cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A284864 Row sums of Sheffer triangle S2[3,2] given by A225466.

Original entry on oeis.org

1, 5, 34, 287, 2839, 31898, 399079, 5480609, 81724300, 1311990425, 22521232003, 411039834356, 7938680371957, 161596770440945, 3454818798460858, 77340712165173755, 1808096791948934755, 44038966942707463946, 1115155787752634260483, 29301563957596710001709
Offset: 0

Views

Author

Wolfdieter Lang, Apr 10 2017

Keywords

Comments

This is a generalization of the Bell sequence A000110 because S2[3,2] is a generalization of the Stirling2 triangle A048993.
For the alternating row sums see A284865.

Crossrefs

Cf. A000110, A225466, A284865, A284859 (case [3,1]).

Programs

  • Mathematica
    T[n_, k_]:=Sum[Binomial[k, j](-1)^(j - k) (2 + 3j)^n/k!, {j, 0, k}]; Table[Sum[T[n, k], {k, 0, n}], {n, 0, 20}] (* Indranil Ghosh, Apr 10 2017 *)
  • PARI
    T(n, k) = sum(j=0, k, binomial(k, j)*(-1)^(j - k)*(2 + 3*j)^n/k!);
    a(n) = sum(k=0, n, T(n, k)); \\ 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)])
    def a(n): return sum([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 10 2017

Formula

a(n) = Sum_{k=0..n} A225466(n, k), n >= 0.
E.g.f.: exp(2*x)*exp((exp(3*x)-1)) (Sheffer property).
a(n) = (1/e)*Sum_{m>=0} (1/m!)*(2 + 3*m)^n, n >= 0, (Dobiński type formula).
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 3^k * a(n-k). - Ilya Gutkovskiy, Jun 21 2022
a(n) ~ 3^n * n^(n + 2/3) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n + 2/3)). - Vaclav Kotesovec, Jun 27 2022

A284865 Alternating row sums of the Sheffer triangle S2[3,2] given by A225466.

Original entry on oeis.org

1, -1, -8, -1, 217, 1196, -3725, -115777, -803150, 3402485, 145172737, 1528780238, -1328359499, -320347469485, -5507171702648, -28294413916213, 915647446089037, 28738067698188692, 369693788462739487, -1233559476327263869
Offset: 0

Views

Author

Wolfdieter Lang, Apr 10 2017

Keywords

Comments

This is a generalization of sequence A000587 because S2[3,2] = A225466 is a generalization of the Stirling2 triangle A048993.
For the row sums see A284864.

Crossrefs

Cf. A000587, A225466, A284864, A284860 (case [3,1]).

Formula

a(n) = Sum_{k=0..n} (-1)^k*A225466(n, k).
E.g.f.: exp(2*x)*exp(1 - exp(3*x)) (Sheffer property).
a(n) = (1/e)*Sum_{m>=0} ((-1)^m / m!)*(2+3*m)^n, n >= 0, (Dobiński type formula).

A282629 Sheffer triangle (exp(x), exp(3*x) - 1). Named S2[3,1].

Original entry on oeis.org

1, 1, 3, 1, 15, 9, 1, 63, 108, 27, 1, 255, 945, 594, 81, 1, 1023, 7380, 8775, 2835, 243, 1, 4095, 54729, 109890, 63180, 12393, 729, 1, 16383, 395388, 1263087, 1151010, 387828, 51030, 2187, 1, 65535, 2816865, 13817034, 18752391, 9658278, 2133054, 201204, 6561, 1, 262143, 19914660, 146620935, 285232185, 210789621, 69502860, 10825650, 767637, 19683
Offset: 0

