A239366 Triangular array read by rows: T(n,k) is the number of palindromic compositions of n having exactly k 1's, n>=0, 0<=k<=n.

1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 3, 0, 3, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 0, 1, 5, 0, 5, 0, 4, 0, 1, 0, 1, 3, 2, 3, 2, 1, 3, 1, 0, 0, 1, 8, 0, 10, 0, 7, 0, 5, 0, 1, 0, 1, 5, 3, 5, 5, 4, 3, 1, 4, 1, 0, 0, 1, 13, 0, 18, 0, 16, 0, 9, 0, 6, 0, 1, 0, 1, 8, 5, 10, 8, 7, 9, 5, 4, 1, 5, 1, 0, 0, 1

Offset: 0

Author

Geoffrey Critzer, Mar 20 2014

Keywords

• nonn
• tabl

Comments

Row sums = 2^floor(n/2).

T(n,0) = A053602(n-1) for n>0, T(n,1) = A079977(n-5) for n>4, T(2n+1,3) = A006367(n-1) for n>0, both bisections of column k=2 contain A010049. - Alois P. Heinz, Mar 21 2014

Examples

			1,
0, 1,
1, 0, 1,
1, 0, 0, 1,
2, 0, 1, 0, 1,
1, 1, 1, 0, 0, 1,
3, 0, 3, 0, 1, 0, 1,
2, 1, 1, 2, 1, 0, 0, 1,
5, 0, 5, 0, 4, 0, 1, 0, 1,
3, 2, 3, 2, 1, 3, 1, 0, 0, 1
There are eight palindromic compositions of 6: T(6,0)=3 because we have: 6, 3+3, 2+2+2.  T(6,2)=3 because we have: 1+4+1, 2+1+1+2, 1+2+2+1.  T(6,4)=1 because we have: 1+1+2+1+1. T(6,6)=1 because we have: 1+1+1+1+1+1.
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened

Programs

Maple

b:= proc(n) option remember;  `if`(n=0, 1, expand(
      add(b(n-j)*`if`(j=1, x^2, 1), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))
    (add(b(i)*`if`(n-2*i=1, x, 1), i=0..n/2)):
seq(T(n), n=0..30);  # Alois P. Heinz, Mar 21 2014

Mathematica

nn=15;Table[Take[CoefficientList[Series[((1+x)*(1-x+x^2+x*y-x^2*y))/(1-x^2-x^4-x^2*y^2+x^4*y^2),{x,0,nn}],{x,y}][[n]],n],{n,1,nn}]//Grid

Formula

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