A277949 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^4.

1, 1, 4, 6, 4, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 4, 10, 20, 31, 40, 44, 40, 31, 20, 10, 4, 1, 1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1, 1, 4, 10, 20, 35, 56, 80, 104, 125, 140, 146, 140, 125, 104, 80, 56, 35, 20, 10, 4, 1

Offset: 1

Author

Juan Pablo Herrera P., Nov 05 2016

Keywords

• nonn
• easy
• tabf

Comments

Sum of n-th row is n^4. The n-th row contains 4n-3 entries. Largest coefficients of each row are listed in A005900.

The n-th row is the fourth row of the n-nomial triangle. For example, row 2 (1,4,6,4,1) is the fourth row in the binomial triangle.

T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with four different playing cards. It is also the number of lattice paths from (0,0) to (4,k) using steps (1,0), (1,1), (1,2), ..., (1,n-1).

Examples

			Triangle starts:
1;
1, 4, 6, 4, 1;
1, 4, 10, 16, 19, 16, 10, 4, 1;
1, 4, 10, 20, 31, 40, 44, 40, 31, 20, 10, 4, 1;
1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1;
1, 4, 10, 20, 35, 56, 80, 104, 125, 140, 146, 140, 125, 104, 80, 56, 35, 20, 10, 4, 1.
...
There are T(3,2) = 10 ways to select 2 cards from two sets of four playing cards ABCD, namely, {AA}, {AB}, {AC}, {AD}, {BB}, {BC}, {BD}, {CC}, {CD}, and {DD}.
		

Links

• Juan Pablo Herrera P., Rows n=1..60 of the triangle, flattened

Crossrefs

Cf. A000292, A004737, A005900, A109439, A210440, A277950, A277951.

Mentioned in: A273975

Programs

Mathematica

Table[CoefficientList[Series[((x^n - 1)/(x - 1))^4, {x, 0, 4 n}], x], {n, 6}] // Flatten (* Michael De Vlieger, Nov 10 2016 *)

PARI

row(n) = Vec(((1 - x^n)/(1 - x))^4);
tabf(nn) = for (n=1, nn, print(row(n)));

Formula

T(n,k) = Sum_{i=k-n+1..k} A109439(T(n,i)).

T(n,k) = A000292(k+1) = (k+3)!/(k!*6) if 0 =< k < n,

T(n,k) = ((k+3)*(k+2)*(k+1)-4*(k-n+3)*(k-n+2)*(k-n+1))/6 if n =< k < 2*n,

T(n,k) = ((4*n-1-k)*(4*n-2-k)*(4*n-3-k)-4*(3*n-1-k)*(3*n-2-k)*(3*n-3-k))/6 if 2*n-3 =< k < 3*n-3,

T(n,k) = A000292(4*n-3-k) = (4*n-1-k)!/((4*n-4-k)!*6) if 3*n-3 =< k < 4n-3.