Views

Author

Wolfdieter Lang, Apr 03 2017

Keywords

Comments

For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references).
The e.g.f. for the sequence of column m is (Sheffer property) exp(x)*(exp(3*x) - 1)^m/m!.
This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named 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).
The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(1/3)) and is A284857(n) / A284858(n).
The main diagonal gives A000244.
The row sums give A284859. The alternating row sums give A284860.
The triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 3*m)^n}{m>=0}, as G(n, x) = Sum{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 3*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225117 with row reversion.
The first column k sequences divided by 3^k are A000012, A002450 (with a leading 0), A016223, A021874. For the e.g.f.s and o.g.f.s see below. - Wolfdieter Lang, Apr 09 2017
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,m)*x^m of the Sheffer triangle S2[d,a] satisfy, as special polynomials of the Boas-Buck class, the identity (see the reference, and we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = - n*a*R(d,a;n-1,x) - Sum_{k=0..n-1} binomial(n, k+1)*(-d)^(k+1)*Bernoulli(k+1)*E_x*R(d,a;n-1-k,x), with E_x = x*d/dx (Euler operator).
This entails a recurrence for the sequence of column m, for n > m:
T(d,a;n,m) = (1/(n - m))*[(n/2)*(2*a + d*m)*T(d,a;n-1,m) + m*Sum_{p=m..n-2} binomial(n,p)(-d)^(n-p)*Bernoulli(n-p)*T(d,a;p,m)], with input T(d,a;n,n) = d^n. For the present [d,a] = [3,1] case see the formula and example sections below. - Wolfdieter Lang, Aug 09 2017 (End)
The inverse of this triangular Sheffer matrix S2[3,1] is S1[3,1] with rational elements S1[3,1](n, k) = (-1)^(n-k)*A286718(n, k)/3^k. - Wolfdieter Lang, Nov 15 2018
Named after the American mathematician Isador Mitchell Sheffer (1901-1992). - Amiram Eldar, Jun 19 2021

Examples

			The triangle T(n, m) begins:
  n\m 0      1        2         3         4         5        6        7      8     9
  0:  1
  1:  1      3
  2:  1     15        9
  3:  1     63      108        27
  4:  1    255      945       594        81
  5:  1   1023     7380      8775      2835       243
  6:  1   4095    54729    109890     63180     12393      729
  7:  1  16383   395388   1263087   1151010    387828    51030     2187
  8:  1  65535  2816865  13817034  18752391   9658278  2133054   201204   6561
  9:  1 262143 19914660 146620935 285232185 210789621 69502860 10825650 767637 19683
  ...
------------------------------------------------------------------------------------
Nontrivial recurrence for m=0 column from z-sequence: T(4,0) = 4*(1*1 + 63*(-1/6) + 108*(11/54) + 27*(-49/108)) = 1.
Recurrence for m=2 column from a-sequence: T(4, 2) = (4/2)*(1*63*3 + 2*108*(3/2) + 3*27*(-3/6)) = 945.
Recurrence for row polynomial R(3, x) (Meixner type): ((3*x + 1) + 3*x*d_x)*(1 + 15*x + 9*x^2) = 1 + 63*x + 108*x^2 + 27*x^3.
E.g.f. and o.g.f. of n = 1 powers {(1 + 3*m)^1}_{m>=0} A016777: E(1, x) = exp(x) * (T(1, 0) + T(1, 1)*x) = exp(x)*(1+3*x). O.g.f.: G(1, x) = T(1, 0)*0!/(1-x) + T(1, 1)*1!*x/(1-x)^2 = (1+2*x)/(1-x)^2.
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (1/2)*(2*(2 + 3*2)*T(3, 2) + 2*6*(-3)^2*bernoulli(2)*T(2, 2)) = (1/2)*(16*108 + 12*9*(1/6)*9) = 945. - _Wolfdieter Lang_, Aug 09 2017
		

References

  • Ralph P. Boas, Jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[m, k] (-1)^(k - m) (1 + 3 k)^n/m!, {k, 0, m}], {n, 0, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 08 2017 *)
  • PARI
    T(n, m) = sum(k=0, m, binomial(m, k) * (-1)^(k - m) * (1 + 3*k)^n/m!);
    for(n=0, 9, for(m=0, n, print1(T(n, m),", ");); print();) \\ Indranil Ghosh, Apr 08 2017

