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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.
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%I #59 Dec 11 2023 07:17:25

%S 1,132,4719,81796,884884,6852768,41314284,204951252,869562265,

%T 3245256300,10880587575,33309352440,94307358288,249485071616,

%U 621856804272,1470540624696,3318218562009,7179339254516,14955909351383,30104651175324,58733021049780,111358254207200

%N 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.

%C Starting with 132 = A000108(6), the sequence is the Hankel transform of the Catalan numbers shifted to remove the first 6 terms.

%H G. C. Greubel, <a href="/A335857/b335857.txt">Table of n, a(n) for n = 0..1000</a>

%H Jishe Feng, <a href="https://arxiv.org/abs/2010.06586">The explicit formula of Hankel determinant with Catalan elements</a>, arXiv:2010.06586 [math.GM], 2020.

%H <a href="/index/Rec#order_16">Index entries for linear recurrences with constant coefficients</a>, signature (16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1).

%F 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).

%F From _G. C. Greubel_, Dec 17 2021: (Start)

%F a(n) = binomial(n+5, 5)*binomial(2*n+7, 3)*binomial(2*n+9, 7)/1260.

%F 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)

%F a(n) = -a(-6-n) for all n in Z. - _Michael Somos_, Jun 27 2023

%F a(n) ~ n^15/4465125. - _Stefano Spezia_, Dec 09 2023

%e a(1) = 132 because 132 is the determinant of the 1 X 1 matrix [132].

%e a(2) = 4719 because 4719 is the determinant of the matrix

%e [ 132 429 ]

%e [ 429 1430 ].

%e a(3) = 81796 because 81796 is the determinant of the matrix

%e [ 132, 429, 1430 ]

%e [ 429, 1430, 4862 ]

%e [ 1430, 4862, 16796 ].

%e G.f. = 1 + 132*x + 4719*x^2 + 81796*x^3 + 884884*x^4 + ... - _Michael Somos_, Jun 27 2023

%t 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 *)

%t a[ n_] := Binomial[n+5, 4]*Binomial[2*n+7, 3]*Binomial[2*n+9, 8]/1575; (* _Michael Somos_, Jun 27 2023 *)

%o (PARI)

%o H(seq)={my(n=(#seq+1)\2); matdet(matrix(n,n,i,j,seq[i+j-1]))}

%o 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

%o (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 */

%o (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

%Y Hankel transforms of the Catalan numbers shifted by k for k=0..5 are A000012, A000012, A000027, A000330, A006858, A091962.

%Y Cf. A000108 (Catalan numbers).

%K nonn,easy

%O 0,2

%A _Feng Jishe_, Oct 12 2020