%I #78 Nov 27 2023 03:16:37

%S 1,6,720,3628800,1316818944000,52563198423859200000,

%T 327312129899898454671360000000,

%U 428017682605583614976547335700480000000000,152240508705590071980086429193304853792686080000000000000

%N a(n) = Product_{k=0..n} ((2k+2)(2k+3))^(n-k).

%C Hankel transform of A000698(n+1).

%C The sequence 1,1,6,720,... with general term Product_{k=0..n, ((2k+1)(2k+0^k))^(n-k)} is the Hankel transform of A112934. - _Paul Barry_, Dec 04 2009

%C a(n) is also the determinant of the n X n matrix M(i,j) = i^(2*j)*sinh(2*j*arccsch(i))/(2*sqrt(i^2+1)), with i and j from 1 to n, which is the same matrix generated by sequences of length n by the linear recurrences with kernel { 2*(k^2 + z), -k^4 }, and initial conditions { 1, 2*(k^2 + z) }, with k from 1 to n, and z = 2. Regardless of the value of z, for every n, the determinant of the n X n matrix of polynomials generated gives always a(n) as result. - _Federico Provvedi_, Feb 01 2021

%H Seiichi Manyama, <a href="/A168467/b168467.txt">Table of n, a(n) for n = 0..28</a>

%F G.f.: Q(0)/(2*x) -1/x, where Q(k) = 1 + 1/(1 -(2*k+1)!*x/((2*k+1)!*x + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Sep 17 2013

%F a(n) = Product_{k=1..n} (2*k+1)!. - _Vladimir Reshetnikov_, Sep 06 2016

%F a(n) ~ A^(-1/2) * 2^(n^2 + 3*n + 53/24) * exp((-3/2)*n^2 + (-5/2)*n + 1/24) * n^(n^2 + (5/2)*n + 35/24) * Pi^((n+1)/2), where A = A074962 is the Glaisher-Kinkelin constant. - _Vladimir Reshetnikov_, Sep 06 2016

%F a(n) = A000178(2*n + 1) / A098694(n). - _Vaclav Kotesovec_, Oct 28 2017

%F a(n) = A202768(n)*A000142(n). - _Federico Provvedi_, Feb 01 2021

%e From _Federico Provvedi_, Apr 01 2021: (Start)

%e From both formulas in the comment above and in particular with z=2 from the linear recurrences, the determinant of the 5 X 5 matrix: ( (1,6,35,204,1189), (1,12,128,1344,14080),(1,22,403,7084,123205), (1,36,1040,28224,749824), (1,54,2291,89964,3426181) ) = 1316818944000 = a(5).

%e For a generic z, the determinant doesn't change as shown in this example, where the determinant of the 3 X 3 square matrix:

%e ( ( 1, 2*(z+1), (2*z + 1)*(2*z+3) ),

%e ( 1, 2*(z+4), 4*(z+6)*(z+2) ),

%e ( 1, 2*(z+9), (2*z + 9)(2*z + 27)) ) = 720 = a(3). (End)

%t Table[2^(n^2 + 2*n + 23/24) Glaisher^(3/2) Pi^(-n/2 - 3/4) BarnesG[n + 2] BarnesG[n + 5/2]/E^(1/8), {n, 0, 10}] (* _Vladimir Reshetnikov_, Sep 06 2016 *)

%t Table[Product[((2k+2)(2k+3))^(n-k),{k,0,n}],{n,0,10}] (* _Harvey P. Dale_, Dec 26 2019 *)

%t Table[Det@Table[LinearRecurrence[{2*k^2,-k^4},{1, 2*k^2},n], {k, 1, n}], {n,1,20}] (* Federico Provvedi, Feb 01 2021 *)

%t Det@Expand@Array[(#1^(2 #2))/(4 Sqrt[1 + #1^2])((Sqrt[1+1/#1^2]+1/#1)^(2 #2)-(Sqrt[1+1/#1^2]-1/#1)^(2 #2))&,{#,#}]&/@Range[20] (* Federico Provvedi, Apr 01 2021 *)

%o (Python)

%o from math import prod

%o def A168467(n): return prod(((m:=k+1<<1)*(m+1))**(n-k) for k in range(1,n+1))*3**n<<n # _Chai Wah Wu_, Nov 26 2023

%Y Cf. A000178, A000698, A098694, A112934, A306635, A001109, A099157, A246645, A202768, A000142.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Nov 26 2009