

A099514


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 + 2*z^2)^(n[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.


1



1, 1, 1, 1, 2, 2, 1, 3, 5, 4, 1, 4, 9, 13, 4, 1, 5, 14, 28, 18, 12, 1, 6, 20, 50, 49, 56, 8, 1, 7, 27, 80, 105, 161, 56, 32, 1, 8, 35, 119, 195, 366, 210, 200, 16, 1, 9, 44, 168, 329, 721, 581, 732, 160, 80, 1, 10, 54, 228, 518, 1288, 1337, 2045, 780, 640, 32, 1, 11, 65, 300, 774, 2142, 2716, 4824, 2674, 2884, 432, 192
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OFFSET

0,5


COMMENTS

Row sums form A099515. 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.: ((1a*xc*x^2*y^2) + b*x*y)/((1a*xc*x^2*y^2)^2  x*(b*x*y)^2).


LINKS

Table of n, a(n) for n=0..77.


FORMULA

G.f.: (1x+x*y2*x^2*y^2)/((1x)^24*x^2*y^2+3*x^3*y^2+4*x^4*y^4).


EXAMPLE

Rows begin:
[1],
[1,1],
[1,2,2],
[1,3,5,4],
[1,4,9,13,4],
[1,5,14,28,18,12],
[1,6,20,50,49,56,8],
[1,7,27,80,105,161,56,32],
[1,8,35,119,195,366,210,200,16],
[1,9,44,168,329,721,581,732,160,80],...
and can be derived from coefficients of (1+z+2*z^2)^n:
[1],
[1,1,2],
[1,2,5,4,4],
[1,3,9,13,18,12,8],
[1,4,14,28,49,56,56,32,16],
[1,5,20,50,105,161,210,200,160,80,32],...
by shifting each column k down by [k/2] rows.


MATHEMATICA

With[{m = 11}, CoefficientList[CoefficientList[Series[(1x+x*y2*x^2*y^2)/ ((1x)^24*x^2*y^2+3*x^3*y^2+4*x^4*y^4), {x, 0 , m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 17 2020 *)


PROG

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


CROSSREFS

Cf. A099509, A099510, A099512, A099515.
Sequence in context: A257005 A160232 A026300 * A228352 A303911 A205575
Adjacent sequences: A099511 A099512 A099513 * A099515 A099516 A099517


KEYWORD

nonn,tabl


AUTHOR

Paul D. Hanna, Oct 20 2004


EXTENSIONS

a(50..51) corrected by Georg Fischer, Feb 17 2020


STATUS

approved



