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A225468
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Triangle read by rows, S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
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11
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1, 2, 1, 4, 7, 1, 8, 39, 15, 1, 16, 203, 159, 26, 1, 32, 1031, 1475, 445, 40, 1, 64, 5187, 12831, 6370, 1005, 57, 1, 128, 25999, 107835, 82901, 20440, 1974, 77, 1, 256, 130123, 888679, 1019746, 369061, 53998, 3514, 100, 1
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OFFSET
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0,2
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COMMENTS
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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.)
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)
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LINKS
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FORMULA
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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.
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
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EXAMPLE
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[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
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MAPLE
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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);
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MATHEMATICA
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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 *)
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PROG
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(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)]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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