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Number of ways of being dealt an n-point Bridge hand.
1

%I #33 May 16 2023 09:48:15

%S 1,4,10,24,51,96,170,284,445,664,952,1304,1716,2184,2680,3176,3650,

%T 4064,4380,4584,4658,4584,4380,4064,3650,3176,2680,2184,1716,1304,952,

%U 664,439,268,148,64,25,4

%N Number of ways of being dealt an n-point Bridge hand.

%C 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.

%F a(n) = [y^n][x^13]((Product_{k=1..4} 1 + x*y^k)^4/(1-x)). - _Andrew Howroyd_, Aug 23 2019

%o (PARI) Vecrev(polcoef(prod(k=1, 4, (1 + x*y^k + O(x*x^13))^4)/(1-x), 13)) \\ _Andrew Howroyd_, Aug 22 2019

%Y Cf. A363058.

%K nonn,fini,full

%O 0,2

%A _Donal LYONS_, Aug 17 2019