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A108267
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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.
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9
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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
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OFFSET
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0,5
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COMMENTS
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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
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)
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LINKS
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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.
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FORMULA
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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).
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)
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EXAMPLE
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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 + ...).
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MAPLE
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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
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MATHEMATICA
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T[n_, k_] := Coefficient[(1 - x)^(n + 1)*
Sum[Binomial[n + n*j + j, n*j + j]*x^j, {j, 0, n}], x, k];
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PROG
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(PARI) T(n, k)=polcoeff((1-x)^(n+1)*sum(j=0, n, binomial(n+n*j+j, n*j+j)*x^j), k)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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