A269941 Triangle read by rows, the coefficients of the partial P-polynomials.

1, 0, -1, 0, -1, 1, 0, -1, 2, -1, 0, -1, 1, 2, -3, 1, 0, -1, 2, 2, -3, -3, 4, -1, 0, -1, 1, 2, 2, -1, -6, -3, 6, 4, -5, 1, 0, -1, 2, 2, 2, -3, -3, -6, -3, 4, 12, 4, -10, -5, 6, -1, 0, -1, 1, 2, 2, 2, -3, -3, -6, -6, -3, 1, 12, 6, 12, 4, -10, -20, -5, 15, 6, -7, 1

Offset: 0

Author

Peter Luschny, Mar 08 2016

Keywords

• sign
• tabf

Comments

For the definition of the partial P-polynomials see the link 'P-transform'. The triangle of coefficients of the inverse partial P-polynomials is A269942.

Examples

			[[1]],
[[0], [-1]],
[[0], [-1], [1]],
[[0], [-1], [2], [-1]],
[[0], [-1], [1, 2], [-3], [1]],
[[0], [-1], [2, 2], [-3, -3], [4], [-1]],
[[0], [-1], [1, 2, 2], [-1, -6, -3], [6, 4], [-5], [1]],
[[0], [-1], [2, 2, 2], [-3, -3, -6, -3], [4, 12, 4], [-10, -5], [6], [-1]]
Replacing the sublists by their sums reduces the triangle to a signed version of the triangle A097805.
		

Links

• Peter Luschny, The P-transform, 2016.
• Peter Luschny, The Partition Transform, A SageMath Jupyter Notebook, GitHub, 2016/2022.
• Marko Riedel, Answer to Question 4943578, Mathematics Stack Exchange, 2024.
• Peter Taylor, Answer to Question 474483, MathOverflow, 2024.

Crossrefs

Cf. A097805, A268441, A268442, A269942.

Programs

Maple

PTrans := proc(n, f, nrm:=NULL) local q, p, r, R;
if n = 0 then return [1] fi; R := [seq(0,j=0..n)];
for q in combinat:-partition(n) do
   p := [op(ListTools:-Reverse(q)),0]; r := p[1]+1;
   mul(binomial(p[j], p[j+1])*f(j)^p[j], j=1..nops(q));
   R[r] := R[r]-(-1)^r*% od;
if nrm = NULL then R else [seq(nrm(n,k)*R[k+1],k=0..n)] fi end:
A269941_row := n -> seq(coeffs(p), p in PTrans(n, n -> x[n])):
seq(lprint(A269941_row(n)), n=0..8);

Sage

def PtransMatrix(dim, f, norm = None, inverse = False, reduced = False):
    i = 1; F = [1]
    if reduced:
        while i <= dim: F.append(f(i)); i += 1
    else:
        while i <= dim: F.append(F[i-1]*f(i)); i += 1
    C = [[0 for k in range(m+1)] for m in range(dim)]
    C[0][0] = 1
    if inverse:
        for m in (1..dim-1):
            C[m][m] = -C[m-1][m-1]/F[1]
            for k in range(m-1, 0, -1):
                C[m][k] = -(C[m-1][k-1]+sum(F[i]*C[m][k+i-1]
                          for i in (2..m-k+1)))/F[1]
    else:
        for m in (1..dim-1):
            C[m][m] = -C[m-1][m-1]*F[1]
            for k in range(m-1, 0, -1):
                C[m][k] = -sum(F[i]*C[m-i][k-1] for i in (1..m-k+1))
    if norm == None: return C
    for m in (1..dim-1):
        for k in (1..m): C[m][k] *= norm(m,k)
    return C
def PMultiCoefficients(dim, norm = None, inverse = False):
    def coefficient(p):
        if p <= 1: return [p]
        return SR(p).fraction(ZZ).numerator().coefficients()
    f = lambda n: var('x'+str(n))
    P = PtransMatrix(dim, f, norm, inverse)
    return [[coefficient(p) for p in L] for L in P]
print(flatten(PMultiCoefficients(9)))