login
A099602
Triangle, read by rows, such that row n equals the inverse binomial transform of column n of the triangle of trinomial coefficients (A027907), omitting leading zeros.
5
1, 1, 1, 1, 2, 1, 2, 5, 4, 1, 1, 5, 8, 5, 1, 3, 13, 22, 18, 7, 1, 1, 9, 26, 35, 24, 8, 1, 4, 26, 70, 101, 84, 40, 10, 1, 1, 14, 61, 131, 160, 116, 49, 11, 1, 5, 45, 171, 363, 476, 400, 215, 71, 13, 1, 1, 20, 120, 363, 654, 752, 565, 275, 83, 14, 1, 6, 71, 356, 1017, 1856, 2282, 1932
OFFSET
0,5
COMMENTS
Row sums form A099603, where A099603(n) = Fibonacci(n+1)*2^floor((n+1)/2). Central coefficients of even-indexed rows form A082759, where A082759(n) = Sum_{k=0..n} binomial(n,k)*trinomial(n,k). Antidiagonal sums form A099604.
Matrix inverse equals triangle A104495, which is generated from self-convolutions of the Catalan sequence (A000108).
FORMULA
G.f.: (1 + (y+1)*x - (y+1)*x^2)/(1 - (y+1)*(y+2)*x^2 + (y+1)^2*x^4).
EXAMPLE
Rows begin:
1;
1, 1;
1, 2, 1;
2, 5, 4, 1;
1, 5, 8, 5, 1;
3, 13, 22, 18, 7, 1;
1, 9, 26, 35, 24, 8, 1;
4, 26, 70, 101, 84, 40, 10, 1;
1, 14, 61, 131, 160, 116, 49, 11, 1;
5, 45, 171, 363, 476, 400, 215, 71, 13, 1;
1, 20, 120, 363, 654, 752, 565, 275, 83, 14, 1;
...
The binomial transform of row 2 = column 2 of A027907: BINOMIAL[1,2,1] = [1,3,6,10,15,21,28,36,45,55,...].
The binomial transform of row 3 = column 3 of A027907: BINOMIAL[2,5,4,1] = [2,7,16,30,50,77,112,156,210,...].
The binomial transform of row 4 = column 4 of A027907: BINOMIAL[1,5,8,5,1] = [1,6,19,45,90,161,266,414,615,...].
The binomial transform of row 5 = column 5 of A027907: BINOMIAL[3,13,22,18,7,1] = [3,16,51,126,266,504,882,1452,...].
PROG
(PARI) {T(n, k)=polcoeff(polcoeff((1+(y+1)*x-(y+1)*x^2)/(1-(y+1)*(y+2)*x^2+(y+1)^2*x^4)+x*O(x^n), n, x)+y*O(y^k), k, y)}
(PARI) {T(n, k)=(matrix(n+1, n+1, i, j, if(i>=j, polcoeff(polcoeff( (1+x*y/(1+x))/(1+x-y^2*(1-(1+4*x+O(x^i))^(1/2))^2/4+O(y^j)), i-1, x), j-1, y)))^-1)[n+1, k+1]}
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, Oct 25 2004
STATUS
approved