OFFSET
1,1
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..50
Andrew Howroyd, Formula for the number of paths
Eric Weisstein's World of Mathematics, Complete Tripartite Graph
Eric Weisstein's World of Mathematics, Graph Path
MATHEMATICA
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!*(i + p)!*(j + p)!*2^(k - i - j)*Binomial[p + i + j - 1, k - 1](1 + n - k); a[n_]:=3*Sum[Sum[Sum[Sum[c[n, k, i, j, p], {p, k - i - j, n}], {j, 0, k - i}], {i, 0, k}], {k, n}]/2; Table[a[n], {n, 12}] (* Indranil Ghosh, Aug 14 2017, after PARI *)
PROG
(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!*(i+p)!*(j+p)!*2^(k-i-j)*binomial(p+i+j-1, k-1)*(1+n-k)}
a(n)={3*(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) )))))/2} \\ Andrew Howroyd, Aug 13 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)*factorial(i + p)*factorial(j + p)*2**(k - i - j)*binomial(p + i + j - 1, k - 1)*(1 + n - k)
def a(n): return 3*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)])/2
print([a(n) for n in range(1, 13)]) # Indranil Ghosh, Aug 14 2017, after PARI
CROSSREFS
KEYWORD
nonn
AUTHOR
Eric W. Weisstein, Jul 27 2017
EXTENSIONS
Terms a(5) and beyond from Andrew Howroyd, Aug 13 2017
STATUS
approved