Formula

A nontrivial recurrence for the column m=0 entries T(n, 0) = 1 from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)*Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+1) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 3*k)^n/m!, 0 <= m <= n.
E.g.f. of triangle: exp(z)*exp(x*(exp(3*z)-1)) (Sheffer type).
E.g.f. for sequence of column m is exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
From Wolfdieter Lang, Apr 09 2017: (Start)
Standard three-term recurrence: T(n, m) = 0 if n < m, T(n,-1) = 0, T(0, 0) = 1, T(n, m) = 3*T(n-1, m-1) + (1+3*m)*T(n-1, m) for n >= 1. From the T(n, m) formula. Compare with the recurrence of S2[3,2] given in A225466.
The o.g.f. for sequence of column m is 3^m*x^m/Product_{j=0..m} (1 - (1+3*j)*x). (End)
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^k*Stirling2(k, m), 0 <= m <= n. - Wolfdieter Lang, Apr 13 2017
Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*((n/2)*(2 + 3*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n,p)*(-3)^(n-p)*Bernoulli(n-p)*T(p, m)), for n > m >= 0, with input T(m, m) = 3^m. See a comment above. - Wolfdieter Lang, Aug 09 2017

A286718 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
Offset: 0

Views

Author

Wolfdieter Lang, May 18 2017

Keywords

Comments

This is a generalization of the unsigned Stirling1 triangle A132393.
In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/A002209(k+1).
This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018

Examples

			The triangle T(n, k) begins:
n\k        0        1        2       3      4     5    6  7 8 ...
O:         1
1:         1        1
2:         4        5        1
3:        28       39       12       1
4:       280      418      159      22      1
5:      3640     5714     2485     445     35     1
6:     58240    95064    45474    9605   1005    51    1
7:   1106560  1864456   959070  227969  28700  1974   70  1
8:  24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
From _Wolfdieter Lang_, Aug 09 2017: (Start)
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Crossrefs

S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
Column sequences for k = 0..4: A007559, A024216(n-1), A286721(n-2), A382984, A382985.
Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
Row sums: A008544. Alternating row sums: A000007.
Beta sequence: A002208(n+1)/A002209(n+1).

Programs

  • Mathematica
    T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[, -1] = 0; T[0, 0] = 1; T[n, k_] /; nJean-François Alcover, Jun 20 2018 *)

Formula

Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;x,n) with the rising factorial
risefac(d,a;x,n) := Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}{n-k}(a_0,a_1,...,a{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017

A225468 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.

Original entry on oeis.org

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

Views

Author

Peter Luschny, May 16 2013

Keywords

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)

Examples

			[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
		

Crossrefs

Cf. A048993 (m=1), A039755 (m=2), A225469 (m=4).

Programs

  • 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 *)
  • 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)]

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

A225467 Triangle read by rows, T(n, k) = 4^k*S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 3, 4, 9, 40, 16, 27, 316, 336, 64, 81, 2320, 4960, 2304, 256, 243, 16564, 63840, 54400, 14080, 1024, 729, 116920, 768496, 1071360, 485120, 79872, 4096, 2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384, 6561, 5758240, 101417920, 322850304
Offset: 0

Views

Author

Peter Luschny, May 08 2013

Keywords

Comments

The definition of the Stirling-Frobenius subset numbers of order m is in A225468.
This is the Sheffer triangle (exp(3*x), exp(4*x) - 1). See also the P. Bala link under A225469, the Sheffer triangle (exp(3*x),(1/4)*(exp(4*x) - 1)), which is named there exponential Riordan array S_{(4,0,3)}. - Wolfdieter Lang, Apr 13 2017

Examples

			[n\k][  0,      1,       2,        3,        4,       5,      6,     7]
[0]     1,
[1]     3,      4,
[2]     9,     40,      16,
[3]    27,    316,     336,       64,
[4]    81,   2320,    4960,     2304,      256,
[5]   243,  16564,   63840,    54400,    14080,    1024,
[6]   729, 116920,  768496,  1071360,   485120,   79872,   4096,
[7]  2187, 821356, 8921136, 19144384, 13502720, 3777536, 430080, 16384.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see the Maple program): T(4, 2) = 4*T(3, 1) + (4*2+3)*T(3, 2) = 4*316 + 11*336 = 4960.
Boas-Buck recurrence for column k = 2, and n = 4: T(4, 2) = (1/2)*(2*(6 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(28*336 + 12*16*(1/6)*16) = 4960. (End)
		

Crossrefs

Cf. A048993 (m=1), A154537 (m=2), A225466 (m=3). A225469 (scaled).
Cf. Columns: A000244, 4*A016138, 16*A018054. A225118.

