A335857 a(n) is the determinant of the n X n Hankel matrix A with A(i,j) = A000108(i+j+6) for 0<=i,j<=n-1.
1, 132, 4719, 81796, 884884, 6852768, 41314284, 204951252, 869562265, 3245256300, 10880587575, 33309352440, 94307358288, 249485071616, 621856804272, 1470540624696, 3318218562009, 7179339254516, 14955909351383, 30104651175324, 58733021049780, 111358254207200
Offset: 0
Examples
a(1) = 132 because 132 is the determinant of the 1 X 1 matrix [132].
a(2) = 4719 because 4719 is the determinant of the matrix
[ 132 429 ]
[ 429 1430 ].
a(3) = 81796 because 81796 is the determinant of the matrix
[ 132, 429, 1430 ]
[ 429, 1430, 4862 ]
[ 1430, 4862, 16796 ].
G.f. = 1 + 132*x + 4719*x^2 + 81796*x^3 + 884884*x^4 + ... - _Michael Somos_, Jun 27 2023
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Jishe Feng, The explicit formula of Hankel determinant with Catalan elements, arXiv:2010.06586 [math.GM], 2020.
- Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 24.
- Index entries for linear recurrences with constant coefficients, signature (16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1).
Crossrefs
Programs
-
Mathematica
Table[Binomial[n+5, 5]*Binomial[2*n+7, 3]*Binomial[2*n+9, 7]/1260, {n,0,30}] (* G. C. Greubel, Dec 17 2021 *) a[ n_] := Binomial[n+5, 4]*Binomial[2*n+7, 3]*Binomial[2*n+9, 8]/1575; (* Michael Somos, Jun 27 2023 *) -
PARI
H(seq)={my(n=(#seq+1)\2); matdet(matrix(n,n,i,j,seq[i+j-1]))} a(n, k=6)={H(vector(2*n, i, my(t=i+k-1); binomial(2*t,t)/(t+1)))} \\ Andrew Howroyd, Nov 26 2020 -
PARI
{a(n) = prod(k=1, 5, (n+k)^min(6-k, k)) * prod(k=1, 4, (2*n+2*k+1)^min(5-k, k))/285768000}; /* Michael Somos, Jun 27 2023 */ -
Sage
[binomial(n+5, 5)*binomial(2*n+7, 3)*binomial(2*n+9, 7)/1260 for n in (0..30)] # G. C. Greubel, Dec 17 2021
Formula
a(n) = 2^5/(5 * 7!* 9!)*(n + 1)*(n + 2)^2*(n + 3)^3 *(n + 4)^2*(n + 5)*(2*n + 3)*(2*n + 5)^2*(2*n + 7)^2*(2*n + 9).
From G. C. Greubel, Dec 17 2021: (Start)
a(n) = binomial(n+5, 5)*binomial(2*n+7, 3)*binomial(2*n+9, 7)/1260.
G.f.: (1 + 116*x + 2727*x^2 + 21572*x^3 + 70328*x^4 + 103376*x^5 + 70328*x^6 + 21572*x^7 + 2727*x^8 + 116*x^9 + x^10)/(1 - x)^16. (End)
a(n) = -a(-6-n) for all n in Z. - Michael Somos, Jun 27 2023
a(n) ~ n^15/4465125. - Stefano Spezia, Dec 09 2023
Comments