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Triangle T(n,k)=Sum_{i=0..n} |stirling1(n,n-i)|*binomial(i,k), k=0..n-1.
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%I #25 Jan 14 2024 12:35:27

%S 1,2,1,6,7,2,24,46,29,6,120,326,329,146,24,720,2556,3604,2521,874,120,

%T 5040,22212,40564,39271,21244,6084,720,40320,212976,479996,598116,

%U 444849,197380,48348,5040,362880,2239344,6023772,9223012,8788569

%N Triangle T(n,k)=Sum_{i=0..n} |stirling1(n,n-i)|*binomial(i,k), k=0..n-1.

%C Sum_{k=0..n-1} T(n,k)=(2*n-1)!!.

%C Alternating row sums = 1. - _Gerald McGarvey_, Aug 06 2006

%C Essentially triangle given by [1,1,2,2,3,3,4,4,5,5,6,6,...] DELTA [0,1,1,2,2,3,3,4,4,5,5,...] = [1;1,0;2,1,0;6,7,2,0;24,46,29,6,0;...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Nov 20 2006

%H G. C. Greubel, <a href="/A059364/b059364.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H Grzegorz Rzadkowski and M. Urlinska, <a href="http://arxiv.org/abs/1612.06635">A Generalization of the Eulerian Numbers</a>, arXiv preprint arXiv:1612.06635 [math.CO], 2016-2017.

%F For n>1, T(n,k) = (n-1)*T(n-1,k-1) + n*T(n-1,k) (assuming any T(i,j) outside the triangle = 0). - _Gerald McGarvey_, Aug 06 2006

%e [1],

%e [2, 1],

%e [6, 7, 2],

%e [24, 46, 29, 6],

%e [120, 326, 329, 146, 24],

%e [720, 2556, 3604, 2521, 874, 120], ...

%e 2+1=3!!, 6+7+2=5!!, 24+46+29+6=7!!, 120+326+329+146+24=9!!.

%t Table[Sum[Abs[StirlingS1[n, n - j]]*Binomial[j, k], {j, 0, n}], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* _G. C. Greubel_, Jan 08 2017 *)

%o (PARI) T(n,k)=if(n<1,0,n!*polcoeff(polcoeff((1-x-x*y+x*O(x^n))^(-1/(1+y)),n),k))

%o (Sage)

%o def A059364(n,k): return add(stirling_number1(n,n-i)*binomial(i,k) for i in (0..n))

%o for n in (1..5): [A059364(n,k) for k in (0..n-1)] # _Peter Luschny_, May 12 2013

%Y Cf. A001147, A059340.

%K easy,nonn,tabl

%O 1,2

%A _Vladeta Jovovic_, Jan 28 2001