A363041 - OEIS

%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