|
|
A302612
|
|
a(n) = (n+1)*(n^4-4*n^3+11*n^2-8*n+12)/12.
|
|
5
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
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.
|
|
LINKS
|
|
|
FORMULA
|
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)
|
|
EXAMPLE
|
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.
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|