OFFSET
0,2
COMMENTS
The definition of the Stirling-Frobenius subset numbers: S_m(n, k) = (sum_{j=0..n} binomial(j, n-k)*A_m(n, j)) / (m^k*k!) where A_m(n, j) are the generalized Eulerian numbers. For m = 1 this gives the classical Stirling set numbers A048993. (See the links for details.)
From Peter Bala, Jan 27 2015: (Start)
Exponential Riordan array [ exp(2*z), 1/3*(exp(3*z) - 1)].
Triangle of connection constants between the polynomial basis sequences {x^n}n>=0 and { n!*3^n*binomial((x - 2)/3,n) }n>=0. An example is given below.
This triangle is the particular case a = 3, b = 0, c = 2 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)
LINKS
Vincenzo Librandi, Rows n = 0..50, flattened
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 8.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.
Peter Luschny, Generalized Eulerian polynomials.
Peter Luschny, The Stirling-Frobenius numbers.
Shi-Mei Ma, Toufik Mansour, and Matthias Schork, Normal ordering problem and the extensions of the Stirling grammar, Russian Journal of Mathematical Physics, 2014, 21(2), arXiv 1308.0169 p. 12.
FORMULA
T(n, k) = (sum_{j=0..n} binomial(j, n-k)*A_3(n, j)) / (3^k*k!) with A_3(n,j) = A225117.
For a recurrence see the Maple program.
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = sum {i = 0..n} (-1)^(n+i)*3^(i-k)*binomial(n,i)*Stirling2(i+1,k+1).
E.g.f.: exp(2*z)*exp(x/3*(exp(3*z) - 1)) = 1 + (2 + x)*z + (4 + 7*x + x^2)*z^2/2! + ....
T(n,k) = 1/(3^k*k!)*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*(3*j + 2)^n.
O.g.f. for n-th diagonal: exp(-2*x/3)*sum {k >= 0} (3*k + 2)^(k + n - 1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k: 1/( (1 - 2*x)*(1 - 5*x)...(1 - (3*k + 2)*x ). (End)
E.g.f. column k: exp(2*x)*(exp(3*x - 1)/3^k, k >= 0. See the Bala link for the S(3,0,2) exponential Riordan aka Sheffer triangle. - Wolfdieter Lang, Apr 10 2017
EXAMPLE
[n\k][ 0, 1, 2, 3, 4, 5, 6]
[0] 1,
[1] 2, 1,
[2] 4, 7, 1,
[3] 8, 39, 15, 1,
[4] 16, 203, 159, 26, 1,
[5] 32, 1031, 1475, 445, 40, 1,
[6] 64, 5187, 12831, 6370, 1005, 57, 1.
Connection constants: Row 3: [8, 39, 15, 1] so
x^3 = 8 + 39*(x - 2) + 15*(x - 2)*(x - 5) + (x - 2)*(x - 5)*(x - 8). - Peter Bala, Jan 27 2015
MAPLE
SF_S := 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_S(n-1, k-1, m) + (m*(k+1)-1)*SF_S(n-1, k, m) end:
seq(print(seq(SF_S(n, k, 3), k=0..n)), n = 0..5);
MATHEMATICA
EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/(k!*m^k); Table[ SFS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
PROG
(Sage)
@CachedFunction
def EulerianNumber(n, k, m) :
if n == 0: return 1 if k == 0 else 0
return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m) + (m*k+1)*EulerianNumber(n-1, k, m)
def SF_S(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/ (factorial(k)*m^k)
for n in (0..6): [SF_S(n, k, 3) for k in (0..n)]
CROSSREFS
KEYWORD
AUTHOR
Peter Luschny, May 16 2013
STATUS
approved