A164652 -id:A164652 - cp's OEIS frontend

A363041 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).

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, 777, 0, 126, 0, 1, 1, 0, 1555, 0, 2835, 0, 210, 0, 1, 0, 341, 0, 14575, 0, 8547, 0, 330, 0, 1, 1, 0, 14421, 0, 91960, 0, 22407, 0, 495, 0, 1

Offset: 1

Peter Bala, May 14 2023

A companion triangle to the triangle of Hultman numbers A164652.
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.
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.
Do these numbers have a combinatorial interpretation?

			Triangle begins
     k = 1     2     3     4     5     6     7     8     9    10
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
n = 1:   1
    2:   0     1
    3:   1     0     1
    4:   0     5     0     1
    5:   1     0    15     0     1
    6:   0    21     0    35     0     1
    7:   1     0   161     0    70     0     1
    8:   0    85     0   777     0   126     0     1
    9:   1     0  1555     0  2835     0   210     0     1
   10:   0   341     0 14575     0  8547     0   330     0     1
   ...
Matrix product (|A008275|)^-1 * A164652 * A008277 begins
  / 1             \ /1         \ /1           \   /1           \
  |-1   1         | |0 1       | |1  1        |   |0 1         |
  | 1  -3  1      | |1 0  1    | |1  3  1     | = |0 0 1       |
  |-1   7 -6   1  | |0 5  0 1  | |1  7  6  1  |   |0 1 0  1    |
  | 1 -15 25 -10 1| |8 0 15 0 1| |1 15 25 10 1|   |0 0 5  0 1  |
  | ...           | |...       | |...         |   |0 1 0 15 0 1|
  |               | |          | |            |   |...         |

Row sums give A363042.

Cf. A008275, A008277, A164652, A129825.

A362041:= (n, k)-> `if`(n-k mod 2 = 0, Stirling2(n+1,k)/binomial(k+1,2), 0):
for n from 1 to 10 do seq(A362041(n,k), k = 1..n) od;

T(n,k) = if ((n-k) % 2, 0, stirling(n+1, k, 2)/binomial(k+1, 2)); \\ Michel Marcus, May 23 2023

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.)
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.
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.

A237996 Triangular array read by rows. T(n,k) is the number of even permutations of {1,2,...,n} that have exactly k cycles, n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 0, 11, 0, 1, 0, 24, 0, 35, 0, 1, 0, 0, 274, 0, 85, 0, 1, 0, 720, 0, 1624, 0, 175, 0, 1, 0, 0, 13068, 0, 6769, 0, 322, 0, 1, 0, 40320, 0, 118124, 0, 22449, 0, 546, 0, 1, 0, 0, 1026576, 0, 723680, 0, 63273, 0, 870, 0, 1

Offset: 0

Geoffrey Critzer, Feb 16 2014

Row sums = A001710(n).

If a permutation of {1, 2, ..., n} is written as a product of m disjoint cycles (where the fixed points of the permutation are viewed as 1-cycles) then the parity of the permutation is (-1)^(n-m). It is an even permutation if the number of cycles of even length is even (possibly zero), and it is an odd permutation if the number of cycles of even length is odd. - Peter Bala, Jun 25 2024

			Triangle begins:
  1;
  0,     1;
  0,     0,       1;
  0,     2,       0,      1;
  0,     0,      11,      0,      1;
  0,    24,       0,     35,      0,     1;
  0,     0,     274,      0,     85,     0,     1;
  0,   720,       0,   1624,      0,   175,     0,   1;
  0,     0,   13068,      0,   6769,     0,   322,   0,   1;
  0, 40320,       0, 118124,      0, 22449,     0, 546,   0, 1;
  0,     0, 1026576,      0, 723680,     0, 63273,   0, 870, 0, 1;
  ...

