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a(n) = (2^n)!/2^(2^n-1).
7

%I #55 Jul 06 2023 06:01:16

%S 1,1,3,315,638512875,122529844256906551386796875,

%T 13757108753595648665519665029568345104465749222289382342659100341796875

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

%C a(n) is also the number of knockout tournament seedings with 2^n teams. - _Alexander Karpov_, Aug 09 2015

%C From _Zhujun Zhang_, Jun 17 2019: (Start)

%C 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.

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

%H Jianing Song, <a href="/A067667/b067667.txt">Table of n, a(n) for n = 0..8</a>

%H Mark R. Brown, <a href="https://doi.org/10.1137/0207026">Implementation and analysis of binomial queue algorithms</a>, SIAM Journal on Computing, 1978, 7(3):298-319.

%H Alexander Karpov, <a href="http://www.uni-heidelberg.de/md/awi/forschung/dp600.pdf">A theory of knockout tournament seedings</a>, Heidelberg University, AWI Discussion Paper Series, No. 600.

%H Zhujun Zhang, <a href="https://www.researchgate.net/publication/333261503_A_Note_on_Counting_Binomial_Heaps">A Note on Counting Binomial Heaps</a>, ResearchGate, June 2019.

%F From _Alexander Karpov_, Aug 09 2015: (Start)

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

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

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

%F (End)

%F a(n) = A049606(2^n). - _Zhujun Zhang_, Jun 16 2019

%F a(n) = Product_{odd k < 2^n} k^(n - floor(log_2(k))). - _Harry Richman_, May 18 2023

%t Table[(2^n)! / 2^(2^n - 1), {n, 6}] (* _Vincenzo Librandi_, Aug 10 2015 *)

%o (Magma) [Factorial(2^n)/2^(2^n-1): n in [1..6]]; // _Vincenzo Librandi_, Aug 10 2015

%o (PARI) a(n) = (2^n)!/2^(2^n-1) \\ _Jianing Song_, Jul 15 2021

%Y Cf. A049606, A069954.

%K nonn

%O 0,3

%A _Benoit Cloitre_, Feb 04 2002

%E a(0) prepended by _Jianing Song_, Jul 15 2021