A302612 a(n) = (n+1)*(n^4-4*n^3+11*n^2-8*n+12)/12.

1, 2, 6, 20, 65, 186, 462, 1016, 2025, 3730, 6446, 10572, 16601, 25130, 36870, 52656, 73457, 100386, 134710, 177860, 231441, 297242, 377246, 473640, 588825, 725426, 886302, 1074556, 1293545, 1546890, 1838486, 2172512, 2553441, 2986050, 3475430, 4026996

Offset: 0

Author

Bryan T. Ek, Apr 10 2018

Keywords

• nonn
• easy

Comments

The limit as q->1^- of the unimodal polynomial [q^(n*k+n+4)-q^(n*k+n+3)+q^(n*k+n+1)-q^(n*k+4)-q^((n-1)*k+n+3)+q^((n-1)*k+3)+q^(k+n+1)-q^(k+1)-q^n+q^3-q+1]/[(1-q)^2(1-q^2)(1-q^n)] after making the simplification k=n. This unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <=2. See G_2(n,k) from arXiv:1711.11252.

As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984.

Examples

			For n=4, G_2(4,4)=q^16+q^15+2*q^14+3*q^13+5*q^12+5*q^11+6*q^10+6*q^9+7*q^8+6*q^7+6*q^6+5*q^5+5*q^4+3*q^3+2*q^2+q+1 (using the formula in the comments). Then substituting q=1 yields 65.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Bryan Ek, q-Binomials and related symmetric unimodal polynomials, arXiv:1711.11252 [math.CO], 2017-2018.
• Bryan Ek, Unimodal Polynomials and Lattice Walk Enumeration with Experimental Mathematics, arXiv:1804.05933 [math.CO], 2018.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000984, A002522, A302644, A302645, A302646.

Programs

PARI

Vec((1 - 4*x + 9*x^2 - 6*x^3 + 10*x^4) / (1 - x)^6 + O(x^40)) \\ Colin Barker, Apr 11 2018

Formula

From Colin Barker, Apr 11 2018: (Start)

G.f.: (1 - 4*x + 9*x^2 - 6*x^3 + 10*x^4) / (1 - x)^6.

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5.

(End)

Extensions

More terms from Colin Barker, Apr 11 2018