A060739 a(n) = (-1)^(n(n-1)/2) * Product_{k=0,...,n-1} (n+k-1)!/((k!)^2 * (n-1-k)!).
1, 1, -2, -36, 7200, 17640000, -560105280000, -239102222768640000, 1408147589778024775680000, 116620600756651855983415296000000, -137839975629646325813680872620851200000000, -2352568589682795058651211199786427114330521600000000
Offset: 0
Examples
Here is the matrix A_4 for n=4: [1, 1, 1, 1; 1, 1/2, 1/3, 1/4; 1, 1/3, 1/6, 1/10; 1, 1/4, 1/10, 1/20]; a(4) = 7200 because det(A_4) = 1/7200
Links
- Harry J. Smith, Table of n, a(n) for n=0,...,43
- T. M. Richardson, The Reciprocal Pascal Matrix, arXiv preprint arXiv:1405.6315 [math.CO], 2014.
- Doron Zeilberger, Reverend Charles to the aid of Major Percy and Fields-Medalist Enrico, arXiv:1405.6315 [math.CO], 2014.
- Doron Zeilberger, Reverend Charles to the aid of Major Percy and Fields-Medalist Enrico, Amer. Math. Monthly 103 (1996), 501-502.
Crossrefs
Cf. A001700. [Alexander Adamchuk, Nov 14 2009]
Programs
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Maple
A060739 := n->(-1)^(n*(n-1)/2) * mul( (n+k-1)!/((k!)^2 * (n-1-k)!), k=0..n-1);
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Mathematica
a[n_] := (-1)^(n (n - 1)/2)*Product[ Multinomial[k, k, n - 1 - k], {k, 0, n - 1}]; Table[a[n], {n, 0, 11}] (* Jean-François Alcover, Dec 08 2011, after first formula *)
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PARI
for(n=1,15,print1(1/matdet(matrix(n,n,i,j,i/(j+i))),",")) \\ See Cloitre's comment
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PARI
{ for (n=0, 43, if (n<2, a=1, a=(-1)^(n\2)/matdet(matrix(n-1, n-1, i, j, i/(j+i)))); write("b060739.txt", n, " ", a); ) } \\ Harry J. Smith, Jul 10 2009
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Sage
def A060739(n): return (-1)^(n//2)*A163085(2*(n-1))/factorial(n-1) if n > 0 else 1 [A060739(i) for i in (0..11)] # Peter Luschny, Sep 18 2012
Formula
If Multinomial[a, b, c] denotes the multinomial coefficient (a+b+c)! / (a! * b! * c!) (which is an integer) then : a(n) = (-1)^(n(n-1)/2) * Product k=0, ..., n-1 Multinomial[k, k, n-1-k] = (-1)^(n(n-1)/2) * product k=0, ..., n-1 (n+k-1)!/((k!)^2 * (n-1-k)!)
a(n) = (-1)^(n(n-1)/2) * Product[ Binomial[2k,k]^2/2, {k,1,n-1} ]. [Alexander Adamchuk, Nov 14 2009]
|a(n)| = A163085(2*(n-1))/(n-1)! for n > 0. - Peter Luschny, Sep 18 2012
|a(n)| ~ A^3 * 2^(2*n^2 - 3*n + 5/12) * exp(n - 1/4) / (Pi^n * n^(n - 1/4)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, May 19 2020
Comments