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%I #45 Jun 17 2023 13:19:38
%S 1,1,0,1,-1,0,1,-3,2,0,1,-6,11,-6,0,1,-10,35,-50,24,0,1,-15,85,-225,
%T 274,-120,0,1,-21,175,-735,1624,-1764,720,0,1,-28,322,-1960,6769,
%U -13132,13068,-5040,0
%N Triangle of Stirling numbers of 1st kind, S(n, n-k), n >= 0, 0 <= k <= n.
%C Triangle is the matrix product of the binomial coefficients with the Stirling numbers of the first kind.
%C Triangle T(n,k) giving coefficients in expansion of n!*C(x,n) in powers of x. E.g., 3!*C(x,3) = x^3-3*x^2+2*x.
%C The matrix product of binomial coefficients with the Stirling numbers of the first kind results in the Stirling numbers of the first kind again, but the triangle is shifted by (1,1).
%C Essentially [1,0,1,0,1,0,1,0,...] DELTA [0,-1,-1,-2,-2,-3,-3,-4,-4,...] where DELTA is the operator defined in A084938; mirror image of the Stirling-1 triangle A048994. - _Philippe Deléham_, Dec 30 2006
%C From _Doudou Kisabaka_, Dec 18 2009: (Start)
%C The sum of the entries on each row of the triangle, starting on the 3rd row, equals 0. E.g., 1+(-3)+2+0 = 0.
%C The entries on the triangle can be computed as follows. T(n,r) = T(n-1,r) - (n-1)*T(n-1,r-1). T(n,r) = 0 when r equals 0 or r > n. T(n,r) = 1 if n==1. (End)
%H Reinhard Zumkeller, <a href="/A054654/b054654.txt">Rows n = 0..125 of triangle, flattened</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PochhammerSymbol.html">Pochhammer Symbol</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/RisingFactorial.html">Rising Factorial</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FallingFactorial.html">FallingFactorial</a>
%F n!*binomial(x, n) = Sum_{k=0..n} T(n, k)*x^(n-k).
%F (In Maple notation:) Matrix product A*B of matrix A[i,j]:=binomial(j-1,i-1) with i = 1 to p+1, j = 1 to p+1, p=8 and of matrix B[i,j]:=stirling1(j,i) with i from 1 to d, j from 1 to d, d=9.
%F T(n, k) = (-1)^k*Sum_{j=0..k} E2(k, j)*binomial(n+j-1, 2*k), where E2(k, j) are the second-order Eulerian numbers A340556. - _Peter Luschny_, Feb 21 2021
%e Matrix begins:
%e 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
%e 0, 1, -1, 2, -6, 24, -120, 720, -5040, ...
%e 0, 0, 1, -3, 11, -50, 274, -1764, 13068, ...
%e 0, 0, 0, 1, -6, 35, -225, 1624, -13132, ...
%e 0, 0, 0, 0, 1, -10, 85, -735, 6769, ...
%e 0, 0, 0, 0, 0, 1, -15, 175, -1960, ...
%e 0, 0, 0, 0, 0, 0, 1, -21, 322, ...
%e 0, 0, 0, 0, 0, 0, 0, 1, -28, ...
%e 0, 0, 0, 0, 0, 0, 0, 0, 1, ...
%e ...
%e Triangle begins:
%e 1;
%e 1, 0;
%e 1, -1, 0;
%e 1, -3, 2, 0;
%e 1, -6, 11, -6, 0;
%e 1, -10, 35, -50, 24, 0;
%e 1, -15, 85, -225, 274, -120, 0;
%e 1, -21, 175, -735, 1624, -1764, 720, 0;
%e ...
%p a054654_row := proc(n) local k; seq(coeff(expand((-1)^n*pochhammer (-x,n)),x,n-k),k=0..n) end: # _Peter Luschny_, Nov 28 2010
%p seq(seq(Stirling1(n, n-k), k=0..n), n=0..8); # _Peter Luschny_, Feb 21 2021
%t row[n_] := Reverse[ CoefficientList[ (-1)^n*Pochhammer[-x, n], x] ]; Flatten[ Table[ row[n], {n, 0, 8}]] (* _Jean-François Alcover_, Feb 16 2012, after Maple *)
%t Table[StirlingS1[n,n-k],{n,0,10},{k,0,n}]//Flatten (* _Harvey P. Dale_, Jun 17 2023 *)
%o (PARI) T(n,k)=polcoeff(n!*binomial(x,n), n-k)
%o (Haskell)
%o a054654 n k = a054654_tabl !! n !! k
%o a054654_row n = a054654_tabl !! n
%o a054654_tabl = map reverse a048994_tabl
%o -- _Reinhard Zumkeller_, Mar 18 2014
%Y Essentially Stirling numbers of first kind, multiplied by factorials - see A008276.
%Y The Stirling2 counterpart is A106800.
%Y Cf. A054655, A039810, A039814, A126350, A126351, A126353, A340556.
%K tabl,sign,easy,nice
%O 0,8
%A _N. J. A. Sloane_, Apr 18 2000
%E Additional comments from _Thomas Wieder_, Dec 29 2006
%E Edited by _N. J. A. Sloane_ at the suggestion of _Eric W. Weisstein_, Jan 20 2008
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