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%I #27 Oct 08 2017 23:46:42
%S 1,1,1,0,1,1,0,-1,0,1,0,2,-1,-2,1,0,-6,5,5,-5,1,0,24,-26,-15,25,-9,1,
%T 0,-120,154,49,-140,70,-14,1,0,720,-1044,-140,889,-560,154,-20,1,0,
%U -5040,8028,-64,-6363,4809,-1638,294,-27,1,0,40320,-69264,8540,50840,-44835,17913,-3990,510,-35,1
%N Expansion of (1 + y)^(1 + x).
%C Generalized Stirling number triangle of first kind. Row sums are (1,2,2,0,0,0,...) = 2C(2,n) - 2C(1,n) + C(0,n). Inverse is A105794.
%C Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, -1, 0, -2, -1, -3, -2, -4, -3, -5, -4, -6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Aug 23 2006
%F E.g.f.: (1+y)^(1+x); rows have g.f. k!binomial(x+1, k); Columns have g.f. (1+x)log(1+x)^k.
%F If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then T(n,i) = f(n,i,-1), for n=1,2,...; i=0...n. - _Milan Janjic_, Dec 21 2008
%F So T(n,k) = Stirling1(n,k) + n*Stirling1(n-1,k), Stirling1 being the (signed) Stirling numbers of first kind A048994. In terms of lower triangular matrices, 0<= k <= n, T is also the product [Stirling1] * [Pascal] = [A048994] * [A007318], i.e., T(n,k) = Sum_{j=0..n} Stirling1(n,j) * binomial(j,k). - _Giuliano Cabrele_, Jan 19 2009
%F This is the triangle of connection constants for expressing the basis of falling factorial polynomials x_(k) := x*(x-1)*...*(x-k+1) in terms of the polynomial sequence (x-1)^n, that is, x_(n) = Sum_{k = 0..n} T(n,k)*(x-1)^k. - _Peter Bala_, Jul 10 2013
%F From _Wolfdieter Lang_, Jun 19 2017: (Start)
%F Triangle T is the (infinite) matrix product of A048994 (Stirling1) and A007318 (Pascal): T(n,k) = Sum_{m=k..n} Stirling1(n, m)*Pascal(m, k), n >= k >= 0, and 0 for n < k. Note that the Pascal matrix is Sheffer (e^t, t) of the Appell type.
%F T is the Sheffer (aka exponential Riordan) matrix (1+t, log(1+t)).
%F E.g.f. column k: (1+x)*(log(1+x))^k/k!, k >= 0.
%F The a-sequence for T is A027641/A027642 (Bernoulli), and the z-sequence is A033999 (repeat(1,-1) (see a W. Lang link under A006232 for a- and z-sequences for Sheffer matrices, also for references).
%F Therefore the combined recurrence is: T(n, 0) = n*Sum_{j=0..n-1} (-1)^j*T(n-1, j), n >= 1, T(0, 0) = 1, and T(n, m) = (n/m)*Sum_{j=0..n-m} binomial(m-1+j, m-1)*Bernoulli(j)*T(n-1, m-1+j), n >= m >= 1. (End)
%e From _Wolfdieter Lang_, Jun 19 2017: (Start)
%e The Triangle T(n, k) starts
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e 0: 1
%e 1: 1 1
%e 2: 0 1 1
%e 3: 0 -1 0 1
%e 4: 0 2 -1 -2 1
%e 5: 0 -6 5 5 -5 1
%e 6: 0 24 -26 -15 25 -9 1
%e 7: 0 -120 154 49 -140 70 -14 1
%e 8: 0 720 -1044 -140 889 -560 154 -20 1
%e 9: 0 -5040 8028 -64 -6363 4809 -1638 294 -27 1
%e 10: 0 40320 -69264 8540 50840 -44835 17913 -3990 510 -35 1
%e ... reformatted
%e Recurrence from a- and z-sequence (see above): T(1, 0) = T(0, 0) = 1; T(1, 1) = (1/1)*(1*T(0, 0) = 1, T(2, 0) = 2*(T(1, 0) - T(1, 1)) = 0, T(2, 1) = (2/1)*(T(1,0) + (-1/2)*T(1, 1)) = 1. T(3, 1) = (3/1)*(0 + (-1/2)*T(2, 1) + (1/6)*T(2, 2)) = -1. (End)
%t t[0, 0] = 1; t[n_, k_] := StirlingS1[n, k] + n*StirlingS1[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 04 2013, after _Giuliano Cabrele_ *)
%Y Cf. A007318, A027641/A027642, A033999, A048994, A094645, A269953.
%K easy,sign,tabl
%O 0,12
%A _Paul Barry_, Apr 20 2005
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