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 23 results. Next

A131689 Triangle of numbers T(n,k) = k!*Stirling2(n,k) = A000142(k)*A048993(n,k) read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 6, 6, 0, 1, 14, 36, 24, 0, 1, 30, 150, 240, 120, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 126, 1806, 8400, 16800, 15120, 5040, 0, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 0, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880
Offset: 0

Views

Author

Philippe Deléham, Sep 14 2007

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,6,...] where DELTA is the operator defined in A084938; another version of A019538.
See also A019538: version with n > 0 and k > 0. - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 21 2014: (Start)
T(n,k) gives the number of (k-1)-dimensional faces in the interior of the first barycentric subdivision of the standard (n-1)-dimensional simplex. For example, the barycentric subdivision of the 1-simplex is o--o--o, with 1 interior vertex and 2 interior edges, giving T(2,1) = 1 and T(2,2) = 2.
This triangle is used when calculating the face vectors of the barycentric subdivision of a simplicial complex. Let S be an n-dimensional simplicial complex and write f_k for the number of k-dimensional faces of S, with the usual convention that f_(-1) = 1, so that F := (f_(-1), f_0, f_1,...,f_n) is the f-vector of S. If M(n) denotes the square matrix formed from the first n+1 rows and n+1 columns of the present triangle, then the vector F*M(n) is the f-vector of the first barycentric subdivision of the simplicial complex S (Brenti and Welker, Lemma 2.1). For example, the rows of Pascal's triangle A007318 (but with row and column indexing starting at -1) are the f-vectors for the standard n-simplexes. It follows that A007318*A131689, which equals A028246, is the array of f-vectors of the first barycentric subdivision of standard n-simplexes. (End)
This triangle T(n, k) appears in the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} (x^k/(1 - x)^(k+2))*T(n, k). See also the Eulerian triangle A008292 with a Mar 31 2017 comment for a rewritten form. For the e.g.f. see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
T(n,k) = the number of alignments of length k of n strings each of length 1. See Slowinski. An example is given below. Cf. A122193 (alignments of strings of length 2) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018
The row polynomials R(n,x) are the Fubini polynomials. - Emanuele Munarini, Dec 05 2020
From Gus Wiseman, Feb 18 2022: (Start)
Also the number of patterns of length n with k distinct parts (or with maximum part k), where we define a pattern to be a finite sequence covering an initial interval of positive integers. For example, row n = 3 counts the following patterns:
(1,1,1) (1,2,2) (1,2,3)
(2,1,2) (1,3,2)
(2,2,1) (2,1,3)
(1,1,2) (2,3,1)
(1,2,1) (3,1,2)
(2,1,1) (3,2,1)
(End)
Regard A048994 as a lower-triangular matrix and divide each term A048994(n,k) by n!, then this is the matrix inverse. Because Sum_{k=0..n} (A048994(n,k) * x^n / n!) = A007318(x,n), Sum_{k=0..n} (A131689(n,k) * A007318(x,k)) = x^n. - Natalia L. Skirrow, Mar 23 2023
T(n,k) is the number of ordered partitions of [n] into k blocks. - Alois P. Heinz, Feb 21 2025

Examples

			The triangle T(n,k) begins:
  n\k 0 1    2     3      4       5        6        7        8        9      10 ...
  0:  1
  1:  0 1
  2:  0 1    2
  3:  0 1    6     6
  4:  0 1   14    36     24
  5:  0 1   30   150    240     120
  6:  0 1   62   540   1560    1800      720
  7:  0 1  126  1806   8400   16800    15120     5040
  8:  0 1  254  5796  40824  126000   191520   141120    40320
  9:  0 1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 0 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... reformatted and extended. - _Wolfdieter Lang_, Mar 31 2017
From _Peter Bala_, Feb 04 2018: (Start)
T(4,2) = 14 alignments of length 2 of 4 strings of length 1. Examples include
  (i) A -    (ii) A -    (iii) A -
      B -         B -          - B
      C -         - C          - C
      - D         - D          - D
There are C(4,1) = 4 alignments of type (i) with a single gap character - in column 1, C(4,2) = 6 alignments of type (ii) with two gap characters in column 1 and C(4,3) = 4 alignments of type (iii) with three gap characters in column 1, giving a total of 4 + 6 + 4 = 14 alignments. (End)
		