Programs

  • 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, 4), 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, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • PARI
    T(n, k) = sum(m=0, k, binomial(k, m)*(-1)^(m - k)*((3 + 4*m)^n)/k!);
    for(n = 0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 13 2017
    
  • Python
    from sympy import binomial, factorial
    def T(n, k): return sum(binomial(k, m)*(-1)**(m - k)*((3 + 4*m)**n)//factorial(k) for m in range(k + 1))
    for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 13 2017
  • 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 A225467(n): return SF_SS(n, k, 4)
    

Formula

T(n, k) = (1/k!)*sum_{j=0..n} binomial(j, n-k)*A_4(n, j) where A_m(n, j) are the generalized Eulerian numbers A225118.
For a recurrence see the Maple program.
T(n, 0) ~ A000244; T(n, 1) ~ A190541.
T(n, n) ~ A000302; T(n, n-1) ~ A002700.
From Wolfdieter Lang, Apr 13 2017: (Start)
T(n, k) = Sum_{m=0..k} binomial(k,m)*(-1)^(m-k)*((3+4*m)^n)/k!, 0 <= k <= n.
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^(n-k)*4^k*Stirling2(k, m), 0 <= m <= n.
E.g.f. exp(3*z)*exp(x*(exp(4*z) - 1)) (Sheffer property).
E.g.f. column k: exp(3*x)*((exp(4*x) - 1)^k)/k!, k >= 0.
O.g.f. column k: (4*x)^k/Product_{j=0..k} (1 - (3 + 4*j)*x), k >= 0.
(End)
Boas-Buck recurrence for column sequence k: T(n, k) = (1/(n - k))*((n/2)*(6 + 4*k)*T(n-1, k) + k*Sum_{p=k..n-2} binomial(n, p)*(-4)^(n-p)*Bernoulli(n-p)*T(p, k)), for n > k >= 0, with input T(k, k) = 4^k. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017

A225472 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.

Original entry on oeis.org

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

Views

Author

Peter Luschny, May 17 2013

Keywords

Comments

The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).

Examples

			[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.
		

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.

Programs

  • 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 *)
  • 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)]

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

A290319 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 4*x)^(-1/4), (-1/4)*log(1 - 4*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 5, 6, 1, 45, 59, 15, 1, 585, 812, 254, 28, 1, 9945, 14389, 5130, 730, 45, 1, 208845, 312114, 122119, 20460, 1675, 66, 1, 5221125, 8011695, 3365089, 633619, 62335, 3325, 91, 1, 151412625, 237560280, 105599276, 21740040, 2441334, 158760, 5964, 120, 1, 4996616625, 7990901865, 3722336388, 823020596, 102304062, 7680414, 355572, 9924, 153, 1, 184874815125, 300659985630, 145717348221, 34174098440, 4608270890, 386479380, 20836578, 722760, 15585, 190, 1
Offset: 0

Views

Author

Wolfdieter Lang, Aug 08 2017

Keywords

Comments

This generalization of the unsigned Stirling1 triangle A132393 is called here |S1hat[4,1]|.
The signed matrix S1hat[4,1] with elements (-1)^(n-k)*|S1hat[4,1]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[4,1] with elements S2[4,1](n, k)/d^k, where S2[4,1] is Sheffer (exp(x), exp(4*x) - 1), given in A285061. See also the P. Bala link below for the scaled and signed version s_{(4,0,1)}.
For the general |S1hat[d,a]| case see a comment in A286718.

