A108267 Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)*Sum_{j=0..n} binomial(n + n*j + j, n*j + j)*x^j.

1, 1, 1, 1, 7, 1, 1, 31, 31, 1, 1, 121, 381, 121, 1, 1, 456, 3431, 3431, 456, 1, 1, 1709, 26769, 60691, 26769, 1709, 1, 1, 6427, 193705, 848443, 848443, 193705, 6427, 1, 1, 24301, 1343521, 10350421, 19610233, 10350421, 1343521, 24301, 1

Offset: 0

Author

Paul D. Hanna, May 29 2005 and May 31 2005

Keywords

• nonn
• tabl

Comments

G.f. of row n divided by (1-x)^(n+1) equals g.f. of row n of table A060543.

Matrix product of this triangle with Pascal's triangle (A007318) equals A108291.

Seeing each row as a polynomial, all roots seem to be negative reals. - F. Chapoton, Nov 01 2022

From Thomas Anton, Jan 05 2023: (Start)

Consider the set [m] := {1, 2, 3, ..., m} ordered cyclically, and then mapped into itself via f. Let us consider a in [m] as the (a-1)th m-th root of unity e^(2*Pi*i*(a-1)/m). Then f may be extended to a continuous map f':S^1 -> S^1 as follows:

For a immediately before b in the cyclic order, map the interval between a and b to S^1 so that a point in it moving clockwise at constant speed has a value moving clockwise at constant speed, and the map travels the shortest distance possible given this condition.

T(n, k) gives the number of f for m = n-1 such that f(1) = 1 and f' has degree k. This is trivially one n-th of the number of f with degree k when f(1) is arbitrary.

Equivalent to having degree k is that there are k values a immediately before b in the cyclic order such that f(a) > f(b) (in the standard order of N).

If we change things so that a immediately before b satisfies f(a) = f(b) corresponds to a full rotation (this is equivalent to using the condition f(a) >= f(b) in the last paragraph), then T(n, k) is the number of f with degree k+1.

T(n, k) is the (k+1)*(n-1)th (n-1)-nomial coefficient of power n - 1.

(End)

Examples

			Triangle begins:
  1;
  1,    1;
  1,    7,      1;
  1,   31,     31,      1;
  1,  121,    381,    121,      1;
  1,  456,   3431,   3431,    456,      1;
  1, 1709,  26769,  60691,  26769,   1709,    1;
  1, 6427, 193705, 848443, 848443, 193705, 6427, 1;
  ...
G.f. of row 3: (1 + 31*x + 31*x^2 + x^3) = (1-x)^4*(1 + 35*x + 165*x^2 + 455*x^3 + ... + C(4*j+3,4*j)*x^j + ...).
		

Links

• M. Bayer, B. Goeckner, S. J. Hong, T. McAllister, M. Olsen, C. Pinckney, J. Vega and M. Yip, Lattice polytopes from Schur and symmetric Grothendieck polynomials, Electronic Journal of Combinatorics, Volume 28, Issue 2 (2021). See Proposition 53 and Table 1.
• Tanay Wakhare, Iterated Entropy Derivatives and Binary Entropy Inequalities, arXiv:2312.14743 [cs.IT], 2023.
• Tanay Wakhare, Two Studies of Constraints in High Dimensions: Entropy Inequalities and the Randomized Symmetric Binary Perceptron, Master's Thesis, MIT (2024). See p. 22.
• Raphael Yuster, Almost k-union closed set systems, arXiv:2302.12276 [math.CO], 2023, p. 8.

Crossrefs

Cf. A108267, A000169, A048775, A060543, A108291, A108292, A056856.

Programs

Maple

p := n -> (1-x)^(n+1)*add(binomial(n + n*j + j, n*j + j)*x^j, j = 0..n):
seq(print(seq(coeff(p(n), x, k), k = 0..n)), n = 0..8); # Peter Luschny, Nov 02 2022

Mathematica

T[n_, k_] := Coefficient[(1 - x)^(n + 1)*
     Sum[Binomial[n + n*j + j, n*j + j]*x^j, {j, 0, n}], x, k];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 23 2021 *)

PARI

T(n,k)=polcoeff((1-x)^(n+1)*sum(j=0,n,binomial(n+n*j+j,n*j+j)*x^j),k)

Formula

T(n, 1) = A048775(n) = binomial(2*n + 1, n + 1) - (n + 1).

Sum_{k=0..n} T(n, k) = A000169(n) = (n + 1)^n.

Sum_{k=0..n} T(n, k)*2^k = A108292(n).

From Thomas Anton, Jan 05 2023: (Start)

T(n, k) = Sum_{i=0..k} (-1)^i*binomial(n + 1, i)*binomial(n+(n+1)*(k-i), n).

T(n, k) = T(n, n-k).

(End)