Crossrefs

Case m=1 of the polynomials defined in A278073.
Cf. A000142 (diagonal), A000670 (row sums), A000012 (alternating row sums), A210029 (central terms).
Cf. A008292, A028246 (o.g.f. and e.g.f. of sums of powers).
A version for partitions is A116608, or by maximum A008284.
A version for compositions is A235998, or by maximum A048004.
Classes of patterns:
- A000142 = strict
- A005649 = anti-run, complement A069321
- A019536 = necklace
- A032011 = distinct multiplicities
- A060223 = Lyndon
- A226316 = (1,2,3)-avoiding, weakly A052709, complement A335515
- A296975 = aperiodic
- A345194 = alternating, up/down A350354, complement A350252
- A349058 = weakly alternating
- A351200 = distinct runs
- A351292 = distinct run-lengths

Programs

  • Julia
    function T(n, k)
        if k < 0 || k > n return 0 end
        if n == 0 && k == 0 return 1 end
        k*(T(n-1, k-1) + T(n-1, k))
    end
    for n in 0:7
        println([T(n, k) for k in 0:n])
    end
    # Peter Luschny, Mar 26 2020
    
  • Maple
    A131689 := (n,k) -> Stirling2(n,k)*k!: # Peter Luschny, Sep 17 2011
    # Alternatively:
    A131689_row := proc(n) 1/(1-t*(exp(x)-1)); expand(series(%,x,n+1)); n!*coeff(%,x,n); PolynomialTools:-CoefficientList(%,t) end:
    for n from 0 to 9 do A131689_row(n) od; # Peter Luschny, Jan 23 2017
  • Mathematica
    t[n_, k_] := k!*StirlingS2[n, k]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
    T[n_, k_] := If[n <= 0 || k <= 0, Boole[n == 0 && k == 0], Sum[(-1)^(i + k) Binomial[k, i] i^(n + k), {i, 0, k}]]; (* Michael Somos, Jul 08 2018 *)
  • PARI
    {T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(k + i) * binomial(k, i) * i^n))};
    /* Michael Somos, Jul 08 2018 */
    
  • SageMath
    @cached_function
    def F(n): # Fubini polynomial
        R. = PolynomialRing(ZZ)
        if n == 0: return R(1)
        return R(sum(binomial(n, k)*F(n - k)*x for k in (1..n)))
    for n in (0..9): print(F(n).list()) # Peter Luschny, May 21 2021

Formula

T(n,k) = k*(T(n-1,k-1) + T(n-1,k)) with T(0,0)=1. Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A000629(n), A033999(n), A000007(n), A000670(n), A004123(n+1), A032033(n), A094417(n), A094418(n), A094419(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. [corrected by Philippe Deléham, Feb 11 2013]
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A000670(n), A122704(n) for x=-1, 0, 1, 2 respectively. - Philippe Deléham, Oct 09 2007
Sum_{k=0..n} (-1)^k*T(n,k)/(k+1) = Bernoulli numbers A027641(n)/A027642(n). - Peter Luschny, Sep 17 2011
G.f.: F(x,t) = 1 + x*t + (x+x^2)*t^2/2! + (x+6*x^2+6*x^3)*t^3/3! + ... = Sum_{n>=0} R(n,x)*t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x. - Philippe Deléham, Feb 11 2013
T(n,k) = [t^k] (n! [x^n] (1/(1-t*(exp(x)-1)))). - Peter Luschny, Jan 23 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See also Bala, Example E8. - Peter Bala, Jan 08 2018

A032033 Stirling transform of A032031.

Original entry on oeis.org

1, 3, 21, 219, 3045, 52923, 1103781, 26857659, 746870565, 23365498683, 812198635941, 31055758599099, 1295419975298085, 58538439796931643, 2848763394161128101, 148537065755389540539, 8261178848690959117605, 488177936257344615487803, 30544839926043868901604261
Offset: 0

Views

Author

Keywords

Comments

Also "AIJ" (ordered, indistinct, labeled) transform of 3,3,3,3...
Third row of array A094416 (generalized ordered Bell numbers).

