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 13 results. Next

A075515 Fifth column of triangle A075498.

Original entry on oeis.org

1, 45, 1260, 28350, 563031, 10333575, 179866170, 3016747800, 49263275061, 788796913905, 12445575859080, 194186867360850, 3004103990159091, 46168557763591035, 705914973500103990, 10750288516418083500
Offset: 0

Views

Author

Wolfdieter Lang, Oct 02 2002

Keywords

Comments

The e.g.f. given below is Sum_{m=0..4} A075513(5,m)*exp(3*(m+1)*x)/4!.

Crossrefs

Formula

a(n) = A075498(n+5, 5) = (3^n)*S2(n+5, 5) with S2(n, m) := A008277(n, m) (Stirling2).
a(n) = Sum_{m=0..4} A075513(5, m)*exp((m+1)*3)^n/4!.
G.f.: 1/Product_{k=1..5} (1 - 3*k*x).
E.g.f.: (d^5/dx^5)(((exp(3*x)-1)/3)^5)/5! = (exp(3*x) - 64*exp(6*x) + 486*exp(9*x) - 1024*exp(12*x) + 625*exp(15*x))/4!.

A075516 Sixth column of triangle A075498.

Original entry on oeis.org

1, 63, 2394, 71442, 1848987, 43615341, 964942308, 20385709344, 416206043253, 8280505692459, 161494678323342, 3101091077181006, 58823743379417199, 1104995938593100617, 20595841868175915096
Offset: 0

Views

Author

Wolfdieter Lang, Oct 02 2002

Keywords

Comments

The e.g.f. given below is Sum_{m=0..5} A075513(6,m)*exp(3*(m+1)*x)/5!.

Crossrefs

Formula

a(n) = A075498(n+6, 6) = (3^n)*S2(n+6, 6) with S2(n, m) := A008277(n, m) (Stirling2).
a(n) = Sum_{m=0..5} A075513(6, m)*((m+1)*3)^n/5!.
G.f.: 1/Product_{k=1..6} (1 - 3*k*x).
E.g.f.: (d^6/dx^6)(((exp(3*x)-1)/3)^6)/6! = (-exp(3*x) + 160*exp(6*x) - 2430*exp(9*x) + 10240*exp(12*x) - 15625*exp(15*x) + 7776*exp(18*x))/5!.

A075906 Seventh column of triangle A075498.

Original entry on oeis.org

1, 84, 4158, 158760, 5182947, 152457228, 4166544096, 107883135360, 2681751885813, 64597295294532, 1518037879508514, 34979886546859800, 793401360863472999, 17766424516726033596, 393690756719422620612
Offset: 0

Views

Author

Wolfdieter Lang, Oct 02 2002

Keywords

Comments

The e.g.f. given below is Sum_{m=0..6} A075513(7,m)*exp(3*(m+1)*x)/6!.

Crossrefs

Cf. A075516.

Formula

a(n) = A075498(n+7, 7) = (3^n)*S2(n+7, 7) with S2(n, m) := A008277(n, m) (Stirling2).
a(n) = Sum_{m=0..6} A075513(7, m)*((m+1)*3)^n/6!.
G.f.: 1/Product_{k=1..7} (1 - 3*k*x).
E.g.f.: (d^7/dx^7)(((exp(3*x)-1)/3)^7)/7! = (exp(3*x) - 384*exp(6*x) + 10935*exp(9*x) - 81920*exp(12*x) + 234375*exp(15*x) - 279936*exp(18*x) + 117649*exp(21*x))/6!.

A004212 Shifts one place left under 3rd-order binomial transform.

Original entry on oeis.org

1, 1, 4, 19, 109, 742, 5815, 51193, 498118, 5296321, 60987817, 754940848, 9983845261, 140329768789, 2087182244308, 32725315072135, 539118388883449, 9304591246975030, 167804098493079547, 3155000165773280893
Offset: 0

Views

Author

Keywords

Comments

Equals the eigensequence of triangle A027465, the cube of Pascal's triangle. - Gary W. Adamson, Apr 10 2009
Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+3 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=3, otherwise F(k+1)=F(k); see example and Fxtbook link. - Joerg Arndt, Apr 30 2011

