A167986 Triangle T(n,k) = Number of k-cycles on the graph of an n-orthoplex. n>=2, k>=3.
0, 1, 8, 15, 24, 16, 32, 102, 288, 640, 960, 744, 80, 370, 1584, 5920, 18240, 43080, 69120, 56256, 160, 975, 5664, 30080, 141120, 564120, 1835520, 4542336, 7580160, 6385920, 280, 2121, 15624, 108080, 684480, 3876600, 19138560, 79805376
Offset: 2
Examples
T(3,3) = 8, because in dimension n=3, the cross-polytope is the octahedron, which has 8 3-cycles in its graph. Triangle starts 0, 1; 8, 15, 24, 16; 32, 102, 288, 640, 960, 744; 80, 370, 1584, 5920, 18240, 43080, 69120, 56256; ... In terms of cycle polynomials: 0*x^3 + 1*x^4; 8*x^3 + 15*x^4 + 24*x^5 + 16*x^6; 32*x^3 + 102*x^4 + 288*x^5 + 640*x^6 + 960*x^7 + 744*x^8; ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 2..871
- Eric Weisstein's World of Mathematics, Cross Polytope
- Eric Weisstein's World of Mathematics, Cocktail Party Graph
- Eric Weisstein's World of Mathematics, Cycle Polynomial
Crossrefs
Programs
-
Magma
b:= func< n,k,j | (-1)^j*Binomial(n,j)*Binomial(2*(n-j),k-2*j)*2^(j-1)*Factorial(k-j-1) >; A167986:= func< n,k | (&+[b(n,k,j): j in [0..Floor(k/2)]]) >; [A167986(n,k): k in [3..2*n], n in [2..10]]; // G. C. Greubel, Jan 17 2023
-
Mathematica
T[n_, k_]:= Sum[(-1)^j*Binomial[n, j]*Binomial[2*(n-j), k-2*j]*2^j*(k - j - 1)!/2, {j, 0, Floor[k/2]}]; Table[T[n, k], {n,2,7}, {k,3,2*n}]//Flatten (* Jean-François Alcover, Oct 08 2017, after Andrew Howroyd *) Table[Binomial[2n, k]*Gamma[k]*HypergeometricPFQ[{(1-k)/2, -k/2}, {1 - k, 1/2 -n}, -2]/2, {n,7}, {k,3,2n}]//Flatten (* Eric W. Weisstein, Mar 25 2020 *)
-
PARI
a(n,k)=sum(j=0,k\2, (-1)^j*binomial(n,j)*binomial(2*(n-j),k-2*j)*2^j*(k-j-1)!)/2; for (n=2,6,for (k=3,2*n, print1(a(n,k), ","));print); \\ Andrew Howroyd, May 09 2017
-
SageMath
def A167986(n,k): return simplify(binomial(2*n, k)*gamma(k)*hypergeometric([(1-k)/2, -k/2], [1-k, 1/2-n], -2)/2) flatten([[A167986(n,k) for k in range(3,2*n+1)] for n in range(2,11)]) # G. C. Greubel, Jan 17 2023
Formula
T(n,k) = Sum_{j=0..floor(k/2)} (-1)^j*binomial(n,j)*binomial(2*(n-j),k-2*j)*2^j*(k-j-1)!/2. - Andrew Howroyd, May 09 2017
Comments