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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!.
5

%I #38 Sep 08 2022 08:45:46

%S 1,1,2,2,4,6,6,12,18,24,24,48,72,96,120,120,240,360,480,600,720,720,

%T 1440,2160,2880,3600,4320,5040,5040,10080,15120,20160,25200,30240,

%U 35280,40320,40320,80640,120960,161280,201600,241920,282240,322560,362880

%N 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!.

%C Note that the last term of the n-th row is the factorial of (n+1) = (n+1)! = A000142(n+1).

%C 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

%C From _Dennis P. Walsh_, May 18 2020: (Start)

%C 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.

%C 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).

%C 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)

%H Reinhard Zumkeller, <a href="/A162608/b162608.txt">Rows n=0..150 of triangle, flattened</a>

%F From _Robert Israel_, Jul 03 2015: (Start)

%F T(n,k) = n!*k, k = 1 .. n+1.

%F T(n+1,k) = (n+1)*T(n,k).

%F T(n,k+1) = T(n,k)+T(n,1). (End)

%e Triangle begins:

%e 1;

%e 1, 2;

%e 2, 4, 6;

%e 6, 12, 18, 24;

%e 24, 48, 72, 96, 120;

%e 120, 240, 360, 480, 600, 720;

%e 720, 1440, 2160, 2880, 3600, 4320, 5040;

%e 5040, 10080, 15120, 20160, 25200, 30240, 35280, 40320;

%e 40320, 80640, 120960, 161280, 201600, 241920, 282240, 322560, 362880;

%e 362880,725760,1088640,1451520,1814400,2177280,2540160,2903040,3265920,3628800;

%e ...

%e Observation: It appears that rows sums = A001710(n+2).

%t Table[k n!, {n, 0, 8}, {k, n + 1}] // Flatten (* _Michael De Vlieger_, Jul 03 2015 *)

%o (Haskell)

%o a162608 n k = a162608_tabl !! n !! k

%o a162608_row n = a162608_tabl !! n

%o a162608_tabl = map fst $ iterate f ([1], 1) where

%o f (row, n) = (row' ++ [head row' + last row'], n + 1) where

%o row' = map (* n) row

%o -- _Reinhard Zumkeller_, Mar 09 2012

%o (Magma) /* As triangle */ [[Factorial(n)*k: k in [1..n+1]]: n in [0.. 15]]; // _Vincenzo Librandi_, Jul 04 2015

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

%Y Cf. A178883. - _Alford Arnold_, Sep 28 2010

%K easy,nonn,tabl

%O 0,3

%A _Omar E. Pol_, Jul 22 2009