Examples

			Restricted growth strings: a(0)=1 corresponds to the empty string, a(1)=1 to [0],
a(2)=3 to [00], [01], [02], and [03], a(3) = 19 to
        RGS           F
01:  [ 0 0 0 ]    [ 0 0 0 ]
02:  [ 0 0 1 ]    [ 0 0 0 ]
03:  [ 0 0 2 ]    [ 0 0 0 ]
04:  [ 0 0 3 ]    [ 0 0 3 ]
05:  [ 0 1 0 ]    [ 0 0 0 ]
06:  [ 0 1 1 ]    [ 0 0 0 ]
07:  [ 0 1 2 ]    [ 0 0 0 ]
08:  [ 0 1 3 ]    [ 0 0 3 ]
09:  [ 0 2 0 ]    [ 0 0 0 ]
10:  [ 0 2 1 ]    [ 0 0 0 ]
11:  [ 0 2 2 ]    [ 0 0 0 ]
12:  [ 0 2 3 ]    [ 0 0 3 ]
13:  [ 0 3 0 ]    [ 0 3 3 ]
14:  [ 0 3 1 ]    [ 0 3 3 ]
15:  [ 0 3 2 ]    [ 0 3 3 ]
16:  [ 0 3 3 ]    [ 0 3 3 ]
17:  [ 0 3 4 ]    [ 0 3 3 ]
18:  [ 0 3 5 ]    [ 0 3 3 ]
19:  [ 0 3 6 ]    [ 0 3 6 ]
- _Joerg Arndt_, Apr 30 2011
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A075498 (row sums).
Cf. A027465. - Gary W. Adamson, Apr 10 2009
Cf. A004211 (RGS where s(k)<=F(k)+2), A004213 (s(k)<=F(k)+4), A005011 (s(k)<=F(k)+5), A000110 (s(k)<=F(k)+1). - Joerg Arndt, Apr 30 2011
Cf. A009235.

Programs

  • Mathematica
    Table[Sum[StirlingS2[n,k] 3^(-k+n),{k,n}],{n,20}] (* Vincenzo Librandi, May 21 2012 *)
    Table[3^n BellB[n, 1/3], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(3^(n-k)*binomial(n-1,k-1)*a(k-1),k,1,n); /* Vladimir Kruchinin, Nov 28 2011 */
  • PARI
    x='x+O('x^66); /* that many terms */
    egf=exp(intformal(exp(3*x))); /* =  1 + x + 2*x^2 + 19/6*x^3 + 109/24*x^4 + ... */
    /* egf=exp(1/3*(exp(3*x)-1)) */ /* alternative computation */
    Vec(serlaplace(egf)) /* show terms */ /* Joerg Arndt, Apr 30 2011 */
    

Formula

a_n = Sum_{k=0..n} 3^(n-k)*Stirling2(n, k). - Emeric Deutsch, Feb 11 2002
E.g.f.: exp((exp(3*x)-1)/3).
O.g.f. A(x) satisfies A'(x)/A(x) = e^(3*x).
E.g.f.: exp(Integral_{t = 0..x} exp(3*t)). - Joerg Arndt, Apr 30 2011
O.g.f.: Sum_{k>=0} x^k/Product_{j=1..k} (1-3*j*x). - Joerg Arndt, Apr 30 2011
Hankel transform is A000178(n)*3^C(n+1,2). - Paul Barry, Mar 31 2008
Define f_1(x), f_2(x), ... such that f_1(x)=e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n) = e^(-1/2)*3^{n-1}*f_n(1/3). - Milan Janjic, May 30 2008
a(n) = the upper left term in M^n, M = the following infinite square production matrix:
1, 3, 0, 0, 0, ...
1, 1, 3, 0, 0, ...
1, 2, 1, 3, 0, ...
1, 3, 3, 1, 3, ...
... (in which a diagonal of (3,3,3,...) is appended to the right of Pascal's triangle). - Gary W. Adamson, Jul 29 2011
G.f. satisfies A(x) = 1+x/(1-3*x)*A(x/(1-3*x)). a(n) = Sum_{k=1..n} 3^(n-k)*binomial(n-1,k-1)*a(k-1), n > 0, a(0)=1. - Vladimir Kruchinin, Nov 28 2011 [corrected by Ilya Gutkovskiy, May 02 2019]
From Peter Bala, May 16 2012: (Start)
Recurrence equation: a(n+1) = Sum_{k = 0..n} 3^(n-k)*C(n,k)*a(k). Written umbrally this is a(n+1) = (a + 3)^n (expand the binomial and replace a^k with a(k)). More generally, a*f(a) = f(a+3) holds umbrally for any polynomial f(x). An inductive argument then establishes the umbral recurrence a*(a-3)*(a-6)*...*(a-3*(n-1)) = 1 with a(0) = 1. Compare with the Bell numbers B(n) = A000110(n), which satisfy the umbral recurrence B*(B-1)*...*(B-(n-1)) = 1 with B(0) = 1.
Touchard's congruence holds: for prime p not equal to 3, a(p+k) == (a(k) + a(k+1)) (mod p) for k = 0,1,2,... (adapt the proof of Theorem 10.1 in Gessel). In particular, a(p) == 2 (mod p) for prime p <> 3. (End)
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-x*3*k)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 16 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 3*x^2*(k+1)/( 3*x^2*(k+1) - (1-x-3*x*k)*(1-4*x-3*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
a(n) ~ 3^n * n^n * exp(n/LambertW(3*n) - 1/3 - n) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^n). - Vaclav Kotesovec, Jul 15 2021
a(n) = exp(-1/3)*Sum_{n >= 0} (3*n)^k/(n!*3^n). - Peter Bala, Jun 29 2024