Crossrefs

Programs

  • Maple
    b:= proc(n, m) option remember;
         `if`(n=0, 3^m*m!, m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..20);  # Alois P. Heinz, Aug 04 2021
  • Mathematica
    a[n_] := PolyLog[-n, 3/4]/4; a[0] = 1; Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Nov 14 2011 *)
    t = 30; Range[0, t]! CoefficientList[Series[1/(4 - 3 Exp[x]), {x, 0, t}], x] (* Vincenzo Librandi, Mar 16 2014 *)
  • PARI
    a(n)=ceil(polylog(-n,3/4)/4) \\ Charles R Greathouse IV, Jul 14 2014
    
  • PARI
    my(N=25,x='x+O('x^N)); Vec(serlaplace(1/(4 - 3*exp(x)))) \\ Joerg Arndt, Jan 15 2024

Formula

E.g.f.: 1/(4-3*exp(x)).
a(n) = 3 * A050352(n), n > 0.
a(n) = Sum_{k=0..n} Stirling2(n, k) * (3^k) * k!.
a(n) = (1/4) * Sum_{k>=0} k^n * (3/4)^k. - Karol A. Penson, Jan 25 2002
a(n) = Sum_{k=0..n} A131689(n,k)*3^k. - Philippe Deléham, Nov 03 2008
G.f. A(x)=B(x)/x, where B(x)=x+3*x^2+21*x^3+... = Sum_{n>=1} b(n)*x^n satisfies 4*B(x)-x = 3*B(x/(1-x)), and b(n)=3*Sum_{k=1..n-1} binomial(n-1,k-1)*b(k), b(1)=1. - Vladimir Kruchinin, Jan 27 2011
a(n) = log(4/3)*Integral_{x = 0..inf} (floor(x))^n * (4/3)^(-x) dx. - Peter Bala, Feb 14 2015
a(0) = 1; a(n) = 3 * Sum_{k=1..n} binomial(n,k) * a(n-k). - Ilya Gutkovskiy, Jan 17 2020
a(0) = 1; a(n) = 3 * a(n-1) - 4 * Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023
a(n) = (3/4) * Sum_{k=0..n} 4^k * (-1)^(n-k) * k! * Stirling2(n,k) for n > 0. - Seiichi Manyama, Jun 01 2025

A094417 Generalized ordered Bell numbers Bo(4,n).

Original entry on oeis.org

1, 4, 36, 484, 8676, 194404, 5227236, 163978084, 5878837476, 237109864804, 10625889182436, 523809809059684, 28168941794178276, 1641079211868751204, 102961115527874385636, 6921180217049667005284, 496267460209336700111076, 37807710659221213027893604
Offset: 0

Views

Author

Ralf Stephan, May 02 2004

Keywords

Comments

Fourth row of array A094416, which has more information.

Crossrefs

Programs

  • Magma
    m:=20; R:=LaurentSeriesRing(RationalField(), m); b:=Coefficients(R!(1/(5 - 4*Exp(x)))); [Factorial(n-1)*b[n]: n in [1..m]]; // Bruno Berselli, Mar 17 2014
    
  • Maple
    a:= proc(n) option remember;
          `if`(n=0, 1, 4* add(binomial(n, k) *a(k), k=0..n-1))
        end:
    seq(a(n), n=0..20);
  • Mathematica
    max = 16; f[x_] := 1/(5-4*E^x); CoefficientList[Series[f[x], {x, 0, max}], x]*Range[0, max]! (* Jean-François Alcover, Nov 14 2011, after g.f. *)
  • PARI
    my(N=25,x='x+O('x^N)); Vec(serlaplace(1/(5 - 4*exp(x)))) \\ Joerg Arndt, Jan 15 2024
  • SageMath
    def A094416(n,k): return sum(factorial(j)*n^j*stirling_number2(k,j) for j in range(k+1)) # array
    def A094417(k): return A094416(4,k)
    [A094417(n) for n in range(31)] # G. C. Greubel, Jan 12 2024
    

