A362041 -id:A362041 - 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.

A362844 a(n) is the largest k < A360768(n) such that rad(k) = rad(A360768(n)) and n mod k != 0, where rad(n) = A007947(n).

Original entry on oeis.org

12, 18, 24, 36, 40, 48, 54, 45, 50, 60, 72, 56, 80, 96, 98, 90, 84, 75, 108, 63, 120, 100, 144, 126, 150, 147, 162, 112, 132, 160, 192, 196, 135, 156, 180, 176, 175, 200, 168, 198, 240, 216, 252, 270, 204, 234, 250, 288, 294, 208, 228, 280, 242, 300, 297, 225, 336, 324, 224, 264, 320, 375, 306, 276

Offset: 1

Michael De Vlieger, May 19 2023

nonn

Largest nondivisor less than m = A360768(n) that shares the same squarefree kernel as m.

a(n) is in A126706, not a permutation of A126706.

			A360768(1) = 18; the smallest nondivisor k < 18 such that rad(k) = rad(18) = 6 is a(1) = 12.
A360768(2) = 24; the smallest nondivisor k < 24 such that rad(k) = rad(24) = 6 is a(2) = 18.
A360768(5) = 50; the smallest nondivisor k < 50 such that rad(k) = rad(50) = 10 is a(5) = 40.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^12, showing records in red.

Cf. A007947, A126706, A360768, A362041.

rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]]; s = Select[Select[Range[414],  Nor[SquareFreeQ[#], PrimePowerQ[#]] &], #1/#2 >= #3 & @@ {#1, Times @@ #2, #2[[2]]} & @@ {#, FactorInteger[#][[All, 1]]} &]; Table[Function[r, SelectFirst[Range[m - 1, 1, -1], r == rad[#] &] ][rad[m]], {m, s}]

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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