A362365 The sum of the coefficients of x^k in the expansion of (x + x^2 + x^3 + x^4 + x^5 + x^6)^n with k divisible by 4.

1, 9, 55, 322, 1946, 11664, 69980, 419912, 2519416, 15116544, 90699280, 544195552, 3265173536, 19591041024, 117546246080, 705277476992, 4231664861056, 25389989167104, 152339935002880, 914039610015232, 5484237660094976, 32905425960566784, 197432555763399680

Offset

1,2

Comments

a(n) is the number of ways that the sum of n labeled 6-sided dice is divisible by 4. This is important for the game called Mahjong, where the remainder of the sum of n randomly rolled dice when divided by 4 determines the starting player. Usually n=3.

Links

Paolo Xausa, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (4,10,12).

Formula

a(n) = (1/4)*6^n + (2^(n/2-1))*cos(3*Pi*n/4).

a(n) = (1/4)*(6^n + (-1+i)^n + (-1-i)^n), where i is the imaginary unit.

a(n) = (1/4)*(A000400(n) + 2*A009116(n)).

a(n) = 4*a(n-1) + 10*a(n-2) + 12*a(n-3).

G.f.: x*(1 + 5*x + 9*x^2)/((1 - 6*x)*(1 + 2*x + 2*x^2)).

E.g.f.: (1/4)*exp(6*x) + cos(x)/(2*exp(x)) - 3/4.

Limit_{n->oo} a(n)/6^n = 1/4.

Example

For n=2, (x + x^2 + x^3 + x^4 + x^5 + x^6)^2 = x^2 + 2*x^3 + 3*x^4 + 4*x^5 + 5*x^6 + 6*x^7 + 5*x^8 + 4*x^9 + 3*x^10 + 2*x^11 + x^12. So a(2) = 3 + 5 + 1 = 9.

Mathematica

LinearRecurrence[{4, 10, 12}, {1, 9, 55}, 30] (* Paolo Xausa, Aug 30 2024 *)

Prog

(MATLAB) an=(6^n+(-1+1i)^n+(-1-1i)^n)/4
(PARI) a(n)=polcoef(lift(Mod(x+x^2+x^3+x^4+x^5+x^6, 1-x^4)^n), 0) \\ Andrew Howroyd, Apr 17 2023

Crossrefs

Cf. A000400, A009116.

Sequence in context: A244650 A097790 A356339 * A183805 A037578 A096191

Adjacent sequences: A362362 A362363 A362364 * A362366 A362367 A362368

Keyword

easy,nonn

Author

Yui Chit Chan, Apr 17 2023

Status

approved