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A094645 Triangle of generalized Stirling numbers of the first kind. 11

%I #59 Mar 24 2024 13:07:27

%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,-5040,

%U -8028,-64,6363,4809,1638,294,27,1,0,-40320,-69264,-8540,50840,44835,17913,3990,510,35,1

%N Triangle of generalized Stirling numbers of the first kind.

%C From _Wolfdieter Lang_, Jun 20 2011: (Start)

%C The row polynomials s(n,x) := Sum_{j=0..n} T(n,k)*x^k satisfy risefac(x-1,n) = s(n,x), with the rising factorials risefac(x-1,n) := Product_{j=0..n-1} (x-1+j), n >= 1, risefac(x-1,0) = 1. Compare with the formula risefac(x,n) = s1(n,x), with the row polynomials s1(n,x) of A132393 (unsigned Stirling1).

%C This is the lower triangular Sheffer array with e.g.f.

%C T(x,z) = (1-z)*exp(-x*log(1-z)) (the rewritten e.g.f. from the formula section). See the W. Lang link under A006232 for Sheffer matrices and the Roman reference. In the notation which indicates the column e.g.f.s this is Sheffer (1-z,-log(1-z)). In the umbral notation (cf. Roman) this is called Sheffer for (exp(t),1-exp(-t)).

%C The row polynomials satisfy s(n,x) = (x+n-1)*s(n-1,x), s(0,x)=1, and s(n,x) = (x-1)*s1(n-1,x), n >= 1, s1(0,x) = 1, with the unsigned Stirling1 row polynomials s1(n,x).

%C The row polynomials also satisfy

%C s(n,x) - s(n,x-1) = n*s(n-1,x), n > 1, s(0,x) = 1

%C (from the Meixner identity, see the Meixner reference given at A060338).

%C The row polynomials satisfy as well (from corollary 3.7.2. p. 50 of the Roman reference)

%C s(n,x) = (x-1)*s(n-1,x+1), n >= 1, s(0,n) = 1.

%C The exponential convolution identity is

%C s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,y)*s1(n-k,x),

%C n >= 0, with symmetry x <-> y.

%C The row sums are 1 for n=0 and 0 otherwise, and the alternating row sums are 1,-2,2, followed by zeros, with e.g.f. (1-x)^2.

%C The Sheffer a-sequence Sha(n) = A164555(n)/A027642(n) with e.g.f. x/(1-exp(-x)), and the z-sequence is Shz(n) = -1 with e.g.f. -exp(x).

%C The inverse Sheffer matrix is ((-1)^(n-k))*A105794(n,k) with e.g.f. exp(z)*exp(x*(1-exp(-z))). (End)

%C Triangle T(n,k), read by rows, given by (-1, 1, 0, 2, 1, 3, 2, 4, 3, 5, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Jan 16 2012

%C Also coefficients of t in t*(t-1)*Sum[(-1)^(n+m) t^(m-1) StirlingS1[n,m], {m,n}] in which setting t^k equal to k gives n!, from this follows that the dot product of row n with [0,...,n] equals (n-1)!. - _Wouter Meeussen_, May 15 2012

%D S. Roman, The Umbral Calculus, Academic Press, New York, 1984.

%H M. W. Coffey and M. C. Lettington, <a href="http://arxiv.org/abs/1510.05402">On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat</a>, arXiv:1510.05402 [math.NT], 2015.

%H Igor Victorovich Statsenko, <a href="https://aeterna-ufa.ru/sbornik/IN-2024-02-2.pdf#page=15">On the ordinal numbers of triangles of generalized special numbers</a>, Innovation science No 2-2, State Ufa, Aeterna Publishing House, 2024, pp. 15-19. In Russian.

%F E.g.f.: (1-y)^(1-x).

%F Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 respectively. - _Philippe Deléham_, Nov 13 2007

%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 From _Wolfdieter Lang_, Jun 20 2011: (Start)

%F T(n,k) = |S1(n-1,k-1)| - |S1(n-1,k)|, n >= 1, k >= 1, with |S1(n,k)| = A132393(n,k) (unsigned Stirling1).

%F Recurrence: T(n,k) = T(n-1,k-1) + (n-2)*T(n-1,k) if n >= k >= 0; T(n,k) = 0 if n < k; T(n,-1) = 0; T(0,0) = 1.

%F E.g.f. column k: (1-x)*((-log(1-x))^k)/k!. (End)

%F T(n,k) = Sum_{i=0..n} binomial(n,i)*(n-i)!*Stirling1(i,k)*TC(m,n,i) where TC(m,n,k) = Sum_{i=0..n-k} binomial(n+1,n-k-i)*Stirling2(i+m+1,i+1)*(-1)^i, m = 1 for n >= 0. See A130534, A370518 for m=0 and m=2. - _Igor Victorovich Statsenko_, Feb 27 2024

%e Triangle begins

%e 1;

%e -1, 1;

%e 0, -1, 1;

%e 0, -1, 0, 1;

%e 0, -2, -1, 2, 1;

%e 0, -6, -5, 5, 5, 1;

%e 0, -24, -26, 15, 25, 9, 1;

%e ...

%e Recurrence:

%e -2 = T(4,1) = T(3,0) + (4-2)*T(3,1) = 0 + 2*(-1).

%e Row polynomials:

%e s(3,x) = -x+x^3 = (x-1)*s1(2,x) = (x-1)*(x+x^2).

%e s(3,x) = (x-1)*s(2,x+1) = (x-1)*(-(x+1)+(x+1)^2).

%e s(3,x) - s(3,x-1) = -x+x^3 -(-(x-1)+(x-1)^3) = 3*(-x+x^2) = 3*s(2,x).

%p A094645_row := n -> seq((-1)^(n-k)*coeff(expand(pochhammer(x-n+2, n)), x, k), k=0..n): seq(print(A094645_row(n)), n=0..6); # _Peter Luschny_, May 16 2013

%t t[n_, k_] /; n >= k >= 0 := t[n, k] = t[n-1, k-1] + (n-2)*t[n-1, k]; t[n_, k_] /; n < k = 0; t[_, -1] = 0; t[0, 0] = 1; Flatten[ Table[ t[n, k], {n, 0, 10}, {k, 0, n}] ] (* _Jean-François Alcover_, Sep 29 2011, after recurrence *);

%t Table[CoefficientList[t*(t-1)*Sum[(-1)^(n+m)*t^(m-1)*StirlingS1[n,m],{m,n}],t],{n,1,7}] (* _Wouter Meeussen_, May 15 2012 *)

%Y Cf. A049444, A049458, A094646, A132393, A105794.

%K easy,sign,tabl,changed

%O 0,12

%A _Vladeta Jovovic_, May 17 2004

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Last modified March 28 18:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)