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

0,3

Comments

Let A_n be the matrix of size n X n defined by: A_n[i,j] = 1/(binomial coefficient i+j-2 over i-1) = 1/C(i+j-2,i-1) where 1 <= i,j <= n. The diagonals of this matrix are the reciprocals of the entries in the Pascal triangle. Then a(n) = 1/det(A_n) = det((A_n)^(-1)).

From the formula for a(n) it follows that the determinant of (A_n)^(-1) is an integer. By inspecting the values of (A_n)^(-1) for small values of n it looks like (A_n)^(-1) is actually a matrix of integers but I do not have a proof of this fact.

Let M_n be the n X n matrix with M_n(i,j)=i/(i+j); then |a(n-1)|=1/det(M_n). - Benoit Cloitre, Apr 21 2002

Also related to the multinomial coefficients (i+j)!/i!/j! : abs(a(n))=(1/detQ_n-1) where Q_n is the n X n matrix q(i,j)=i!j!/(i+j)! - Benoit Cloitre, May 30 2002

From Alexander Adamchuk, Nov 14 2009: (Start)

Also a(n) = (-1)^(n(n-1)/2) * Product[ Binomial[2k,k]^2/2, {k,1,n-1} ].

It is simpler definition of a(n).

It follows from the observation that Sqrt[ Abs[ a(n+1)/a(n)/2 ] ] = {1, 3, 10, 35, 126, 462, ...} = C(2n+1, n+1) = A001700. (End)

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.

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

Example

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

Maple

A060739 := n->(-1)^(n*(n-1)/2) * mul( (n+k-1)!/((k!)^2 * (n-1-k)!), k=0..n-1);

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 *)

Prog

(PARI) for(n=1, 15, print1(1/matdet(matrix(n, n, i, j, i/(j+i))), ", ")) \\ See Cloitre's comment
(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
(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

Crossrefs

Cf. A005249, A067689.

Cf. A001700. [Alexander Adamchuk, Nov 14 2009]

Sequence in context: A047832 A004003 A369676 * A333209 A224733 A264953

Adjacent sequences: A060736 A060737 A060738 * A060740 A060741 A060742

Keyword

easy,sign,nice

Author

Noam Katz (noamkj(AT)hotmail.com), Apr 25 2001

Status

approved