%I
%S 1,1,1,5,6,1,45,59,15,1,585,812,254,28,1,9945,14389,5130,730,45,1,
%T 208845,312114,122119,20460,1675,66,1,5221125,8011695,3365089,633619,
%U 62335,3325,91,1,151412625,237560280,105599276,21740040,2441334,158760,5964,120,1,4996616625,7990901865,3722336388,823020596,102304062,7680414,355572,9924,153,1,184874815125,300659985630,145717348221,34174098440,4608270890,386479380,20836578,722760,15585,190,1
%N Triangle read by rows: T(n, k)is the Sheffer triangle ((1  4*x)^(1/4), (1/4)*log(1  4*x)). A generalized Stirling1 triangle.
%C This generalization of the unsigned Stirling1 triangle A132393 is called here S1hat[4,1].
%C The signed matrix S1hat[4,1] with elements (1)^(nk)*S1hat[4,1](n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[4,1] with elements S2[4,1](n, k)/d^k, where S2[4,1] is Sheffer (exp(x), exp(4*x)  1), given in A285061. See also the P. Bala link below for the scaled and signed version s_{(4,0,1)}.
%C For the general S1hat[d,a] case see a comment in A286718.
%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.
%F Recurrence: T(n, k) = T(n1, k1) + (4*n  3)*T(n1, 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): (1  4*z)^{(x + 1)/4}.
%F E.g.f. of column k is (1  4*x)^(1/4)*((1/4)*log(1  4*x))^k/k!.
%F Recurrence for row polynomials is R(n, x) = (x+1)*R(n1, x+4), with R(0, x) = 1. Row polynomial R(n, x) = risefac(4,1;x,n) with the rising factorial risefac(d,a;x,n) :=Product_{j=0..n1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
%F T(n, k) = sigma^{(n)}_{nk}(a_0, a_1, ..., a_{n1}) with the elementary symmetric functions with indeterminates a_j = 1 + 4*j.
%F T(n, k) = Sum_{j=0..nk} binomial(nj, k)*S1(n, nj)*4^j, with the unsigned Stirling1 triangle S1 = A132393.
%F BoasBuck type recurrence for column sequence k: T(n, k) = (n!/(n  k)) * Sum_{p=k..n1} 4^(n1p)*(1 + 4*k*beta(n1p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1), beginning with {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718.  _Wolfdieter Lang_, Aug 11 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: 5 6 1
%e 3: 45 59 15 1
%e 4: 585 812 254 28 1
%e 5: 9945 14389 5130 730 45 1
%e 6: 208845 312114 122119 20460 1675 66 1
%e 7: 5221125 8011695 3365089 633619 62335 3325 91 1
%e 8: 151412625 237560280 105599276 21740040 2441334 158760 5964 120 1
%e ...
%e From _Wolfdieter Lang_, Aug 11 2017: (Start)
%e Recurrence: T(4, 2) = T(3, 1) + (16  3)*T(3, 2) = 59 + 13*15 = 254.
%e BoasBuck recurrence for column k=2 and n=4:
%e T(4, 2) = (4!/2)*(4*(1 + 8*(5/12))*T(2, 2)/2! + 1*(1 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(2*13/3 + 5*15/3!) = 254.  _Wolfdieter Lang_, Aug 11 2017
%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 S1hat[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2] and [4,3] is A132393, A028338, A286718, A225470 and A225471, respectively.
%Y Column sequences for k = 0, 1: A007696, A024382.
%Y Row sums: A001813. Alternating row sums: A000007.
%K nonn,easy,tabl
%O 0,4
%A _Wolfdieter Lang_, Aug 08 2017
