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%I #25 Aug 12 2022 19:10:03
%S 1,-1,1,-2,-1,1,0,-7,0,1,0,-14,-13,2,1,120,-46,-65,-15,5,1,0,516,-356,
%T -165,-5,9,1,5040,1392,266,-1421,-280,28,14,1,0,46320,3772,-5740,
%U -3871,-280,98,20,1,0,215280,212724,-26272,-31437,-7791,126,222,27,1
%N Triangle read by rows: T(n,k) gives n! times the coefficient of x^k in the polynomial that describes the number of permutations on x letters with major index n.
%C This n-th row describes a polynomial that eventually agrees with the n-th column of A008302.
%C Conjecture: For each m, T(n,n-m) is a polynomial of degree 2m whose leading coefficient is abs(A290030(m)/A053657(m+1)).
%H Peter Kagey, <a href="/A342645/b342645.txt">Rows n = 0..100, flattened</a>
%H Mike Earnest, <a href="https://math.stackexchange.com/a/4065913/121988">Does the number of permutations in S_n with major index equal to k, satisfy a degree k polynomial?</a>, Mathematics Stack Exchange answer.
%F Conjectures:
%F T(n,n) = 1.
%F T(n,n-1) = (-3n + n^2)/2.
%F T(n,n-2) = (-2n + 21n^2 - 22n^3 + 3n^4)/24.
%F T(n,n-3) = (96n - 134n^2 + 13n^3 + 37n^4 - 13n^5 + n^6)/48.
%e n\k | 0 1 2 3 4 5 6 7 8 9
%e ----+--------------------------------------------------------------
%e 0 | 1;
%e 1 | -1, 1;
%e 2 | -2, -1, 1;
%e 3 | 0, -7, 0, 1;
%e 4 | 0, -14, -13, 2, 1;
%e 5 | 120, -46, -65, -15, 5, 1;
%e 6 | 0, 516, -356, -165, -5, 9, 1;
%e 7 | 5040, 1392, 266, -1421, -280, 28, 14, 1;
%e 8 | 0, 46320, 3772, -5740, -3871, -280, 98, 20, 1;
%e 9 | 0, 215280, 212724, -26272, -31437, -7791, 126, 222, 27, 1;
%e For n = 4, the polynomial that describes the 4th column of A008302 is
%e A008302(x,4) = (-14x -13x^2 +2x^3 + x^4)/4! = Sum_{j=0..4} (T(j,4)*x^j)/4!.
%t A008302T[0, 0] := 1; A008302T[-1, k_] := 0;
%t A008302T[n_, k_] := (A008302T[n, k] = If[0 <= k <= n*(n - 1)/2, A008302T[n, k - 1] + A008302T[n - 1, k] - A008302T[n - 1, k - n], 0]);
%t A342645Row[n_] := (A342645Row[n] = Expand[n!*InterpolatingPolynomial[Table[{m, A008302T[m, n]}, {m, n, 2*n + 2}], x]]);
%t A342645T[n_, k_] := Coefficient[A342645Row[n], x, k];
%Y Cf. A008302, A053657, A290030.
%K sign,tabl
%O 0,4
%A _Peter Kagey_, Mar 17 2021
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