2^(n-1), i)^2)  \\ Michel Marcus, Jun 09 2013

from math import comb
def A037293(n): return comb(1<Chai Wah Wu, Dec 11 2024

a(n) = A001405(2^n). - Labos Elemer, Apr 11 2001
a(n) ~ 2^(2^n - n/2 + 1/2)/ sqrt(Pi). - Vaclav Kotesovec, Nov 13 2014
a(n) = A000984(2^(n-1)) = binomial(2^n,2^(n-1)) = (2^n)!/((2^(n-1))!)^2 for n > 0. - Martin Renner, Jan 16 2017
a(n) = (2^(2^n)*(2^n + 2)*(1/2*(2^n + 1))!)/(sqrt(Pi)*(2^n + 1)*(1/2*(2^n + 2))!) = (2^(2^n)*(2^n + 2)*Gamma((2^n+3)/2))/(sqrt(Pi)*(2^n + 1)*Gamma(2^(n-1)+2)) for n > 0. - Alexander R. Povolotsky, Nov 19 2022
a(n) = 2 * A069954(n-1) for n > 0. - Tilman Piesk, Oct 11 2024

More terms from Erich Friedman

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

Original entry on oeis.org

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

Offset: 0

Benoit Cloitre, Feb 04 2002

nonn

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)

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.

Cf. A049606, A069954.

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

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

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

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

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

A158817 a(n) = (binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2)))/2^n.

Original entry on oeis.org

1, 8, 800, 18783360, 28634752192836096, 187118328452563147377366903401859072, 22533823529098462258163079522899558155141642796614195116180863201125539840

Offset: 2

Tanya Khovanova and Leonid Makar-Limanov, Mar 27 2009

nonn

G. C. Greubel, Table of n, a(n) for n = 2..11
Gyula O. H. Katona and Leonid Makar-Limanov, A problem for abelian groups, Asian-Eur. J. Math. 1 (2008), no. 2, 237--241. (Reviewer: Thomas Britz) 20K01 (05B40 94B65).

Cf. A069954.

Table[(Binomial[2^n, 2^(n-1)] -Binomial[2^(n-1), 2^(n-2)])/2^n, {n, 2, 12}]

[( binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2)) )/2^n for n in (2..12)] # G. C. Greubel, Dec 22 2021

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

cp's OEIS Frontend

A037293 a(n) = Sum_{i=0..2^(n-1)} binomial(2^(n-1), i)^2.

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

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A158817 a(n) = (binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2)))/2^n.

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