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 A290319 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. 1
 1, 1, 1, 5, 6, 1, 45, 59, 15, 1, 585, 812, 254, 28, 1, 9945, 14389, 5130, 730, 45, 1, 208845, 312114, 122119, 20460, 1675, 66, 1, 5221125, 8011695, 3365089, 633619, 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS This generalization of the unsigned Stirling1 triangle A132393 is called here |S1hat[4,1]|. The signed matrix S1hat[4,1] with elements (-1)^(n-k)*|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)}. For the general |S1hat[d,a]| case see a comment in A286718. LINKS Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017. FORMULA Recurrence: T(n, k) = T(n-1, k-1) + (4*n - 3)*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. 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}. E.g.f. of column k is (1 - 4*x)^(-1/4)*((-1/4)*log(1 - 4*x))^k/k!. Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, 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..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)). T(n, k) = sigma^{(n)}_{n-k}(a_0, a_1, ..., a_{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 4*j. T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*4^j, with the unsigned Stirling1 triangle |S1| = A132393. Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(1 + 4*k*beta(n-1-p))*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 EXAMPLE The triangle T(n, k) begins: n\k         0         1         2        3       4      5    6   7  8 ... O:          1 1:          1         1 2:          5         6         1 3:         45        59        15        1 4:        585       812       254       28       1 5:       9945     14389      5130      730      45      1 6:     208845    312114    122119    20460    1675     66    1 7:    5221125   8011695   3365089   633619   62335   3325   91   1 8:  151412625 237560280 105599276 21740040 2441334 158760 5964 120  1 ... From Wolfdieter Lang, Aug 11 2017: (Start) Recurrence: T(4, 2) = T(3, 1) + (16 - 3)*T(3, 2) = 59 + 13*15 = 254. Boas-Buck recurrence for column k=2 and n=4: 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 CROSSREFS 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. |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. Column sequences for k = 0, 1: A007696, A024382. Row sums: A001813. Alternating row sums: A000007. Sequence in context: A195823 A105577 A054655 * A321630 A086745 A086646 Adjacent sequences:  A290316 A290317 A290318 * A290320 A290321 A290322 KEYWORD nonn,easy,tabl AUTHOR Wolfdieter Lang, Aug 08 2017 STATUS approved

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Last modified November 17 16:08 EST 2019. Contains 329241 sequences. (Running on oeis4.)