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 1-8 of 8 results.

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.

Links

Crossrefs

Cf. A000012, A000244, A006232/A006233, A016777, A024036, A111577, A225117, A225466, A284857, A284858, A284859, A284860, A284861, A286718.

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

A111577 Galton triangle T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 12, 1, 1, 85, 105, 22, 1, 1, 341, 820, 325, 35, 1, 1, 1365, 6081, 4070, 780, 51, 1, 1, 5461, 43932, 46781, 14210, 1596, 70, 1, 1, 21845, 312985, 511742, 231511, 39746, 2926, 92, 1, 1, 87381, 2212740, 5430405, 3521385, 867447, 95340, 4950, 117, 1
Offset: 1

Views

Author

Gary W. Adamson, Aug 07 2005

Keywords

Comments

In triangles of analogs to Stirling numbers of the second kind, the multipliers of T(n-1,k) in the recurrence are terms in arithmetic sequences: in Pascal's triangle A007318, the multiplier = 1. In triangle A008277, the Stirling numbers of the second kind, the multipliers are in the set (1,2,3...). For this sequence here, the multipliers are from A016777.
Riordan array [exp(x), (exp(3x)-1)/3]. - Paul Barry, Nov 26 2008
From Peter Bala, Jan 27 2015: (Start)
Working with an offset of 0, this is the triangle of connection constants between the polynomial basis sequences {x^n}, n>=0 and {n!*3^n*binomial((x - 1)/3,n)}, n>=0. An example is given below.
Call this array M and let P denote Pascal's triangle A007318, then P * M = A225468, P^2 * M = A075498. Also P^(-1) * M is a shifted version of A075498.
This triangle is the particular case a = 3, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)
Named after the English scientist Francis Galton (1822-1911). - Amiram Eldar, Jun 13 2021
This is the array of (r, β)-Stirling numbers for r = 1 and β = 3. See Corcino. - Peter Bala, Feb 26 2025

Examples

			T(5,3) = T(4,2) + 7*T(4,3) = 21 + 7*12 = 105.
The triangle starts in row n = 1 as:
  1;
  1,  1;
  1,  5,   1;
  1, 21,  12,  1;
  1, 85, 105, 22, 1;
Connection constants: Row 4: [1, 21, 12, 1] so
x^3 = 1 + 21*(x - 1) + 12*(x - 1)*(x - 4) + (x - 1)*(x - 4)*(x - 7). - _Peter Bala_, Jan 27 2015
From _Peter Bala_, Feb 26 2025: (Start)
The array factorizes as
/1                \     /1               \/1              \/1             \
|1   1            |     |1   1           ||0  1           ||0  1          |
|1   5    1       |  =  |1   4   1       ||0  1   1       ||0  0  1       | ...
|1  21   12   1   |     |1  13   7   1   ||0  1   4  1    ||0  0  1  1    |
|1  85  105  22  1|     |1  44  34  10  1||0  1  13  7  1 ||0  0  1  4  1 |
|...              |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - x), x/(1 - 3*x)). Cf. A193843. (End)

Links

Crossrefs

Cf. A008277, A039755, A075498, A225468.
Cf. A282629, A284861, A225117.

Programs

  • Maple
    A111577 := proc(n,k) option remember; if k = 1 or k = n then 1; else procname(n-1,k-1)+(3*k-2)*procname(n-1,k) ; fi; end:
seq( seq(A111577(n,k),k=1..n), n=1..10) ; # R. J. Mathar, Aug 22 2009
  • Mathematica
    T[, 1] = 1; T[n, n_] = 1;
T[n_, k_] := T[n, k] = T[n-1, k-1] + (3k-2) T[n-1, k];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] (* Jean-François Alcover, Jun 13 2019 *)