J. Riordan, Introduction to Combinatorial Analysis, Wiley, 1958, page 87, problem # 20.

Alois P. Heinz, Rows n = 0..140, flattened
Wikipedia, Parity of a permutaion

Cf. A008275, A130534, A132393, A164652.

with(combinat):
b:= proc(n, i, t) option remember; expand(`if`(n=0, t, `if`(i<1,
      0, add(x^j*multinomial(n, n-i*j, i$j)*(i-1)!^j/j!*b(n-i*j,
      i-1, irem(t+`if`(irem(i,2)=0, j, 0), 2)), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2, 1)):
seq(T(n), n=0..12);  # Alois P. Heinz, Mar 09 2015
# Alternative:
A132393 := (n, k) -> abs(Stirling1(n, k)):
T := (n, k) -> ifelse((n::even and k::even) or (n::odd and k::odd), A132393(n, k),
0): seq(seq(T(n, k), k = 0..n), n = 0..9);  # Peter Luschny, Jun 26 2024

nn=11;a=Log[((1+x)/(1-x))^(1/2)];b=Log[1/(1-x^2)^(1/2)];Table[Take[(Range[0,nn]!CoefficientList[Series[Exp[y a]Cosh[y b] ,{x,0,nn}],{x,y}])[[n]],n],{n,1,nn}]//Grid

E.g.f.: exp(y*A(x))*cosh(y*B(x)) where A(x) = log(((1 + x)/(1 - x))^(1/2)) and B(x) = log((1/(1-x^2))^(1/2)).
From Peter Bala, Jun 25 2024: (Start)
If n and k are both even or both odd, then T(n, k) is equal to the Stirling cycle number |s(n, k)| = A132393(n, k), and 0 otherwise.
n-th row polynomial R(n, x) = (1/2)*( x*(x + 1)*...*(x + n + 1) + x*(x - 1)*...*(x - n - 1) ).
For n >= 1, the zeros of R(n, x) are purely imaginary. (End)

A371665 T(n,k) is the number of reduced unicellular hypermonopoles on n points with k hyperedges, where T(n,k), 1 <= k <= floor(n/2), is an array read by rows.

Original entry on oeis.org

1, 0, 1, 8, 0, 0, 36, 0, 180, 0, 49, 0, 1604, 0, 21, 8064, 0, 5144, 0, 0, 112608, 0, 7680, 0, 604800, 0, 604428, 0, 5445, 0, 11799360, 0, 1669052, 0, 1485, 68428800, 0, 91705536, 0, 2610608, 0, 0, 1741669632, 0, 384036016, 0, 2342340, 0, 10897286400, 0, 18071744976, 0, 972895560, 0, 1126125

Offset: 3

Gabor Hetyei, Apr 02 2024

T(n,k) is zero unless k <= n/2. (proven to be correct)

			The table begins:
         1;
         0,          1;
         8,          0;
         0,         36,        0;
       180,          0,       49;
         0,       1604,        0,        21;
      8064,          0,     5144,         0;
         0,     112608,        0,      7680,       0;
    604800,          0,   604428,         0,    5445;
         0,   11799360,        0,   1669052,       0,    1485;
  68428800,          0, 91705536,         0, 2610608,       0;
         0, 1741669632,        0, 384036016,       0, 2342340, 0;

Robert Cori and Gábor Hetyei, On reduced unicellular hypermonopoles, arXiv:2403.19569 [math.CO], 2024.

Cf. A060593, A164652.

proc(n, k)
    local i;
    coeff(expand(add(combinat:-binomial(n, i)*(-x)^i*(pochhammer(x, n - i + 1) - pochhammer(x - n + i, n - i + 1))/((n - i)*(n - i + 1)), i = 0 .. n - 1)), x, k);
end proc

T(n,k) = Sum_{i=0..k-1} (-1)^i binomial(n,i)*a(n-1-i,k-i) where the a(n,k) are the Hultman numbers from A164652.

T(2*m+1,1) = (2*m)! / (m+1) = A060593(m) for m >= 1.

cp's OEIS Frontend

A363041 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).

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A237996 Triangular array read by rows. T(n,k) is the number of even permutations of {1,2,...,n} that have exactly k cycles, n >= 0, 0 <= k <= n.

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A371665 T(n,k) is the number of reduced unicellular hypermonopoles on n points with k hyperedges, where T(n,k), 1 <= k <= floor(n/2), is an array read by rows.

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