A099509 Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 6, 7, 1, 1, 5, 10, 16, 6, 3, 1, 6, 15, 30, 19, 16, 1, 1, 7, 21, 50, 45, 51, 10, 4, 1, 8, 28, 77, 90, 126, 45, 30, 1, 1, 9, 36, 112, 161, 266, 141, 126, 15, 5, 1, 10, 45, 156, 266, 504, 357, 393, 90, 50, 1, 1, 11, 55, 210, 414, 882, 784, 1016, 357

Offset

0,5

Comments

Row sums form absolute values of A078039. In general if T(n,k) = coefficient of z^k in (a + b*z + c*z^2)^(n-[k/2]), then the resulting number triangle will have the o.g.f.: ((1-a*x-c*x^2*y^2) + b*x*y)/((1-a*x-c*x^2*y^2)^2 - x*(b*x*y)^2).

Links

Table of n, a(n) for n=0..74.

Formula

G.f.: (1-x+x*y-x^2*y^2)/((1-x)^2-2*x^2*y^2+x^3*y^2+x^4*y^4).

Example

Rows begin:
[1],
[1,1],
[1,2,1],
[1,3,3,2],
[1,4,6,7,1],
[1,5,10,16,6,3],
[1,6,15,30,19,16,1],
[1,7,21,50,45,51,10,4],
[1,8,28,77,90,126,45,30,1],
[1,9,36,112,161,266,141,126,15,5],...
and can be derived from coefficients of (1+z+z^2)^n:
[1],
[1,1,1],
[1,2,3,2,1],
[1,3,6,7,6,3,1],
[1,4,10,16,19,16,10,4,1],
[1,5,15,30,45,51,45,30,15,5,1],...
by shifting each column k down by [k/2] rows.

Prog

(PARI) T(n, k)=if(n<k || k<0, 0, polcoeff((1+z+z^2+z*O(z^k))^(n-k\2), k, z))

Crossrefs

Cf. A027907, A078039, A099510.

Sequence in context: A175424 A144401 A034929 * A153859 A363394 A368343

Adjacent sequences: A099506 A099507 A099508 * A099510 A099511 A099512

Keyword

nonn,tabl

Author

Paul D. Hanna, Oct 20 2004

Status

approved