OFFSET
0,2
COMMENTS
The Stirling-Frobenius cycle numbers are defined in A225470.
Triangle T(n,k), read by rows, given by (3, 4, 7, 8, 11, 12, 15, 16, 19, 20, ... (A014601)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015
LINKS
Peter Luschny, Generalized Eulerian polynomials.
Peter Luschny, The Stirling-Frobenius numbers.
FORMULA
For a recurrence see the Sage program.
From Wolfdieter Lang, May 29 2017: (Start)
This is the Sheffer triangle (1/(1 - 4*x)^{-3/4}, -(1/4)*log(1-4*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(4,0,3)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 4*z)^{-(3+x)/4}.
E.g.f. of column k: (1-4*x)^(-3/4)*(-(1/4)*log(1-4*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+3)*R(n-1,x+4), with R(0, x) = 1.
R(n, x) = risefac(4,3;x,n) := Product_{j=0..(n-1)} (x + (3 + 4*j)). (See the P. Bala link, eq. (16) for the signed s_{4,0,3} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*3^(n-k-j)*4^j, with S1p(n, m) = A132393(n, m).
T(n, k) = sigma[4,3]^{(n)}_{n-k}, with the elementary symmetric functions sigma[4,3]^{(n)}_m of degree m in the n numbers 3, 7, 11, ..., 3+4*(n-1), with sigma[4,3]^{(n)}_0 := 1. (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(3 + 8*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
[n\k][ 0, 1, 2, 3, 4, 5, 6 ]
[0] 1,
[1] 3, 1,
[2] 21, 10, 1,
[3] 231, 131, 21, 1,
[4] 3465, 2196, 446, 36, 1,
[5] 65835, 45189, 10670, 1130, 55, 1,
[6] 1514205, 1105182, 290599, 36660, 2395, 78, 1.
...
From Wolfdieter Lang, Aug 11 2017: (Start)
Recurrence: T(4, 2) = T(3, 1) + (4*4 - 1)*T(3, 2) = 131 +15*21 = 446.
Boas-Buck recurrence for column k=2 and n=4: T(4, 2) = (4!/2)*(4*(3+8*(5/12)) *T(2, 2)/2! + 1*(3 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(4*(19/3)/2 + 7*21/3!) = 446.
(End)
MATHEMATICA
T[0, 0] = 1; T[n_, k_] := Sum[Binomial[n - j, k]*Abs[StirlingS1[n, n - j]]* 3^(n - k - j)*4^j, {j, 0, n - k}];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Wolfdieter Lang *)
PROG
(Sage)
@CachedFunction
def SF_C(n, k, m):
if k > n or k < 0 : return 0
if n == 0 and k == 0: return 1
return SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m)
for n in (0..8): [SF_C(n, k, 4) for k in (0..n)]
CROSSREFS
KEYWORD
AUTHOR
Peter Luschny, May 17 2013
STATUS
approved