A288875 Triangle read by rows. The rows give the coefficients of the numerator polynomials for the o.g.f.s of the diagonal sequences of triangle A028338.
1, 1, 1, 3, 8, 1, 15, 71, 33, 1, 105, 744, 718, 112, 1, 945, 9129, 14542, 5270, 353, 1, 10395, 129072, 300291, 191384, 33057, 1080, 1, 135135, 2071215, 6524739, 6338915, 2033885, 190125, 3265, 1, 2027025, 37237680, 150895836, 204889344, 103829590, 18990320, 1038780, 9824, 1
Offset: 0
Examples
The triangle T(n, m) begins: n\m 0 1 2 3 4 5 6 7 8 ... 0: 1 1: 1 1 2: 3 8 1 3: 15 71 33 1 4: 105 744 718 112 1 5: 945 9129 14542 5270 353 1 6: 10395 129072 300291 191384 33057 1080 1 7: 135135 2071215 6524739 6338915 2033885 190125 3265 1 8: 2027025 37237680 150895836 204889344 103829590 18990320 1038780 9824 1 ...
Links
- Peter Bala, Diagonals of triangles with generating function exp(t*F(x)).
- Wolfdieter Lang, On Generating functions of Diagonal Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Programs
-
Mathematica
De[d_, t_] := Sum[A028338[d+m, m] t^m, {m, 0, d}]; A028338[n_, k_] := SeriesCoefficient[Times @@ Table[x+i, {i, 1, 2n-1, 2}], {x, 0, k}]; P[n_, x_] := De[n, x] (1-x)^(2n+1); T[n_, m_] := Coefficient[P[n, x], x, m]; Table[T[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 24 2017 *) T[n_,m_]:=Sum[(-1)^(i-n+m)*Binomial[2*n+1,n-m-i]*(1/(2^i*i!)*Sum[(-1)^(i-j)*Binomial[i,j]*(2*j+1)^(n+i),{j,0,i}]),{i,0,n-m}];Flatten[Table[T[n,m],{n,0,8},{m,0,n}]] (* Detlef Meya, Dec 18 2023, after Peter Bala from A214406 *)
Formula
T(n, m) = [x^m] P(n, x), with the numerator polynomial of the o.g.f. of the diagonal n (main diagonal n=0) D(n, x) = P(n, x)/(1-x)^(2*n+1). See a comment above.
T(n, m) = Sum_{i=0..n-m} ( (-1)^(i-n+m)*binomial(2*n+1,n-m-i)*(1/(2^i*i!))*Sum_{j=0..i} (-1)^(i-j)*binomial(i,j)*(2*j+1)^(n+i) ). - Detlef Meya, Dec 18 2023, after Peter Bala from A214406.
Comments