Formula

T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k).
E.g.f.: exp(x)*exp((y/3)*(exp(3x)-1)). - Paul Barry, Nov 26 2008
Let f(x) = exp(1/3*exp(3*x) + x). Then, with an offset of 0, the row polynomials R(n,x) are given by R(n,exp(3*x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A105794, A143494 and A154537. - Peter Bala, Mar 01 2012
T(n, k) = 1/(3^k*k!)*Sum_{j=0..k}((-1)^(k-j)*binomial(k,j)*(3*j+1)^n). - Peter Luschny, May 20 2013
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n-1} 3^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
O.g.f. for n-th diagonal: exp(-x/3)*Sum_{k >= 0} (3*k + 1)^(k+n-1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k (with offset 0): 1/( (1 - x)*(1 - 4*x)*...*(1 - (3*k + 1)*x) ). (End)

Extensions

Edited and extended by R. J. Mathar, Aug 22 2009

A016781 a(n) = (3*n+1)^5.

Original entry on oeis.org

1, 1024, 16807, 100000, 371293, 1048576, 2476099, 5153632, 9765625, 17210368, 28629151, 45435424, 69343957, 102400000, 147008443, 205962976, 282475249, 380204032, 503284375, 656356768, 844596301, 1073741824, 1350125107, 1680700000, 2073071593, 2535525376
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

In general the e.g.f. of {(1 + 3*m)^n}{m>=0} is E(n,x) = exp(x)*Sum{m=0..n} A282629(n, m)*x^m, and the o.g.f. is G(n, x) = (Sum_{m=0..n} A225117(n, n-m)*x^m)/(1-x)^(n+1). - Wolfdieter Lang, Apr 02 2017

Links

Crossrefs

Cf. A016777, A016778, A016779, A016780, A225117, A282629.

Programs

  • Magma
    [(3*n+1)^5: n in [0..30]]; // Vincenzo Librandi, Sep 21 2011
  • Mathematica
    (3Range[0,20]+1)^5 (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{1,1024,16807,100000,371293,1048576},30] (* Harvey P. Dale, May 13 2012 *)
  • Maxima
    A016781(n):=(3*n+1)^5$
makelist(A016781(n),n,0,20); /* Martin Ettl, Nov 12 2012 */

Formula

a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, May 13 2012
From Wolfdieter Lang, Apr 02 2017: (Start)
O.g.f.: (1+1018*x+10678*x^2+14498*x^3+2933*x^4+32*x^5)/(1-x)^6.
E.g.f: exp(x)*(1+1023*x+7380*x^2+8775*x^3+2835*x^4+243*x^5). (End)
a(n) = A000584(A016777(n)). - Michel Marcus, Apr 06 2017
Sum_{n>=0} 1/a(n) = 2*Pi^5/(3^6*sqrt(3)) + 121*zeta(5)/3^5. - Amiram Eldar, Mar 29 2022

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.

Original entry on oeis.org

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

Views

Author

Peter Luschny, May 08 2013

Keywords

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)

Examples

			[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 k = 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)

Links

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.

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, 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 *)
  • 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
  • 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)

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 k: T(n, k) = (1/(n - k))*((n/2)*(4 + 3*k)*T(n-1, k) + k*Sum_{p=k..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

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

Links

Crossrefs

Cf. A048993 (m=1), A039755 (m=2), A225469 (m=4).
Cf. A075498, A111577. Columns: A000079, A016127, A016297, A025999. A225466, A225472.

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

A225118 Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 4}(x) in descending order.

Original entry on oeis.org

1, 3, 1, 9, 22, 1, 27, 235, 121, 1, 81, 1996, 3446, 620, 1, 243, 15349, 63854, 40314, 3119, 1, 729, 112546, 963327, 1434812, 422087, 15618, 1, 2187, 806047, 12960063, 37898739, 26672209, 4157997, 78117, 1, 6561, 5705752, 162711868, 840642408, 1151050534
Offset: 0

Views

Author

Peter Luschny, May 02 2013

Keywords

Comments

The row sums equal the quadruple factorial numbers A047053 and the alternating row sums, i.e., sum((-1)^k*T(n,k),k=0..n), are up to a sign A079858. - Johannes W. Meijer, May 04 2013

