A036914 a(n) = binomial(2*n,n)*binomial(3*n,2*n)^4.

1, 162, 303750, 995742720, 4202607543750, 20493770553668412, 109738295483524291584, 627433021349790289920000, 3765656995768668039930646470, 23460102529588600192836492187500, 150552597141762184641565143623272500, 989711604190467147276644388444241920000

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

From Peter Bala, Aug 07 2016: (Start)

Compare with the identities:

Sum_{k = 0..2*n} (-1)^(n+k)*binomial(3*n,k)^2*binomial(3*n - k,n)^2 = binomial(3n,n)^2*binomial(2*n,n) = A275047(n), and

Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = binomial(3*n,n)*binomial(2*n,n) = (3*n)!/n!^3 = A006480(n). (Sprugnoli, Section 2.9, Table 10, p. 123).

Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = A000984(n). (End)

References

• The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972; Eq 21.1, page 72 (see the Formula section).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..250
• R. Sprugnoli, Riordan array proofs of identities in Gould's book

Crossrefs

Cf. A000984, A005809, A006480, A275047.

Programs

Magma

[(n+1)*Binomial(3*n,2*n)^4*Catalan(n): n in [0..30]]; // G. C. Greubel, Jun 22 2022

Maple

seq((3*n)!^4/(n!^6*(2*n)!^3), n = 0..20); # Peter Bala, Aug 07 2016

Mathematica

Table[Binomial[2n, n]Binomial[3n, 2n]^4, {n,0,11}] (* Michael De Vlieger, Aug 07 2016 *)

SageMath

b=binomial; [b(2*n,n)*b(3*n,2*n)^4 for n in (0..30)] # G. C. Greubel, Jun 22 2022

Formula

Sum_{k=0..2*n} (-1)^k*C(3*n, k)^3*C(3*n-k, n)^3 = (-1)^n*C(2*n, n)*C(3*n, 2*n)^4.

From Peter Bala, Aug 07 2016: (Start)

a(n) = (3*n)!^4/(n!^6*(2*n)!^3).

a(n) = A005809(n)^4 * A000984(n) = A005809(n)^3 * A006480(n) = A005809(n)^2 * A275047(n).

a(n) = {[x^n] (1 + x)^(3*n)}^4 * [x^n] (1 + x)^(2*n) = [x^n] G(x)^(162*n), where G(x) = 1 + x + 776*x^2 + 1633370*x^3 + 5060509158*x^4 + 19379170742458*x^5 + 84908023350007787*x^6 + ... appears to have integer coefficients.

exp( Sum_{n >= 1} a(n)*x^n/n ) = F(x)^162, where F(x) = 1 + x + 938*x^2 + 2049791*x^3 + 6487994244*x^4 + 25309359070330*x^5 + 112932966264239483*x^6 + ... appears to have integer coefficients. (End)

a(n) ~ (9/16)*9^(6*n)/((Pi*n)^(5/2)*64^n). - Ilya Gutkovskiy, Aug 07 2016