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A225466 Triangle read by rows, 3^k*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0. 11
1, 2, 3, 4, 21, 9, 8, 117, 135, 27, 16, 609, 1431, 702, 81, 32, 3093, 13275, 12015, 3240, 243, 64, 15561, 115479, 171990, 81405, 13851, 729, 128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187, 256, 390369, 7998111, 27533142, 29893941, 13121514, 2561706 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

The definition of the Stirling-Frobenius subset numbers of order m is in A225468.

From Wolfdieter Lang, Apr 09 2017: (Start)

This is the Sheffer triangle (exp(2*x), exp(3*x) - 1), denoted by S2[3,2]. See also A282629 for S2[3,1]. The stirling2 triangle A048993 is in this notation denoted by S2[1,0].

The a-sequence for this Sheffer triangle has e.g.f. 3*x/log(1+x) and is 3*A006232(n)/A006233(n) (Cauchy numbers of the first kind). For a- and z-sequences for Sheffer triangles see the W. Lang link under A006232, also with references).

The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(2/3)) and gives 2*A284862/A284863.

The first column k sequences divided by 3^k are A000079, A016127, A016297, A025999. For the e.g.f.s and o.g.f.s see below.

The row sums give A284864. The alternating row sums give A284865.

This triangle appears in the o.g.f. G(n, x) of the sequence {(2 + 3*m)^n}_{m>=0}, as G(n, x) = Sum_{k=0..n} T(n, k)*k!*x^k/(1-x)^(k+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0} (2 + 3*m)^n t^m/m! = exp(t)*Sum_{k=0..n} T(n, k)*t^k.

The corresponding Eulerian number triangle is A225117(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, m)*m!, 0 <= k <= n. (End)

LINKS

Vincenzo Librandi, Rows n = 0..50, flattened

Peter Luschny, Eulerian polynomials.

Peter Luschny, The Stirling-Frobenius numbers.

Shi-Mei Ma, Toufik Mansour, Matthias Schork, Normal ordering problem and the extensions of the Stirling grammar, Russian Journal of Mathematical Physics, 2014, 21(2), arXiv:1308.0169 [math.CO], 2013, p. 12.

FORMULA

T(n, k) = (1/k!)*Sum_{j=0..n} binomial(j, n-k)*A_3(n, j) where A_m(n, j) are the generalized Eulerian numbers A225117.

For a recurrence see the Maple program.

T(n, 0) ~ A000079; T(n, 1) ~ A005057; T(n, n) ~ A000244.

From Wolfdieter Lang, Apr 09 2017: (Start)

T(n, k) = Sum_{j=0..k} binomial(k,j)*(-1)^(j-k)*(2 + 3*j)^n/k!, 0 <= k <= n.

E.g.f. of triangle: exp(2*z)*exp(x*(exp(3*z)-1)) (Sheffer type).

E.g.f. for sequence of column k is exp(2*x)*((exp(3*x) - 1)^k)/k! (Sheffer property).

O.g.f. for sequence of column k is 3^k*x^k/Product_{j=0..k} (1 - (2+3*j)*x).

A nontrivial recurrence for the column m=0 entries T(n, 0) = 2^n from the z-sequence given above: T(n,0) = n*Sum_{k=0..n-1} z(k)*T(n-1,k), n >= 1, T(0, 0) = 1.

The corresponding recurrence for columns k >= 1 from the a-sequence is T(n, k) = (n/k)* Sum_{j=0..n-k} binomial(k-1+j, k-1)*a(j)*T(n-1, k-1+j).

Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+2) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.

(End)

Boas-Buck recurrence for column sequence m: T(n, k) = (1/(n - m))*[(n/2)*(4 + 3*m)*T(n-1, k) + m* Sum_{p=m..n-2} binomial(n, p)(-3)^(n-p)*Bernoulli(n-p)*T(p, k)], for n > k >= 0, with input T(k, k) = 3^k. See a comment and references in A282629, An example is given below. - Wolfdieter Lang, Aug 11 2017

EXAMPLE

[n\k][ 0,     1,      2,       3,       4,      5,     6,    7]

[0]    1,

[1]    2,     3,

[2]    4,    21,      9,

[3]    8,   117,    135,      27,

[4]   16,   609,   1431,     702,      81,

[5]   32,  3093,  13275,   12015,    3240,    243,

[6]   64, 15561, 115479,  171990,   81405,  13851,   729,

[7]  128, 77997, 970515, 2238327, 1655640, 479682, 56133, 2187.

...

From Wolfdieter Lang, Aug 11 2017: (Start)

Recurrence (see the Maple program): T(4, 2) = 3*T(3, 1) + (3*2+2)*T(3, 2) = 3*117 + 8*135 = 1431.

Boas-Buck recurrence for column m = 2, and n = 4: T(4,2) = (1/2)*[2*(4 + 3*2)*T(3, 2) + 2*6*(-3)^2*Bernoulli(2)*T(2, 2))] = (1/2)*(20*135 + 12*9*(1/6)*9) = 1431. (End)

MAPLE

SF_SS := 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;

m*SF_SS(n-1, k-1, m) + (m*(k+1)-1)*SF_SS(n-1, k, m) end:

seq(print(seq(SF_SS(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]]); SFSS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/k!; Table[ SFSS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-Fran├žois 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_SS(n, k, m):

    return add(EulerianNumber(n, j, m)*binomial(j, n-k) for j in (0..n))/ factorial(k)

def A225466(n): return SF_SS(n, k, 3)

(PARI) T(n, k) = sum(j=0, k, binomial(k, j)*(-1)^(j - k)*(2 + 3*j)^n/k!);

for(n=0, 10, for(k=0, n, print1(T(n, k), ", "); ); print(); ) \\ Indranil Ghosh, Apr 10 2017

(Python)

from sympy import binomial, factorial

def T(n, k): return sum([binomial(k, j)*(-1)**(j - k)*(2 + 3*j)**n/factorial(k) for j in range(k + 1)])

for n in range(11): print [T(n, k) for k in range(n + 1)] # Indranil Ghosh, Apr 10 2017

CROSSREFS

Cf. A048993 (m=1), A154537 (m=2), A225467 (m=4), A225468.

Cf. A000079, A000244, A005057, A016127, A016297, A025999, A006232/A006233, A225117, A225472, A225468, A282629, A284862/A284863, A284864, A284865.

Sequence in context: A012580 A246391 A303973 * A225472 A176234 A308891

Adjacent sequences:  A225463 A225464 A225465 * A225467 A225468 A225469

KEYWORD

nonn,easy,tabl,changed

AUTHOR

Peter Luschny, May 08 2013

STATUS

approved

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Last modified December 8 09:32 EST 2019. Contains 329862 sequences. (Running on oeis4.)