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%I #40 Nov 16 2018 04:53:43
%S 1,1,1,4,5,1,28,39,12,1,280,418,159,22,1,3640,5714,2485,445,35,1,
%T 58240,95064,45474,9605,1005,51,1,1106560,1864456,959070,227969,28700,
%U 1974,70,1,24344320,42124592,22963996,5974388,859369,72128,3514,92,1,608608000,1077459120,616224492,172323696,27458613,2662569,159978,5814,117,1,17041024000,30777463360,18331744896,5441287980,941164860,102010545,7141953,322770,9090,145,1
%N Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.
%C This is a generalization of the unsigned Stirling1 triangle A132393.
%C In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
%C In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
%C In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
%C From _Wolfdieter Lang_, Aug 09 2017: (Start)
%C The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
%C (E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/A002209(k+1).
%C This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
%C The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - _Wolfdieter Lang_, Nov 15 2018
%D Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
%D Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
%H P. Bala, <a href="/A143395/a143395.pdf">A 3 parameter family of generalized Stirling numbers</a>
%H Wolfdieter Lang, <a href="http://arXiv.org/abs/1707.04451">On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers</a>, arXiv:math/1707.04451 [math.NT], July 2017.
%H Wolfdieter Lang, <a href="https://arxiv.org/abs/1708.01421">On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles</a>, arXiv:1708.01421 [math.NT], August 2017.
%H Peter Luschny, <a href="http://www.luschny.de/math/euler/StirlingFrobeniusNumbers.html">The Stirling-Frobenius numbers.</a>
%F Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
%F E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle) is (1 - 3*z)^{-(x+1)/3}.
%F E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
%F Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
%F Row polynomial R(n, x) = risefac(3,1;x,n) with the rising factorial
%F risefac(d,a;x,n) := Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
%F T(n, k) = sigma^{(n)}_{n-k}(a_0,a_1,...,a_{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 3*j.
%F T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393.
%F Boas-Buck column recurrence (see a comment above): T(n, k) =
%F (n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/A002209(k+1). See an example below. - _Wolfdieter Lang_, Aug 09 2017
%e The triangle T(n, k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 ...
%e O: 1
%e 1: 1 1
%e 2: 4 5 1
%e 3: 28 39 12 1
%e 4: 280 418 159 22 1
%e 5: 3640 5714 2485 445 35 1
%e 6: 58240 95064 45474 9605 1005 51 1
%e 7: 1106560 1864456 959070 227969 28700 1974 70 1
%e 8: 24344320 42124592 22963996 5974388 859369 72128 3514 92 1
%e ...
%e From _Wolfdieter Lang_, Aug 09 2017: (Start)
%e Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
%e Boas-Buck recurrence for column k = 2 and n = 5:
%e T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
%e The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
%e (End)
%t T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[_, -1] = 0; T[0, 0] = 1; T[n_, k_] /; n<k = 0;
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jun 20 2018 *)
%Y S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
%Y S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, respectively.
%Y |S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
%Y Column sequences for k = 0, 1: A007559, A024216.
%Y Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
%Y Row sums: A008544. Alternating row sums: A000007.
%Y Beta sequence: A002208(n+1)/A002209(n+1).
%K nonn,easy,tabl
%O 0,4
%A _Wolfdieter Lang_, May 18 2017
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