login
Determinant of n X n matrix M_(i,j) = binomial(i^2, j).
7

%I #35 Feb 21 2023 11:31:49

%S 1,1,6,360,302400,4572288000,1520925880320000,13153940405570764800000,

%T 3412910854477066178396160000000,

%U 30107378079113824305786648526848000000000

%N Determinant of n X n matrix M_(i,j) = binomial(i^2, j).

%C Closed form can be deduced from the conjecture that the count of triangular semi-standard Young tableaux with shape (n,...,1) and max part n equals 2^((n-1)n/2); see Mathematica line. - _Wouter Meeussen_, Nov 26 2017

%F a(n) = Product_{k = 1..n} (2*k-1)!/(k-1)!.

%F a(n) ~ A^(1/2) * 2^(n^2 + n/2 + 5/24) * n^(n^2/2 + n/2 + 1/24) / exp(3*n^2/4 + n/2 + 1/24), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - _Vaclav Kotesovec_, Jul 11 2015

%F a(n) = s_lambda(1,2,3,...,n) where s is the Schur polynomial in n variables and lambda is the partition (n,n-1,n-2,...,1). - _Leonid Bedratyuk_, Feb 06 2022

%F From _Peter Bala_, Feb 19 2023: (Start)

%F a(n) = Product_{k = 0..n} (n + k)!/(2*k)!. Cf. A266091.

%F a(n) = 2^(n*(n-1)/2) * A057863(n).

%F a(n+1)*a(n-1) = (4*n + 2)*a(n)^2.

%F a(n)^3*a(n+2) + a(n-1)*a(n+1)^3 = (8*n + 8)*(a(n)*a(n + 1))^2 for n >= 1.

%F Conjecture: a(n) = the determinant of the n X n matrix ( binomial(i^2 + z, j) ) for arbitrary complex z. (End)

%t Table[Product[(2*k-1)!/(k-1)!,{k,1,n}],{n,0,10}] (* _Vaclav Kotesovec_, Jul 11 2015 *)

%t Table[Product[ -i+j+n, {i,n}, {j, 1-i+n}], {n,0,10}];

%t Round[Table[Sqrt[Glaisher]/(2^(1/24 - n^2)* E^(1/24 + Derivative[1, 0][Zeta][-1, 1/2 + n])* Pi^(1/4 + n/2)*Gamma[1/2 + n]^(-(1/2) - n)), {n, 16}]] (* see comments *) (* _Wouter Meeussen_, Nov 26 2017 *)

%t Table[Glaisher^(3/2) * 2^(n^2 - 1/24) * BarnesG[n + 3/2] / (E^(1/8) * Pi^(n/2 + 1/4)), {n, 0, 12}] (* _Vaclav Kotesovec_, Mar 24 2019 *)

%o (PARI) a(n)=prod(k=1,n,(2*k-1)!/(k-1)!)

%o (PARI) a(n)=matdet(matrix(n,n,i,j,binomial(i^2,j)))

%Y Cf. A057863, A266091.

%K nonn,easy

%O 0,3

%A _Benoit Cloitre_, Aug 27 2003