Examples

			[0]  1
[1]  3*x   +    1
[2]  9*x^2 +   22*x   +    1
[3] 27*x^3 +  235*x^2 +  121*x   + 1
[4] 81*x^4 + 1996*x^3 + 3446*x^2 + 620*x + 1
...
The triangle T(n, k) begins:
n\k
0:    1
1:    3      1
2:    9     22        1
3:   27    235      121        1
4:   81   1996     3446      620        1
5:  243  15349    63854    40314     3119       1
6:  729 112546   963327  1434812   422087   15618     1
7: 2187 806047 12960063 37898739 26672209 4157997 78117 1
...
row n=8: 6561 5705752 162711868 840642408 1151050534 442372648 39531132 390616 1,
row n=9: 19683 40156777 1955297356 16677432820 39523450714 29742429982 6818184988 367889284 1953115 1.
... - _Wolfdieter Lang_, Apr 12 2017

Links

Crossrefs

Coefficients of A_{n,1}(x) = A008292, coefficients of A_{n,2}(x) = A060187, coefficients of A_{n,3}(x) = A225117. A123125, A225467, A225469, A225473.

Programs

  • Maple
    gf := proc(n, k) local f; f := (x,t) -> x*exp(t*x/k)/(1-x*exp(t*x));
series(f(x,t), t, n+2); ((1-x)/x)^(n+1)*k^n*n!*coeff(%, t, n):
collect(simplify(%), x) end:
seq(print(seq(coeff(gf(n, 4), x, n-k), k=0..n)), n=0..6);
# Recurrence:
P := proc(n,x) option remember; if n = 0 then 1 else
  (n*x+(1/4)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
  expand(%) fi end:
A225117 := (n,k) -> 4^n*coeff(P(n,x),x,n-k):
seq(print(seq(A225117(n,k), k=0..n)), n=0..5);  # Peter Luschny, Mar 08 2014
  • Mathematica
    gf[n_, k_] := Module[{f, s}, f[x_, t_] := x*Exp[t*x/k]/(1-x*Exp[t*x]); s = Series[f[x, t], {t, 0, n+2}]; ((1-x)/x)^(n+1)*k^n*n!*SeriesCoefficient[s, {t, 0, n}]]; Table[Table[SeriesCoefficient[gf[n, 4], {x, 0, n-k}], {k, 0, n}], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
  • Sage
    @CachedFunction
def EB(n, k, x):  # Modified cardinal B-splines
    if n == 1: return 0 if (x < 0) or (x >= 1) else 1
    return k*x*EB(n-1, k, x) + k*(n-x)*EB(n-1, k, x-1)
def EulerianPolynomial(n, k): # Generalized Eulerian polynomials
    R. = ZZ[]
    if x == 0: return 1
    return add(EB(n+1, k, m+1/k)*x^m for m in (0..n))
[EulerianPolynomial(n, 4).coefficients()[::-1] for n in (0..5)]

Formula

G.f. of the polynomials is gf(n, k) = k^n*n!*(1/x-1)^(n+1)[t^n](x*e^(t*x/k)*(1-x*e(t*x))^(-1)) for k = 4; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
From Wolfdieter Lang, Apr 12 2017 : (Start)
E.g.f. of row polynomials (rising powers of x): (1-x)*exp(3*(1-x)*z)/(1-y*exp(4*(1-x)*z)), i.e. e.g.f. of the triangle.
E.g.f. for the row polynomials with falling powers of x (A_{n, 4}(x) of the name): (1-x)*exp((1-x)*z)/(1 - x*exp(4*(1-x)*z)).
T(n, k) = Sum_{j=0..k} (-1)^(k-j) * binomial(n+1,k-j) * (3+4*j)^n, 0 <= k <= n.
Recurrence: T(n, k) = (4*(n-k) + 1)*T(n-1, k-1) + (3 + 4*k)*T(n-1, k), n >= 1, with T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k. (End)
In terms of Euler's triangle = A123125: T(n, k) = Sum_{m=0..n} (binomial(n, m)*3^(n-m)*4^m*Sum_{p=0..k} (-1)^(k-p)*binomial(n-m, k-p)*A123125(m, p)), 0 <= k <= n. - Wolfdieter Lang, Apr 13 2017

A284861 Triangle read by rows: T(n, k) = S2[3,1](n, k)*k! with the Sheffer triangle S2[3,1] = (exp(x), exp(3*x) -1) given in A282629.

