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, 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
Offset: 0
Examples
The triangle T(n, k) begins: n\k 0 1 2 3 4 5 6 7 8 ... 0: 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. (End)
Links
- Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
- Peter Bala, A 3 parameter family of generalized Stirling numbers.
- Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.
Crossrefs
Programs
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Mathematica
FoldList[Join[Table[If[i == 1, 0, #[[i-1]]] + (4*#2 - 3)*#[[i]], {i, Length[#]}], {1}] &, {1}, Range[10]] (* Paolo Xausa, Aug 18 2025 *)
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
Comments