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%I A265017 #21 Aug 20 2021 04:27:06
%S A265017 1,0,1,0,0,2,0,3,0,3,0,5,0,0,4,0,7,8,0,0,5,0,25,12,0,0,0,6,0,36,16,15,
%T A265017 0,0,0,7,0,81,20,21,0,0,0,0,8,0,107,74,27,24,0,0,0,0,9,0,316,102,33,
%U A265017 32,0,0,0,0,0,10,0,427,222,39,40,35,0,0,0,0,0,11
%N A265017 Total sum T(n,k) of number of lambda-parking functions of partitions lambda of n into distinct parts with smallest part k; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
%H A265017 Alois P. Heinz, Rows n = 0..100, flattened
%H A265017 Richard P. Stanley, Parking Functions, 2011.
%e A265017 Triangle T(n,k) begins:
%e A265017 00 : 1;
%e A265017 01 : 0, 1;
%e A265017 02 : 0, 0, 2;
%e A265017 03 : 0, 3, 0, 3;
%e A265017 04 : 0, 5, 0, 0, 4;
%e A265017 05 : 0, 7, 8, 0, 0, 5;
%e A265017 06 : 0, 25, 12, 0, 0, 0, 6;
%e A265017 07 : 0, 36, 16, 15, 0, 0, 0, 7;
%e A265017 08 : 0, 81, 20, 21, 0, 0, 0, 0, 8;
%e A265017 09 : 0, 107, 74, 27, 24, 0, 0, 0, 0, 9;
%e A265017 10 : 0, 316, 102, 33, 32, 0, 0, 0, 0, 0, 10;
%e A265017 11 : 0, 427, 222, 39, 40, 35, 0, 0, 0, 0, 0, 11;
%e A265017 12 : 0, 869, 286, 153, 48, 45, 0, 0, 0, 0, 0, 0, 12;
%p A265017 p:= l-> (n-> n!*LinearAlgebra[Determinant](Matrix(n, (i, j)
%p A265017 -> (t->`if`(t<0, 0, l[i]^t/t!))(j-i+1))))(nops(l)):
%p A265017 g:= (n, i, l)-> `if`(i*(i+1)/2n, 0, g(n-i, i-1, [i, l[]])))):
%p A265017 T:= n-> (f-> seq(coeff(f, x, i), i=0..n))(g(n$2, [])):
%p A265017 seq(T(n), n=0..16);
%t A265017 p[l_] := With[{n = Length[l]}, n!*Det[Table[Function[t,
%t A265017 If[t < 0, 0, l[[i]]^t/t!]][j - i + 1], {i, n}, {j, n}]]];
%t A265017 g[n_, i_, l_] := If[i(i+1)/2 < n, 0, If[n == 0, p[l]*x^
%t A265017 If[l == {}, 0, l[[1]]], g[n, i - 1, l] +
%t A265017 If[i > n, 0, g[n - i, i - 1, Prepend[l, i]]]]];
%t A265017 T[n_] := If[n == 0, {1}, CoefficientList[g[n, n, {}], x]];
%t A265017 Table[T[n], {n, 0, 16}] // Flatten (* _Jean-François Alcover_, Aug 20 2021, after _Alois P. Heinz_ *)
%Y A265017 Row sums give A265016.
%Y A265017 Column k=0 gives A000007.
%Y A265017 Main diagonal gives A028310, first lower diagonal is A000004.
%Y A265017 T(2n+1,n) gives A005563.
%Y A265017 T(2n+2,n) gives A028347(n+2).
%Y A265017 T(2n+3,n) gives A028560.
%K A265017 nonn,tabl
%O A265017 0,6
%A A265017 _Alois P. Heinz_, Nov 30 2015
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