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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.
+10
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
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
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. 9.
Peter Luschny, Eulerian polynomials.
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 [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-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_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
KEYWORD
nonn,easy,tabl
AUTHOR
Peter Luschny, May 08 2013
STATUS
approved
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.
+10
11
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
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 equals P * A111577 = P^(-1) * A075498, where P is Pascal's triangle A007318.
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.
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.
T(n, 0) ~ A000079; T(n, 1) ~ A016127; T(n, 2) ~ A016297; T(n, 3) ~ A025999;
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024212.
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
Cf. A048993 (m=1), A039755 (m=2), A225469 (m=4).
KEYWORD
nonn,easy,tabl
AUTHOR
Peter Luschny, May 16 2013
STATUS
approved
Triangle read by rows, k!*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
+10
5
1, 2, 3, 4, 21, 18, 8, 117, 270, 162, 16, 609, 2862, 4212, 1944, 32, 3093, 26550, 72090, 77760, 29160, 64, 15561, 230958, 1031940, 1953720, 1662120, 524880, 128, 77997, 1941030, 13429962, 39735360, 57561840, 40415760, 11022480, 256, 390369, 15996222, 165198852
OFFSET
0,2
COMMENTS
The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).
FORMULA
For a recurrence see the Maple program.
T(n, 0) ~ A000079; T(n, 1) ~ A005057; T(n, n) ~ A032031.
From Wolfdieter Lang, Apr 10 2017: (Start)
E.g.f. for sequence of column k: exp(2*x)*(exp(3*x) - 1)^k, k >= 0. From the Sheffer triangle S2[3,2] = A225466 with column k multiplied with k!.
O.g.f. for sequence of column k is 3^k*k!*x^k/Product_{j=0..k} (1 - (2+3*j)*x), k >= 0.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k, j)*(2+3*j)^n, 0 <= k <= n.
Three term recurrence (see the Maple program): T(n, k) = 0 if n < k , T(n, -1) = 0, T(0,0) = 1, T(n, k) = 3*k*T(n-1, k-1) + (2 + 3*k)*T(n-1, k) for n >= 1, k=0..n.
For the column scaled triangle (with diagonal 1s) see A225468, and the Bala link with (a,b,c) = (3,0,2), where Sheffer triangles are called exponential Riordan triangles.
(End)
The e.g.f. of the row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k is exp(2*z)/(1 - x*(exp(3*z) - 1)). - Wolfdieter Lang, Jul 12 2017
EXAMPLE
[n\k][0, 1, 2, 3, 4, 5, 6 ]
[0] 1,
[1] 2, 3,
[2] 4, 21, 18,
[3] 8, 117, 270, 162,
[4] 16, 609, 2862, 4212, 1944,
[5] 32, 3093, 26550, 72090, 77760, 29160,
[6] 64, 15561, 230958, 1031940, 1953720, 1662120, 524880.
MAPLE
SF_SO := 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*k*SF_SO(n-1, k-1, m) + (m*(k+1)-1)*SF_SO(n-1, k, m) end:
seq(print(seq(SF_SO(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]]); SFSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]; Table[ SFSO[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_SO(n, k, m):
return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))
for n in (0..6): [SF_SO(n, k, 3) for k in (0..n)]
CROSSREFS
Cf. A131689 (m=1), A145901 (m=2), A225473 (m=4).
Cf. A225466, A225468, columns: A000079, 3*A016127, 3^2*2!*A016297, 3^3*3!*A025999.
KEYWORD
nonn,easy,tabl
AUTHOR
Peter Luschny, May 17 2013
STATUS
approved

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