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)
		

Crossrefs

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

A075497 Stirling2 triangle with scaled diagonals (powers of 2).

Original entry on oeis.org

1, 2, 1, 4, 6, 1, 8, 28, 12, 1, 16, 120, 100, 20, 1, 32, 496, 720, 260, 30, 1, 64, 2016, 4816, 2800, 560, 42, 1, 128, 8128, 30912, 27216, 8400, 1064, 56, 1, 256, 32640, 193600, 248640, 111216, 21168, 1848, 72, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 02 2002

Keywords

Comments

This is a lower triangular infinite matrix of the Jabotinsky type. See the D. E. Knuth reference given in A039692 for exponential convolution arrays.
The row polynomials p(n,x) := Sum_{m=1..n} a(n,m)x^m, n >= 1, have e.g.f. J(x; z)= exp((exp(2*z) - 1)*x/2) - 1.
Subtriangle of (0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, 12, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 13 2013
Also the inverse Bell transform of the double factorial of even numbers Product_ {k=0..n-1} (2*k+2) (A000165). For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015
This is the exponential Riordan array [exp(2*x), (exp(2*x) - 1)/2] belonging to the derivative subgroup of the exponential Riordan group. In the notation of Corcino, this is the triangle of (2, 2)-Stirling numbers of the second kind. A factorization of the array as an infinite product is given in the example section. - Peter Bala, Feb 20 2025

Examples

			Triangle begins:
  [1];
  [2,1];
  [4,6,1]; p(3,x) = x*(4 + 6*x + x^2).
  ...;
Triangle (0, 2, 0, 4, 0, 6, 0, 8, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, ...) begins:
  1
  0,  1
  0,  2,   1
  0,  4,   6,   1
  0,  8,  28,  12,  1
  0, 16, 120, 100, 20, 1. - _Philippe Deléham_, Feb 13 2013
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    6   1       |  =  | 4   4   1      ||0  2   1       ||0  0  1       | ...
| 8   28  12   1   |     | 8  12   6  1   ||0  4   4  1    ||0  0  2  1    |
|16  120 100  20  1|     |16  32  24  8  1||0  8  12  6  1 ||0  0  4  4  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - 2*x)) = P^2, where P denotes Pascal's triangle. See A038207. Cf. A143494. (End)
		

Crossrefs

Row sums are A004211.

Programs

  • Maple
    with(combinat):
    b:= proc(n, i) option remember; expand(`if`(n=0, 1,
           `if`(i<1, 0, add(x^j*multinomial(n, n-i*j, i$j)/j!*add(
            binomial(i, 2*k), k=0..i/2)^j*b(n-i*j, i-1), j=0..n/i))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)):
    seq(T(n), n=1..12);  # Alois P. Heinz, Aug 13 2015
    # Alternatively, giving the triangle in the form displayed in the Example section:
    gf := exp(x*exp(z)*sinh(z)):
    X := n -> series(gf, z, n+2):
    Z := n -> n!*expand(simplify(coeff(X(n), z, n))):
    A075497_row := n -> op(PolynomialTools:-CoefficientList(Z(n), x)):
    seq(A075497_row(n), n=0..9); # Peter Luschny, Jan 14 2018
  • Mathematica
    Table[(2^(n - m)) StirlingS2[n, m], {n, 9}, {m, n}] // Flatten (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    for(n=1, 11, for(m=1, n, print1(2^(n - m) * stirling(n, m, 2),", ");); print();) \\ Indranil Ghosh, Mar 25 2017
  • Sage
    # uses[inverse_bell_transform from A265605]
    multifact_2_2 = lambda n: prod(2*k + 2 for k in (0..n-1))
    inverse_bell_matrix(multifact_2_2, 9) # Peter Luschny, Dec 31 2015
    