Original entry on oeis.org

1, 1, 3, 1, 15, 18, 1, 63, 216, 162, 1, 255, 1890, 3564, 1944, 1, 1023, 14760, 52650, 68040, 29160, 1, 4095, 109458, 659340, 1516320, 1487160, 524880, 1, 16383, 790776, 7578522, 27624240, 46539360, 36741600, 11022480, 1, 65535, 5633730, 82902204, 450057384, 1158993360, 1535798880, 1014068160, 264539520
Offset: 0

Views

Author

Wolfdieter Lang, Apr 09 2017

Keywords

Comments

This is a generalization of triangle A131689(n, k) = Stirling2(n, k)*k!, because S2[3,1] is a generalization of the Stirling2 triangle written as S2[1,0].
This triangle appears in the o.g.f. G(3,1;n,x) of the powers {(1+3*m)^n}{m>=0} as G(3,1;n,x) = Sum{k>=0..n} T(n, k)*x^k / (1-x)^k.
This triangle is also related to the generalized row reversed Euler triangle rEu[3,1] with row polynomial rEu(3,1;n,x) = Sum_{m=0..n} rEu(3,1;n,m)*x^m with rEu(3,1;n,m) = Sum_{j=0..m} (-1)^(m-j)*binomial(n-j, m-j)*T(n, m). This follows from the above given o.g.f. of powers G(3,1;n,x) = rEu(3,1;n,x)/(1-x)^(n+1). The Euler triangle E[3,1] (row reversed rEu[3,1] is given in A225117. See a formula below.
The e.g.f. of the row polynomials R(3,1;n,x) = Sum_{m=0..n} T(n, m)*x^m follows from the e.g.f. of the row polynomials of the Sheffer triangle A282629. See the formula section.
The diagonal sequence is A032031(k) = k!*3^k.
The row sums give unsigned A151919, and the alternating row sums give A122803.
The first column k sequences divided by A032031(k) are A000012, A002450 (with a leading 0), A016223, A021874. For the e.g.f.s and o.g.f.s see below.

Examples

			The triangle T(n, k) begins
n\k 0     1      2       3        4        5        6        7 ...
0:  1
1:  1     3
2:  1    15     18
3:  1    63    216     162
4:  1   255   1890    3564     1944
5:  1  1023  14760   52650    68040    29160
6:  1  4095 109458  659340  1516320  1487160   524880
7:  1 16383 790776 7578522 27624240 46539360 36741600 11022480
...
row n=8: 1 65535 5633730 82902204 450057384 1158993360 1535798880 1014068160 264539520,
row n=9: 1 262143 39829320 879725610 6845572440 25294754520 50042059200 54561276000 30951123840 7142567040,
row n=10: 1 1048575 280378098 9155719980 99549149040 507399658920 1406104706160 2251231315200 2083248720000 1035672220800 214277011200.
------------------------------------------------------------------
T(2, 1) =  -1 + 4^2 = 15 = 2*A225117(2,2) + 1*A225117(2,1) = 2*1 + 1*13.

Links

Crossrefs

Cf. A000012, A002450, A016223, A021874, A032031, A111577, A122803, A131689, |A151919|, A225117, A282629, A019538, A145901.

