OFFSET
0,5
COMMENTS
G.f. of row n divided by (1-x)^(n+1) equals g.f. of row n of table A060543.
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)
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.
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)
EXAMPLE
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 + ...).
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 *)
PROG
(PARI) T(n, k)=polcoeff((1-x)^(n+1)*sum(j=0, n, binomial(n+n*j+j, n*j+j)*x^j), k)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, May 29 2005 and May 31 2005
STATUS
approved