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%I #31 Oct 24 2020 12:56:34
%S 1,16,540,32256,3024000,410572800,76281004800,18598035456000,
%T 5762136335155200,2211729098342400000,1030334000462807040000,
%U 572721601599913328640000,374484928188990947328000000,284562454970932936468070400000
%N a(n) = (n+1)*(2*n+2)!/(n+2).
%C Sequence is inspired by A273889, A274119 and A273983.
%C Since Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b) ≡ 0 mod (n*k+a+b), then define B(n,k,a,b) = (Product_{i=0..n}(i*k+a) - Product_{i=0..n}(-i*k-b))/(n*k+a+b), with n*k+a+b <> 0, n >= 0 and k,a,b are integers, such that B(1,n,2,1) = (Product_{i=0..1}(i*n+2) - Product_{i=0..1}(-i*n-1))/(n+3) = A000012(n) with n >= 0;
%C B(3,n,2,1) = (Product_{i=0..3}(i*n+2) - Product_{i=0..3}(-i*n-1))/(3*n+3) = A100040(n+2) with n >= 0;
%C B(2*n+1,0,2,1) = (Product_{i=0..2*n+1}(2) - Product_{i=0..2*n+1}(-1))/3 = A002450(n+1) with n >= 0;
%C B(2*n+1,2,2,1) = (Product_{i=0..2*n+1}(2*i+2) - Product_{i=0..2*n+1}(-2*i-1))/(4*n+5) = A273983(n+1) with n >= 0;
%C and a(n) is B(2*n+1,1,2,1). - _Hong-Chang Wang_, Jun 17 2016
%H Hong-Chang Wang, <a href="/A274136/b274136.txt">Table of n, a(n) for n = 0..30</a>
%H Hong-Chang Wang, <a href="/A274119/a274119_5.png">Definition of the formula B(n,k,a,b)</a>
%F a(n) = B(2*n+1,1,2,1) = (Product_{i=0..2*n+1}(i+2) - Product_{i=0..2*n+1}(-i-1))/(2*n+4), n >= 0.
%F a(n) = A062779(n)/(2*(n+1)). - _Michel Marcus_, Jun 11 2016
%F a(n) ~ sqrt(Pi)*exp(-2*n)*(48*n*(24*n + 13) + 1177)*4^(n-2)*n^(2*n+1/2)/9. - _Ilya Gutkovskiy_, Jul 07 2016
%e a(0) = (2*3 - 1*2)/4 = 1.
%e a(1) = (2*3*4*5 - 1*2*3*4)/6 = 16.
%e a(2) = (2*3*4*5*6*7 - 1*2*3*4*5*6)/8 = 540.
%t a[n_] := (Product[i + 2, {i, 0, 2*n + 1}] - Product[-i - 1, {i, 0, 2*n + 1}])/(2*n + 4); Table[a[n], {n, 0, 10}] (* _G. C. Greubel_, Jun 19 2016 *)
%t Table[((n+1)(2n+2)!)/(n+2),{n,0,30}] (* _Harvey P. Dale_, Oct 24 2020 *)
%o (Python)
%o # subroutine
%o def B (n, k, a, b):
%o pa = pb = 1
%o for i in range(n+1):
%o pa *= (i*k+a)
%o pb *= (-i*k-b)
%o m = n*k+a+b
%o p = pa-pb
%o if m == 0:
%o return "NaN"
%o else:
%o return p/m
%o # main program
%o for j in range(101):
%o print(str(j)+" "+str(B(2*j+1, 1, 2, 1))) # _Hong-Chang Wang_, Jun 14 2016
%o (PARI) a(n) = ((2*n+1)!-(2*n)!)/(2*(n+1)) \\ _Felix Fröhlich_, Jun 11 2016
%o (PARI) a(n) = (prod(i=0, 2*n+1, i+2)-prod(i=0, 2*n+1, -i-1))/(2*n+4) \\ _Felix Fröhlich_, Jul 05 2016
%Y Cf. A000012, A002450, A062779, A100040, A273889, A273983, A274119.
%K nonn
%O 0,2
%A _Hong-Chang Wang_, Jun 10 2016
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