Programs

  • Mathematica
    Table[Sum[Binomial[k, m] (-1)^(k - m) (1 + 3m)^n, {m, 0, k}], {n, 0, 10}, {k, 0, n}]// Flatten (* Indranil Ghosh, Apr 09 2017 *)
  • PARI
    for(n=0, 10, for(k=0, n, print1(sum(m=0, k, binomial(k, m) * (-1)^(k - m)*(1 + 3*m)^n),", "); ); print();) \\ Indranil Ghosh, Apr 09 2017
  • Python
    # Indranil Ghosh, Apr 09 2017
from sympy import binomial
for n in range(11):
    print([sum([binomial(k, m)*(-1)**(k - m)*(1 + 3*m)**n for m in range(k + 1)]) for k in range(n + 1)])

Formula

E.g.f. of the row polynomials R(n, x) (see a comment above) is exp(z)/(1 - x*(exp(3*z) - 1)). This is the e.g.f. for the triangle.
T(n, k) = Sum_{m=0..k} binomial(k, m)*(-1)^(k-m)*(1+ 3*m)^n, 0 <= k <= n.
T(n, k) = Sum_{m=0..k} binomial(n-m, k-m)*A225117(n,n-m), 0 <= k <= n.
Three term recurrence: 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) + (1+3*k)*T(n-1, k) for n >= 1. See A282629.
The column k sequence has e.g.f. exp(x)*(exp(3*x) - 1)^k (from the Sheffer property of A282629).
The o.g.f. is A032031(k)*x^k/Product_{j=0..k} (1 - (1+3*j)*x).
From Peter Bala, Jan 12 2018: (Start)
n-th row polynomial R(n,x) = (1 + 3*x) o (1 + 3*x) o ... o (1 + 3*x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E14 in the Bala link. Cf. A145901.
R(n,x) = Sum_{k = 0..n} binomial(n,k)*3^k*F(k,x) where F(k,x) is the Fubini polynomial of order k, the k-th row polynomial of A019538. (End)

A337997 Triangle read by rows, generalized Eulerian polynomials evaluated at x = 1.

Original entry on oeis.org

1, 0, 1, 0, 2, 8, 0, 6, 48, 162, 0, 24, 384, 1944, 6144, 0, 120, 3840, 29160, 122880, 375000, 0, 720, 46080, 524880, 2949120, 11250000, 33592320, 0, 5040, 645120, 11022480, 82575360, 393750000, 1410877440, 4150656720
Offset: 0

Views

Author

Peter Luschny, Oct 07 2020

Keywords

Examples

			Polynomial triangle starts:
[0] 1
[1] 0, 1
[2] 0, 1+x, x^2+6*x+1
[3] 0, x^2+4*x+1, x^3+23*x^2+23*x+1, 8*x^3+93*x^2+60*x+1
[4] 0, x^3+11*x^2+11*x+1, x^4+76*x^3+230*x^2+76*x+1, 16*x^4+545*x^3+1131*x^2+251*x+
1, 81*x^4+1996*x^3+3446*x^2+620*x+1
Integer triangle starts:
[0] 1
[1] 0,    1
[2] 0,    2,      8
[3] 0,    6,     48,      162
[4] 0,   24,    384,     1944,     6144
[5] 0,  120,   3840,    29160,   122880,    375000
[6] 0,  720,  46080,   524880,  2949120,  11250000,   33592320
[7] 0, 5040, 645120, 11022480, 82575360, 393750000, 1410877440, 4150656720

Links

Crossrefs

Cf. A225116, A225117, A225118, A337996, A008292, A060187, A173018, A123125, A284861.

Programs

  • Maple
    # Two alternative implementations are given in the link.
GeneralizedEulerianPolynomial := proc(n, k, x) local S;
   if n = 0 then  return 1 fi;
   S := m -> add((-1)^j*binomial(n+1,j)*(k*(m-j)+1)^n*signum(k*(m-j)+1),j=0..n+1);
   add(S(m)*x^m, m=0..n)/2 end:
T := (n, k) -> subs(x=1, GeneralizedEulerianPolynomial(n, k, x)):
for n from 0 to 6 do seq(T(n, k), k=0..n) od;

Formula

The polynomials are defined P(0,0,x)=1 and P(n,k,x) = (1/2)*Sum_{m=0..n} S(m)*x^m where S(m) = Sum_{j=0..n+1}(-1)^j*binomial(n+1,j)*(k*(m-j)+1)^n*signum(k*(m-j)+1).
T(n, k) = P(n, k, 1).
T(n, k) = n!*k^n. - Hugo Pfoertner, Oct 07 2020
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