Original entry on oeis.org
0, 0, 18, 162, 972, 4860, 21870, 91854, 367416, 1417176, 5314410, 19486170, 70150212, 248714388, 870500358, 3013270470, 10331213040, 35126124336, 118550669634, 397493421714, 1324978072380, 4393348345260, 14498049539358
Offset: 0
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[(n^2-n)*3^n: n in [0..25]]; // Vincenzo Librandi, Feb 10 2013
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CoefficientList[Series[18 x^2/(1 - 3 x)^3, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2013 *)
LinearRecurrence[{9,-27,27},{0,0,18},40] (* Harvey P. Dale, Dec 15 2014 *)
A293270
a(n) = n^n*binomial(2*n-1, n).
Original entry on oeis.org
1, 1, 12, 270, 8960, 393750, 21555072, 1413199788, 107961384960, 9418192087590, 923780000000000, 100633991211229476, 12055263261877075968, 1575041416811693275900, 222887966509090352332800, 33962507149515380859375000, 5543988061027763016035205120
Offset: 0
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Join[{1}, Table[n^n Binomial[2 n - 1, n], {n, 1, 16}]]
Join[{1}, Table[(-1)^n n^n Binomial[-n, n], {n, 1, 16}]]
Table[SeriesCoefficient[1/(1 - n x)^n, {x, 0, n}], {n, 0, 16}]
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a(n) = n^n*binomial(2*n-1, n); \\ Altug Alkan, Oct 04 2017
A381930
Irregular triangular array read by rows. T(n,k) is the number of length n words x on {0,1,2} such that I(x) + W_0(x)*W_1(x) + W_0(x)*W_2(x) + W_1(x)*W_2(x) = k where I(x) is the number of inversions in x and W_i(x) is the number of occurrences of the letter i in x for i={0,1,2}, n>=0, 0<=k<=floor(2n^2/3).
Original entry on oeis.org
1, 3, 3, 3, 3, 3, 0, 6, 7, 8, 2, 1, 3, 0, 0, 6, 9, 12, 18, 12, 12, 6, 3, 3, 0, 0, 0, 6, 6, 12, 15, 27, 27, 36, 33, 33, 21, 15, 6, 3, 3, 0, 0, 0, 0, 6, 6, 6, 12, 18, 27, 33, 52, 62, 77, 82, 86, 75, 68, 48, 35, 19, 11, 2, 1
Offset: 0
Triangle T(n,k) begins:
1;
3;
3, 3, 3;
3, 0, 6, 7, 8, 2, 1;
3, 0, 0, 6, 9, 12, 18, 12, 12, 6, 3;
3, 0, 0, 0, 6, 6, 12, 15, 27, 27, 36, 33, 33, 21, 15, 6, 3;
...
T(3,3) = 7 because we have: {0, 1, 0}, {0, 1, 2}, {0, 2, 0}, {1, 0, 1}, {1, 2, 1}, {2, 0, 2}, {2, 1, 2}.
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b:= proc(n, i, j, k) option remember; expand(
`if`(n=0, z^(i*j+i*k+j*k), b(n-1, i+1, j, k)*z^(j+k)+
b(n-1, i, j+1, k)*z^k +b(n-1, i, j, k+1)))
end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0$3)):
seq(T(n), n=0..10); # Alois P. Heinz, Mar 10 2025
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nn = 6; B[n_] := FunctionExpand[QFactorial[n, q]]*q^Binomial[n, 2];e[z_] := Sum[z^n/B[n], {n, 0, nn}];Map[CoefficientList[#, q] &,Table[B[n], {n, 0, nn}] CoefficientList[Series[e[z]^3, {z, 0, nn}],z]] // Grid
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