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A168467
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a(n) = Product_{k=0..n} ((2k+2)(2k+3))^(n-k).
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13
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1, 6, 720, 3628800, 1316818944000, 52563198423859200000, 327312129899898454671360000000, 428017682605583614976547335700480000000000, 152240508705590071980086429193304853792686080000000000000
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OFFSET
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0,2
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COMMENTS
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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
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
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LINKS
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FORMULA
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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
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
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EXAMPLE
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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).
For a generic z, the determinant doesn't change as shown in this example, where the determinant of the 3 X 3 square matrix:
( ( 1, 2*(z+1), (2*z + 1)*(2*z+3) ),
( 1, 2*(z+4), 4*(z+6)*(z+2) ),
( 1, 2*(z+9), (2*z + 9)(2*z + 27)) ) = 720 = a(3). (End)
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MATHEMATICA
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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 *)
Table[Product[((2k+2)(2k+3))^(n-k), {k, 0, n}], {n, 0, 10}] (* Harvey P. Dale, Dec 26 2019 *)
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 *)
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 *)
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PROG
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(Python)
from math import prod
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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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