A225470 Triangle read by rows, s_3(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

1, 2, 1, 10, 7, 1, 80, 66, 15, 1, 880, 806, 231, 26, 1, 12320, 12164, 4040, 595, 40, 1, 209440, 219108, 80844, 14155, 1275, 57, 1, 4188800, 4591600, 1835988, 363944, 39655, 2415, 77, 1, 96342400, 109795600, 46819324, 10206700, 1276009, 95200, 4186, 100, 1

Offset: 0

Author

Peter Luschny, May 16 2013

Keywords

• nonn
• easy
• tabl

Comments

The Stirling-Frobenius subset numbers S_{m}(n,k), for m >= 1 fixed, regarded as an infinite lower triangular matrix, can be inverted by Sum_{k} S_{m}(n,k)*s_{m}(k,j)*(-1)^(n-k) = [j = n]. The inverse numbers s_{m}(k,j), which are unsigned, are the Stirling-Frobenius cycle numbers. For m = 1 this gives the classical Stirling cycle numbers A132393. The Stirling-Frobenius subset numbers are defined in A225468.

Triangle T(n,k), read by rows, given by (2, 3, 5, 6, 8, 9, 11, 12, 14, 15, ... (A007494)) 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

Examples

			Triangle starts:
  [n\k][    0,      1,     2,     3,    4,  5,  6]
  [0]       1,
  [1]       2,      1,
  [2]      10,      7,     1,
  [3]      80,     66,    15,     1,
  [4]     880,    806,   231,    26,    1,
  [5]   12320,  12164,  4040,   595,   40,  1,
  [6]  209440, 219108, 80844, 14155, 1275, 57,  1.
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see Maple program): T(4, 2) = T(3, 1) + (3*4 - 1)*T(3, 2) = 66 + 11*15 = 231.
Boas-Buck type recurrence for column k = 2 and n = 4: T(4, 2) = (4!/2)*(3*(2 + 6*(5/12))*T(2, 2)/2! + 1*(2 + 6*(1/2))*T(3,2)/3!) = (4!/2)*(3*9/4 + 5*15/3!) = 231. (End)
		

Links

• Peter Bala, A 3 parameter family of generalized Stirling numbers
• Peter Luschny, Generalized Eulerian polynomials
• Peter Luschny, The Stirling-Frobenius numbers

Crossrefs

Cf. A225468; A132393 (m=1), A028338 (m=2), A225471(m=4).

Column k=0..4 give A008544, A024395(n-1), A286722(n-2), A383221, A383222.

T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024391; T(n, n-3) ~ A024392.

row sums ~ A032031; alternating row sums ~ A007559.

Cf. A132393.

Programs

Maple

SF_C := proc(n, k, m) option remember;
if n = 0 and k = 0 then return(1) fi;
if k > n or  k < 0 then return(0) fi;
SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m) end:
seq(print(seq(SF_C(n, k, 3), k = 0..n)), n = 0..8);

Mathematica

SFC[0, 0, ] = 1; SFC[n, k_, ] /; (k > n || k < 0) = 0; SFC[n, k_, m_] := SFC[n, k, m] = SFC[n-1, k-1, m] + (m*n-1)*SFC[n-1, k, m]; Table[SFC[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2013, after Maple *)

Formula

For a recurrence see the Maple program.

From Wolfdieter Lang, May 18 2017: (Start)

This is the Sheffer triangle (1/(1 - 3*x)^{-2/3}, -(1/3)*log(1-3*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(3,0,2)}.

E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 3*z)^{-(2+x)/3}.

E.g.f. of column k: (1-3*x)^(-2/3)*(-(1/3)*log(1-3*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+2)*R(n-1,x+3), with R(0, x) = 1.

R(n, x) = risefac(3,2;x,n) := Product_{j=0..(n-1)} (x + (2 + 3*j)). (See the P. Bala link, eq. (16) for the signed s_{3,0,2} row polynomials.)

T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*2^(n-k-j)*3^j, with S1p(n, m) = A132393(n, m). (End)

Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 3^(n-1-p)*(2 + 3*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 {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017