Examples

			The triangle T(n, k) begins:
  n\k         0         1         2        3       4      5    6   7  8 ...
  0:          1
  1:          1         1
  2:          5         6         1
  3:         45        59        15        1
  4:        585       812       254       28       1
  5:       9945     14389      5130      730      45      1
  6:     208845    312114    122119    20460    1675     66    1
  7:    5221125   8011695   3365089   633619   62335   3325   91   1
  8:  151412625 237560280 105599276 21740040 2441334 158760 5964 120  1
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: T(4, 2) = T(3, 1) + (16 - 3)*T(3, 2) = 59 + 13*15 = 254.
Boas-Buck recurrence for column k=2 and n=4:
T(4, 2) = (4!/2)*(4*(1 + 8*(5/12))*T(2, 2)/2! + 1*(1 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(2*13/3 + 5*15/3!) = 254. (End)
		

Crossrefs

S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,1], [3,2] and [4,3] is A132393, A028338, A286718, A225470 and A225471, respectively.
Columns k=0..3 give A007696, A024382(n-1), A383700, A383701.
Row sums: A001813. Alternating row sums: A000007.

Programs

  • Mathematica
    FoldList[Join[Table[If[i == 1, 0, #[[i-1]]] + (4*#2 - 3)*#[[i]], {i, Length[#]}], {1}] &, {1}, Range[10]] (* Paolo Xausa, Aug 18 2025 *)

Formula

Recurrence: T(n, k) = T(n-1, k-1) + (4*n - 3)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle): (1 - 4*z)^{-(x + 1)/4}.
E.g.f. of column k is (1 - 4*x)^(-1/4)*((-1/4)*log(1 - 4*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+4), with R(0, x) = 1. Row polynomial R(n, x) = risefac(4,1;x,n) with the rising factorial risefac(d,a;x,n) :=Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}{n-k}(a_0, a_1, ..., a{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 4*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*4^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(1 + 4*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 with {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

A284862 Numerators of exponential expansion of (3/(2*log(1+x)))*(1 - 1/(1+x)^(2/3)).

Original entry on oeis.org

1, -1, 13, -32, 1666, -13426, 515194, -1432000, 1447711256, -4097653768, 256749458824, -2204786032640, 11533922227138736, -33268276510233104, 577462439822785168, -1674851096410984448, 6621155504764033947008, -34711497070334170000000
Offset: 0

Views

Author

Wolfdieter Lang, Apr 09 2017

Keywords

Comments

This is one half of the numerator of the z-sequence of the Sheffer triangle S2[3,2] given in A225466. See the W. Lang link of A006232 for a- and z- sequences for Sheffer triangles and for references.
The denominators are given in A284863.
The nontrivial recurrence for the column m=0 entries A225466(n, 0) = 2^n from the z-sequence z(n) = 2*a(n)/A284863(n) is: T(n,0) = n*Sum_{k=0..n-1} z(k)*A225466(n-1,k), n >= 1, T(0, 0) = 1.

Examples

			The rationals r(n) begin: 1, -1/3, 13/27, -32/27, 1666/405, -13426/729, 515194/5103, -1432000/2187, 1447711256/295245, -4097653768/98415, 256749458824/649539, ...
The z-sequence is {2*r(n)}, n >= 0.
The nontrivial recurrence for A225466(4, 0) = 16 from this z-sequence is: 16 =  8*(1*8 + (-1/3)*117 + (13/27)*135 + (-32/27)*27).
		

Crossrefs

Formula

a(n) = numerator(r(n)), with the rationals (in lowest terms) r(n) = [x^n/n!] (3/(2*log(1+x)))*(1 - 1/(1+x)^(2/3)).

A284863 Denominators of exponential expansion of (3/(2*log(1+x)))*(1 - 1/(1+x)^(2/3)).

Original entry on oeis.org

1, 3, 27, 27, 405, 729, 5103, 2187, 295245, 98415, 649539, 531441, 241805655, 55801305, 71744535, 14348907, 3658971285, 1162261467, 51526925037, 1162261467, 4027235983155, 575319426165, 3608821855035, 1412147682405, 642527195494275, 77103263459313, 480412641554181, 53379182394909
Offset: 0

Views

Author

Wolfdieter Lang, Apr 09 2017

Keywords

Comments

For the numerators see A284862, also for details.
The z-sequence for the Sheffer triangle A225466 is 2*A284862(n)/a(n).

Examples

			See A284862.
		

Crossrefs

Formula

a(n) = denominator(r(n)), with the rationals (in lowest terms) r(n) = [x^n/n!] (1/log((1+x)^(1/3)))*(1 - 1/(1+x)^(2/3)).
Showing 1-10 of 10 results.