A261125 a(n) = (2^(n-1))!*a(n-1) for n>=1, a(0) = 1.
1, 1, 2, 48, 1935360, 40493130637639680000, 10654991354747516157752604498631700563938508800000000000
Offset: 0
Keywords
Links
- Michel Marcus, Table of n, a(n) for n = 0..9
- Alexander Karpov, A theory of knockout tournament seedings, Heidelberg University, AWI Discussion Paper Series, No. 600.
Programs
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Magma
[n le 1 select n else Self(n-1)*Factorial(2^(n - 1)): n in [1..7]]; // Vincenzo Librandi, Aug 10 2015
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Maple
a:= proc(n) option remember: `if`(n=0, 1, a(n-1)*(2^(n-1))!) end: seq(a(n), n=0..6); # Alois P. Heinz, Feb 04 2023
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Mathematica
RecurrenceTable[{a[1] == 1, a[n] == a[n-1] (2^(n - 1))!}, a, {n, 10}] (* Vincenzo Librandi, Aug 10 2015 *) FoldList[(2^#2)!*#1&,1,Range@6] (* Ivan N. Ianakiev, Aug 10 2015 *)
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PARI
first(m)=my(v=vector(m));v[1]=1;for(i=2,m,v[i]=(2^(i-1))!*v[i-1];);v; \\ Anders Hellström, Aug 10 2015
Formula
a(n) = Product_{j=0..n-1} (2^j)!. - Alois P. Heinz, Feb 04 2023
Extensions
a(0)=1 prepended by Alois P. Heinz, Feb 04 2023
Comments