%I #26 May 23 2023 08:17:48
%S 1,0,1,1,0,1,0,5,0,1,1,0,15,0,1,0,21,0,35,0,1,1,0,161,0,70,0,1,0,85,0,
%T 777,0,126,0,1,1,0,1555,0,2835,0,210,0,1,0,341,0,14575,0,8547,0,330,0,
%U 1,1,0,14421,0,91960,0,22407,0,495,0,1
%N Triangle read by rows: T(n,k) = Stirling2(n+1,k)/binomial(k+1,2) if n-k is even, else 0 (1 <= k <= n).
%C A companion triangle to the triangle of Hultman numbers A164652.
%C The triangle of Hultman numbers can be constructed from the triangle of Stirling cycle numbers ( |A008275(n,k)| )n,k>=1 by removing the triangular number factor n*(n-1)/2 from every other entry in the n-th row (n >= 2) and setting the remaining entries to 0.
%C Here we carry out the analogous construction starting with the triangle of Stirling numbers of the second kind A008277, but now removing the triangular number factor k*(k+1)/2 from every other entry in the k-th column and setting the remaining entries to 0.
%C Do these numbers have a combinatorial interpretation?
%F Let P(n,x) = (1 - x)*(1 - 2*x)*...*(1 - n*x). The g.f. for the k-th column of the triangle is (1/(k*(k + 1)))*x^(k-1)*(1/P(k,x) - 1/P(k,-x)) = (x^k)*(x^k*R(k-1,1/x))/((1 - x^2)*(1 - 4*x^2)*...*(1 - k^2*x^2)), where R(n,x) denotes the n-th row polynomial of A164652. (Since the entries of triangle A164652 are integers, it follows that the entries of the present triangle are also integers.)
%F It appears that the matrix product (|A008275|)^-1 * A164652 * A008277 = I_1 + A363041 (direct sum, where I_1 is the 1 X 1 identity matrix). See the Example section.
%F The sequence of row sums of the inverse array begins [1, 1, 0, -4, 0, 120, 0, -12096, 0, 3024000, 0, -1576143360, 0, 1525620096000, 0, -2522591034163200, 0, 6686974460694528000, 0, -27033456071346536448000, ...], and appears to be essentially A129825.
%e Triangle begins
%e k = 1 2 3 4 5 6 7 8 9 10
%e - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e n = 1: 1
%e 2: 0 1
%e 3: 1 0 1
%e 4: 0 5 0 1
%e 5: 1 0 15 0 1
%e 6: 0 21 0 35 0 1
%e 7: 1 0 161 0 70 0 1
%e 8: 0 85 0 777 0 126 0 1
%e 9: 1 0 1555 0 2835 0 210 0 1
%e 10: 0 341 0 14575 0 8547 0 330 0 1
%e ...
%e Matrix product (|A008275|)^-1 * A164652 * A008277 begins
%e / 1 \ /1 \ /1 \ /1 \
%e |-1 1 | |0 1 | |1 1 | |0 1 |
%e | 1 -3 1 | |1 0 1 | |1 3 1 | = |0 0 1 |
%e |-1 7 -6 1 | |0 5 0 1 | |1 7 6 1 | |0 1 0 1 |
%e | 1 -15 25 -10 1| |8 0 15 0 1| |1 15 25 10 1| |0 0 5 0 1 |
%e | ... | |... | |... | |0 1 0 15 0 1|
%e | | | | | | |... |
%p A362041:= (n, k)-> `if`(n-k mod 2 = 0, Stirling2(n+1,k)/binomial(k+1,2), 0):
%p for n from 1 to 10 do seq(A362041(n,k), k = 1..n) od;
%o (PARI) T(n,k) = if ((n-k) % 2, 0, stirling(n+1, k, 2)/binomial(k+1, 2)); \\ _Michel Marcus_, May 23 2023
%Y Row sums give A363042.
%Y Cf. A008275, A008277, A164652, A129825.
%K nonn,tabl,easy
%O 1,8
%A _Peter Bala_, May 14 2023