Formula

a(n, m) = (2^(n-m)) * Stirling2(n, m).
a(n, m) = (Sum_{p=0..m-1} A075513(m, p)*((p+1)*2)^(n-m))/(m-1)! for n >= m >= 1, else 0.
a(n, m) = 2*m*a(n-1, m) + a(n-1, m-1), n >= m >= 1, else 0, with a(n, 0) := 0 and a(1, 1)=1.
G.f. for m-th column: (x^m)/Product_{k=1..m}(1-2*k*x), m >= 1.
E.g.f. for m-th column: (((exp(2*x)-1)/2)^m)/m!, m >= 1.
The row polynomials in t are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+2*x)*d/dx. Cf. A008277. - Peter Bala, Nov 25 2011
From Peter Bala, Jan 13 2018: (Start)
n-th row polynomial R(n,x)= x o x o ... o x (n factors), where o is the deformed Hadamard product of power series defined in Bala, section 3.1.
R(n+1,x)/x = (x + 2) o (x + 2) o...o (x + 2) (n factors).
R(n+1,x) = x*Sum_{k = 0..n} binomial(n,k)*2^(n-k)*R(k,x).
Dobinski-type formulas: R(n,x) = exp(-x/2)*Sum_{i >= 0} (2*i)^n* (x/2)^i/i!; 1/x*R(n+1,x) = exp(-x/2)*Sum_{i >= 0} (2 + 2*i)^n* (x/2)^i/i!. (End)

A075499 Stirling2 triangle with scaled diagonals (powers of 4).

Original entry on oeis.org

1, 4, 1, 16, 12, 1, 64, 112, 24, 1, 256, 960, 400, 40, 1, 1024, 7936, 5760, 1040, 60, 1, 4096, 64512, 77056, 22400, 2240, 84, 1, 16384, 520192, 989184, 435456, 67200, 4256, 112, 1, 65536, 4177920, 12390400, 7956480, 1779456, 169344, 7392, 144, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 02 2002

Keywords

Comments

This is a lower triangular infinite matrix of the Jabotinsky type. See the Knuth reference given in A039692 for exponential convolution arrays.
The row polynomials p(n,x) := Sum_{m=1..n} a(n,m)x^m, n >= 1, have e.g.f. J(x; z)= exp((exp(4*z) - 1)*x/4) - 1
Also the inverse Bell transform of the quadruple factorial numbers 4^n*n! (A047053) adding 1,0,0,0,... as column 0. For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015

Examples

			[1]; [4,1]; [16,12,1]; ...; p(3,x) = x(16 + 12*x + x^2).
From _Andrew Howroyd_, Mar 25 2017: (Start)
Triangle starts
*     1
*     4      1
*    16     12      1
*    64    112     24      1
*   256    960    400     40     1
*  1024   7936   5760   1040    60    1
*  4096  64512  77056  22400  2240   84   1
* 16384 520192 989184 435456 67200 4256 112 1
(End)
		

Crossrefs

Columns 1-7 are A000302, A016152, A019677, A075907-A075910. Row sums are A004213.

