A178883 -id:A178883 - cp's OEIS frontend

A181416 Irregular table T(n,k) = n*A178883(n,k) read by rows.

1, 2, 4, 6, 12, 18, 24, 32, 16, 72, 96, 120, 120, 120, 180, 180, 480, 600, 720, 576, 576, 288, 648, 1296, 216, 1152, 1728, 3600, 4320, 5040, 3360, 3360, 3360, 3024, 6048, 3024, 3024, 4032, 12096, 4032, 8400, 16800, 30240, 35280, 40320

Offset: 1

Alford Arnold, Oct 17 2010

The row sum of row n is A001286(n).

			In row n=3 the products are (3,3,3) times (2,4,6) yielding (6,12,18) which adds to 36, the third Lah number.
The table starts in row n=1 with row lengths A000041(n) as:
1;
2,4;
6,12,18;
24,32,16,72,96;
120,120,120,180,180,480,600;

Cf. A051683.

T(n,k) = A036042(n,k)*A178883(n,k), 1<=k<= A000041(n).

A162608 Triangle read by rows in which row n lists n+1 terms, starting with n!, such that the difference between successive terms is also equal to n!.

Original entry on oeis.org

1, 1, 2, 2, 4, 6, 6, 12, 18, 24, 24, 48, 72, 96, 120, 120, 240, 360, 480, 600, 720, 720, 1440, 2160, 2880, 3600, 4320, 5040, 5040, 10080, 15120, 20160, 25200, 30240, 35280, 40320, 40320, 80640, 120960, 161280, 201600, 241920, 282240, 322560, 362880

Offset: 0

Omar E. Pol, Jul 22 2009

Note that the last term of the n-th row is the factorial of (n+1) = (n+1)! = A000142(n+1).
Sequence A178883 (with shape A000041) is a "refinement" of Table A162608; as expected, both sequences have row sums A001710(n+2). - Alford Arnold, Sep 28 2010
From Dennis P. Walsh, May 18 2020: (Start)
T(n,k) provides the number of length (n+2) permutations with elements 1 and 2 as cycle-mates in a (k+1)-cycle. We note that 1 and 2 are cycle-mates if they are elements of the same cycle in the permutation.
For example, T(3,2) counts the 12 permutations of length 5 that have 1 and 2 in the same 3 cycle, namely, (1 2 3)(4)(5), (1 3 2)(4)(5), (1 2 3)(4 5), (1 3 2)(4 5), (1 2 4)(3)(5), (1 4 2)(3)(5), (1 2 4)(3 5), (1 4 2)(3 5),(1 2 5)(3)(4), (1 5 2)(3)(4), (1 2 5)(3 4), and (1 5 2)(3 4).
Note that there are binomial(n,k-1) ways to choose the other (k-1) cycle-mates of 1 and 2 in the (k+1)-cycle and then k! different (k+1)-cycles with these elements. Since there are (n+1-k)! ways to permute the remaining elements, we obtain T(n,k) = (n+1-k)!*k!*binomial(n,k-1) = n!*k. (End)

			Triangle begins:
1;
1,     2;
2,     4,     6;
6,     12,    18,     24;
24,    48,    72,     96,     120;
120,   240,   360,    480,    600,    720;
720,   1440,  2160,   2880,   3600,   4320,   5040;
5040,  10080, 15120,  20160,  25200,  30240,  35280,  40320;
40320, 80640, 120960, 161280, 201600, 241920, 282240, 322560, 362880;
362880,725760,1088640,1451520,1814400,2177280,2540160,2903040,3265920,3628800;
...
Observation: It appears that rows sums = A001710(n+2).

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened

Cf. A000142, A001710, A051683, A159797, A162611, A162614, A162622.

Cf. A178883. - Alford Arnold, Sep 28 2010

a162608 n k = a162608_tabl !! n !! k
a162608_row n = a162608_tabl !! n
a162608_tabl = map fst $ iterate f ([1], 1) where
   f (row, n) = (row' ++ [head row' + last row'], n + 1) where
     row' = map (* n) row
-- Reinhard Zumkeller, Mar 09 2012

/* As triangle */ [[Factorial(n)*k: k in [1..n+1]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 04 2015

Table[k n!, {n, 0, 8}, {k, n + 1}] // Flatten (* Michael De Vlieger, Jul 03 2015 *)

From Robert Israel, Jul 03 2015: (Start)
T(n,k) = n!*k, k = 1 .. n+1.
T(n+1,k) = (n+1)*T(n,k).
T(n,k+1) = T(n,k)+T(n,1). (End)

cp's OEIS Frontend

A181416 Irregular table T(n,k) = n*A178883(n,k) read by rows.

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A162608 Triangle read by rows in which row n lists n+1 terms, starting with n!, such that the difference between successive terms is also equal to n!.

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