OFFSET
0,2
COMMENTS
Points are counted in the standard (Milton Work) way, Ace=4, King=3, Queen=2, Jack=1, all other cards = 0. The first term a(0) then is the number of ways of being dealt a "Yarborough" (a hand with 0 points). That consists of 13 cards with 0 Aces, 0 Kings, 0 Queens and 0 Jacks which, all other cards being considered as indistinguishable, can be dealt in 1 way. The final term, a(37), represents a 37-point hand consisting of 4 Aces, 4 Kings, 4 Queens and 1 Jack. This can be dealt in 4 ways (the Jack is of any one of the four suits Clubs, Diamonds, Hearts, Spades). May you be dealt many of them (but not against me). The sequence would of course be symmetric, and sum to 2^16 except for the restriction to 13 cards.
FORMULA
a(n) = [y^n][x^13]((Product_{k=1..4} 1 + x*y^k)^4/(1-x)). - Andrew Howroyd, Aug 23 2019
PROG
(PARI) Vecrev(polcoef(prod(k=1, 4, (1 + x*y^k + O(x*x^13))^4)/(1-x), 13)) \\ Andrew Howroyd, Aug 22 2019
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Donal LYONS, Aug 17 2019
STATUS
approved