A093883 Product of all possible sums of two distinct numbers taken from among first n natural numbers.
1, 3, 60, 12600, 38102400, 2112397056000, 2609908810629120000, 84645606509847871488000000, 82967862872337478796810649600000000, 2781259372192376861719959017613164544000000000
Offset: 1
Keywords
A168467 a(n) = Product_{k=0..n} ((2*k+2)*(2*k+3))^(n-k).
1, 6, 720, 3628800, 1316818944000, 52563198423859200000, 327312129899898454671360000000, 428017682605583614976547335700480000000000, 152240508705590071980086429193304853792686080000000000000
Offset: 0
Comments
Hankel transform of A000698(n+1).
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
Examples
From _Federico Provvedi_, Apr 01 2021: (Start) 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)
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..28
Crossrefs
Programs
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Mathematica
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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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<
Chai Wah Wu, Nov 26 2023
Formula
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) = Product_{k=1..n} (2*k+1)!. - Vladimir Reshetnikov, Sep 06 2016
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
For n > 0, a(n) = n * (2*n+1) * sqrt(BarnesG(2*n)) * Gamma(2*n)^2 / (sqrt(Gamma(n)) * 2^((n-3)/2)). - Vaclav Kotesovec, Nov 27 2024
A110468 a(n) = (2*n + 1)!/(n + 1).
1, 3, 40, 1260, 72576, 6652800, 889574400, 163459296000, 39520825344000, 12164510040883200, 4644631106519040000, 2154334728240414720000, 1193170003333152768000000, 777776389315596582912000000, 589450799582646796969574400000, 513927415886120176107847680000000
Offset: 0
Comments
Convolution of (-1)^n*n! and n! with interpolated zeros suppressed.
Denominator of absolute value of coefficient of 1/(x+n^2) in the partial fraction decomposition of 1/(x+1)*1/(x+4)*..*1/(x+n^2). - Joris Roos (jorisr(AT)gmx.de), Aug 07 2009
With offset = 1: a(n) is the number of permutations of {1,2,...,2n} composed of two cycles of length n. - Geoffrey Critzer, Nov 11 2012
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..200
Programs
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Mathematica
Table[(2n)!/(2n^2),{n,1,20}] (* Geoffrey Critzer, Nov 11 2012 *)
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PARI
for(n=0,50, print1((2*n+1)!/(n+1), ", ")) \\ G. C. Greubel, Aug 28 2017
Formula
E.g.f.: log((1-x)*(1+x))/(-x).
a(n) = (2*n)!*Sum_{k = 0..2*n} (-1)^k/binomial(2*n, k).
a(n) = Sum_{k = 0..2*n} k!*(-1)^k*(2*n-k)!.
Sum_{n>=0} 1/a(n) = e/2. - Franz Vrabec, Jan 17 2008
(n+1)*a(n) + 2*(-n^2)*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Nov 15 2012
a(n) = Product_{i=1..n} (n+1-i)*(n+1+i). - Vaclav Kotesovec, Oct 21 2014
a(n) = A145877(2*n+2, n+1). - Alois P. Heinz, Apr 21 2017
a(n) = A346085(2*n+2, n+1). - Alois P. Heinz, Jul 04 2021
Sum_{n>=0} (-1)^n/a(n) = (cos(1) + sin(1))/2 = (1/2) * A143623. - Amiram Eldar, Feb 08 2022
a(p-1) == 1 (mod p), p a prime. - Peter Bala, Jul 29 2024
Sum_{n>=0} x^(2*n+1)/a(n) = (sinh(x) + x*cosh(x))/2. - Michael Somos, Jul 23 2025
Extensions
Simpler definition from Robert Israel, Jul 20 2006
A203475 a(n) = Product_{1 <= i < j <= n} (i^2 + j^2).
1, 5, 650, 5525000, 5807194900000, 1226800120038480000000, 77092420109247492627600000000000, 2001314057760220784660590245696000000000000000, 28468550112906756205383102673584071297339520000000000000000000
Offset: 1
Keywords
Comments
Each term divides its successor, as in A203476.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..28
Programs
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Magma
[(&*[(&*[j^2 + k^2: k in [1..j]])/(2*j^2): j in [1..n]]): n in [1..20]]; // G. C. Greubel, Aug 28 2023
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Maple
a:= n-> mul(mul(i^2+j^2, i=1..j-1), j=2..n): seq(a(n), n=1..10); # Alois P. Heinz, Jul 23 2017
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Mathematica
f[j_]:= j^2; z = 15; v[n_]:= Product[Product[f[k] + f[j], {j,k-1}], {k,2,n}] Table[v[n], {n,z}] (* A203475 *) Table[v[n+1]/v[n], {n,z-1}] (* A203476 *)
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SageMath
[product(product(j^2+k^2 for k in range(1,j)) for j in range(1,n+1)) for n in range(1,21)] # G. C. Greubel, Aug 28 2023
Formula
a(n) ~ c * 2^(n^2/2) * exp(Pi*n*(n+1)/4 - 3*n^2/2 + n) * n^(n*(n-1) - 3/4), where c = A323755 = sqrt(Gamma(1/4)) * exp(Pi/24) / (2*Pi)^(9/8) = 0.274528350333552903800408993482507428142383783773190451181... - Vaclav Kotesovec, Jan 26 2019
Extensions
Name edited by Alois P. Heinz, Jul 23 2017
Comments
Examples
References
Links
Crossrefs
Programs
Maple
Mathematica
PARI
Formula
Extensions