A234616 Numbers of undirected cycles in the complete tripartite graph K_{n,n,n}.

1, 63, 6705, 1960804, 1271288295, 1541975757831, 3135880743480163, 9904953891455450640, 45915662047529291081589, 299038026557168514632822455, 2642895689915240835222121682301, 30814273315381549790551229559722628

Offset: 1

Author

Eric W. Weisstein, Dec 28 2013

Keywords

• nonn

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..100 (terms 1..50 from Andrew Howroyd)
• Andrew Howroyd, Formula for the number of cycles
• Eric Weisstein's World of Mathematics, Complete Tripartite Graph
• Eric Weisstein's World of Mathematics, Graph Cycle

Crossrefs

Cf. A234365, A234633, A070968.

Cf. A296546 (cycle polynomial coefficients of K_n,n,n).

Programs

Mathematica

Table[(Sum[Binomial[n, k] Binomial[n, i + p] Binomial[n, j + p] Binomial[k, i] Binomial[k - i, j] (k - 1)! (i + p)! (j + p)! 2^(k - i - j) Binomial[p + i + j - 1, k - 1], {k, n}, {i, 0, k}, {j, 0, k - i}, {p, k - i - j, n}] + Sum[Binomial[n, k]^2 k! (k - 1)!, {k, 2, n}])/2 - n^2, {n, 10}] (* Eric W. Weisstein, May 26 2017 *)
Table[(n^2 (HypergeometricPFQ[{1, 1, 1 - n, 1 - n}, {2}, 1] - 3) + Sum[2^(k - i - j) Binomial[k, i] Binomial[k - i, j] Binomial[n, k] Binomial[n, i + p] Binomial[n, j + p] Binomial[i + j + p - 1, k - 1] (k - 1)! (i + p)! (j + p)!, {k, n}, {i, 0, k}, {j, 0, k - i}, {p, k - i - j, n}])/2, {n, 10}] (* Eric W. Weisstein, May 25 2023 *)

PARI

c(n,k,i,j,p) = {binomial(n,k)*binomial(n,i+p)*binomial(n,j+p)*binomial(k,i)*binomial(k-i,j)*(k-1)!*(i+p)!*(j+p)!*2^(k-i-j)*binomial(p+i+j-1,k-1)}
a(n)={(sum(k=1,n,sum(i=0,k,sum(j=0,k-i,sum(p=k-i-j,n, c(n,k,i,j,p) )))) + sum(k=2,n,binomial(n,k)^2*k!*(k-1)!))/2 - n^2} \\ Andrew Howroyd, May 25 2017

Python

from sympy import binomial, factorial
def c(n, k, i, j, p): return binomial(n, k)*binomial(n, i + p)*binomial(n, j + p)*binomial(k, i)*binomial(k - i, j)*factorial(k - 1)*factorial(i + p)*factorial(j + p)*2**(k - i - j)*binomial(p + i + j - 1, k - 1)
def a(n): return (sum([sum([sum([sum([c(n, k, i, j, p) for p in range(k - i - j, n + 1)]) for j in range(k - i + 1)]) for i in range(k + 1)]) for k in range(1, n + 1)]) + sum(binomial(n, k)**2*factorial(k)*factorial(k - 1) for k in range(2, n + 1)))/2 - n**2
print([a(k) for k in range(1, 13)]) # Indranil Ghosh, Aug 14 2017, after PARI

Formula

Row sums of A296546.

a(n) ~ sqrt(3*Pi) * 2^(3*n - 1/2) * n^(3*n - 1/2) / exp(3*n - 3/2). - Vaclav Kotesovec, Feb 17 2024

Extensions

a(7)-a(12) from Andrew Howroyd, May 25 2017