Programs

  • Mathematica
    Table[(4^(n - m)) StirlingS2[n, m], {n, 9}, {m, n}] // Flatten (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    for(n=1, 11, for(m=1, n, print1(4^(n - m) * stirling(n, m, 2),", ");); print();) \\ Indranil Ghosh, Mar 25 2017
  • Sage
    # uses[inverse_bell_transform from A265605]
    # Adds a column 1,0,0,... at the left side of the triangle.
    multifact_4_4 = lambda n: prod(4*k + 4 for k in (0..n-1))
    inverse_bell_matrix(multifact_4_4, 9) # Peter Luschny, Dec 31 2015
    

Formula

a(n, m) = (4^(n-m)) * stirling2(n, m).
a(n, m) = (Sum_{p=0..m-1} A075513(m, p)*((p+1)*4)^(n-m))/(m-1)! for n >= m >= 1, else 0.
a(n, m) = 4m*a(n-1, m) + a(n-1, m-1), n >= m >= 1, else 0, with a(n, 0) := 0 and a(1, 1)=1.
G.f. for m-th column: (x^m)/Product_{k=1..m}(1-4k*x), m >= 1.
E.g.f. for m-th column: (((exp(4x)-1)/4)^m)/m!, m >= 1.

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

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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

A016137 Expansion of 1/((1-3*x)*(1-6*x)).

Original entry on oeis.org

1, 9, 63, 405, 2511, 15309, 92583, 557685, 3352671, 20135709, 120873303, 725416965, 4353033231, 26119793709, 156723545223, 940355620245, 5642176768191, 33853189749309, 203119525916343, 1218718317759525, 7312313393341551, 43873890820402509, 263243376303474663, 1579460351964026805, 9476762394213697311
Offset: 0

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Author

Keywords

Crossrefs

Second column of triangle A075498.

Programs

Formula

a(n) = (3^n)*Stirling2(n+2, 2), n >= 0, with Stirling2(n, m) = A008277(n, m).
a(n) = 2*6^n - 3^n.
E.g.f.: (d^2/dx^2)((((exp(3*x)-1)/3)^2)/2!) = -exp(3*x) + 2*exp(6*x).
With leading zero, this is (6^n - 3^n)/3, the binomial transform of A016127 (with extra leading zero). - Paul Barry, Aug 20 2003
With leading zero, this is the fourth binomial transform of A001045, with a(n) = (2^n-1)(3^n/3 - 0^n/3) = A000225(n)*(A000244(n-1) - 0^n/3). - Paul Barry, Apr 28 2004
a(n) = Sum_{k=0..n} A100852(n,k). - Reinhard Zumkeller, Nov 20 2004
Sum_{k=1..n} 3^(k-1)*3^(n-k)*binomial(n, k). - Zerinvary Lajos, Sep 24 2006
a(n) = 9*a(n-1) - 18*a(n-2), n >= 2. - Vincenzo Librandi, Mar 14 2011

Extensions

More terms added by G. C. Greubel, Nov 14 2024

A017933 Expansion of 1/((1-3x)(1-6x)(1-9x)).

Original entry on oeis.org

1, 18, 225, 2430, 24381, 234738, 2205225, 20404710, 186995061, 1703091258, 15448694625, 139763668590, 1262226050541, 11386154248578, 102632111782425, 924629361662070, 8327306431726821, 74979611075290698
Offset: 0

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Author

Keywords

Crossrefs

Third column of triangle A075498.

Programs

  • Magma
    m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-3*x)*(1-6*x)*(1-9*x)))); /* or */ I:=[1, 18, 225]; [n le 3 select I[n] else 18*Self(n-1)-99*Self(n-2)+162*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jul 02 2013
  • Mathematica
    CoefficientList[Series[1 / ((1 - 3 x) (1 - 6 x) (1 - 9 x)), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 02 2013 *)
    LinearRecurrence[{18,-99,162},{1,18,225},20] (* Harvey P. Dale, Sep 09 2023 *)
  • PARI
    Vec(1/((1-3*x)*(1-6*x)*(1-9*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012
    

Formula

a(n) = (3^n)*Stirling2(n+3, 3), n >= 0, with Stirling2(n, m) = A008277(n, m).
a(n) = (3^n - 8*6^n + 9*9^n)/2.
G.f.: 1/((1-3*x)*(1-6*x)*(1-9*x)).
E.g.f.: (d^3/dx^3)((((exp(3*x)-1)/3)^3)/3!) = (exp(3*x) - 8*exp(6*x) + 9*exp(9*x))/2.
a(0)=1, a(1)=18, a(2)=225; for n > 2, a(n) = 18*a(n-1) - 99*a(n-2) + 162*a(n-3). - Vincenzo Librandi, Jul 02 2013
a(n) = 15*a(n-1) - 54*a(n-2) + 3^n. - Vincenzo Librandi, Jul 02 2013
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