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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

Wolfdieter Lang, Jul 21 2017

Keywords

Comments

The Sheffer triangle A028338 of the type (1/sqrt(1-2*x), -(1/2)*log(1 - 2*x)) is called here |S1hat[2,1]|. The o.g.f. of the sequence of diagonal d, d >= 0 is D(d, t) = Sum_{m=0..d} A028338(d+m, m)*t^m. The e.g.f. of these o.g.f.s is taken as ED(y,t) := Sum_{d >= 0} D(d, t)*y^(d+1)/(d+1)!.
This e.g.f. is found to be ED(y,t) = 1 - sqrt(1 - 2*x(t;y)), where x = x(t;y) is the compositional inverse of y = y(t;x) = x*(1 - t*(-log(1-2*x)/(2*x))) = x + t*log(1-2*x)/2. The o.g.f.s are then D(d, t) = P(d, t)/(1 - t)^(2*d+1), with the row polynomials P(d, t) = Sum_{m=0..d} T(d, m)*t^m, d >= 0.
This computation was inspired by an article of P. Bala (see a link under, e.g., A112007) for Sheffer triangles of the Jabotinsky type (1, F(x)). There Sheffer is called exponential Riordan, and the diagonals are labeled by n = d+1, n >= 1.

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
...
		

Crossrefs

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.