Formula

E.g.f.: 1/(5 - 4*exp(x)).
a(n) = 4 * A050353(n) for n>0.
a(n) = Sum_{k=0..n} A131689(n,k) * 4^k. - Philippe Deléham, Nov 03 2008
E.g.f.: A(x) with A_n = 4 * Sum_{k=0..n-1} C(n,k) * A_k; A_0 = 1. - Vladimir Kruchinin, Jan 27 2011
G.f.: 2/G(0), where G(k)= 1 + 1/(1 - 8*x*(k+1)/(8*x*(k+1) - 1 + 10*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
a(n) = log(5/4)*int {x = 0..inf} (floor(x))^n * (5/4)^(-x) dx. - Peter Bala, Feb 14 2015
a(0) = 1; a(n) = 4 * a(n-1) - 5 * Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023
From Seiichi Manyama, Jun 01 2025: (Start)
a(n) = (-1)^(n+1)/5 * Li_{-n}(5/4), where Li_{n}(x) is the polylogarithm function.
a(n) = (1/5) * Sum_{k>=0} k^n * (4/5)^k.
a(n) = (4/5) * Sum_{k=0..n} 5^k * (-1)^(n-k) * A131689(n,k) for n > 0. (End)

A094419 Generalized ordered Bell numbers Bo(6,n).

Original entry on oeis.org

1, 6, 78, 1518, 39390, 1277646, 49729758, 2258233998, 117196187550, 6842432930766, 443879517004638, 31674687990494478, 2465744921215207710, 207943837884583262286, 18885506918597311159518, 1837699347783655374914958, 190743171535070652261555870, 21035482423625416328497024206
Offset: 0

Views

Author

Ralf Stephan, May 02 2004

Keywords

Comments

Sixth row of array A094416, which has more information.

Crossrefs

Programs

  • Magma
    A094416:= func< n,k | (&+[Factorial(j)*n^j*StirlingSecond(k,j): j in [0..k]]) >;
    A094419:= func< k | A094416(6,k) >;
    [A094419(n): n in [0..30]]; // G. C. Greubel, Jan 12 2024
    
  • Mathematica
    t = 30; Range[0, t]! CoefficientList[Series[1/(7 - 6 Exp[x]),{x, 0, t}], x] (* Vincenzo Librandi, Mar 16 2014 *)
  • PARI
    my(N=25,x='x+O('x^N)); Vec(serlaplace(1/(7-6*exp(x)))) \\ Joerg Arndt, Jan 15 2024
    
  • PARI
    a(n) = (-1)^(n+1)*polylog(-n, 7/6)/7; \\ Seiichi Manyama, Jun 01 2025
  • SageMath
    def A094416(n,k): return sum(factorial(j)*n^j*stirling_number2(k,j) for j in range(k+1)) # array
    def A094419(k): return A094416(6,k)
    [A094419(n) for n in range(31)] # G. C. Greubel, Jan 12 2024
    

Formula

E.g.f.: 1/(7 - 6*exp(x)).
a(n) = Sum_{k=0..n} A131689(n,k) * 6^k. - Philippe Deléham, Nov 03 2008
a(n) ~ n! / (7*(log(7/6))^(n+1)). - Vaclav Kotesovec, Mar 14 2014
a(0) = 1; a(n) = 6 * Sum_{k=1..n} binomial(n,k) * a(n-k). - Ilya Gutkovskiy, Jan 17 2020
a(0) = 1; a(n) = 6 * a(n-1) - 7 * Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 17 2023
From Seiichi Manyama, Jun 01 2025: (Start)
a(n) = (-1)^(n+1)/7 * Li_{-n}(7/6), where Li_{n}(x) is the polylogarithm function.
a(n) = (1/7) * Sum_{k>=0} k^n * (6/7)^k.
a(n) = (6/7) * Sum_{k=0..n} 7^k * (-1)^(n-k) * A131689(n,k) for n > 0. (End)

A094416 Array read by antidiagonals: generalized ordered Bell numbers Bo(r,n).

Original entry on oeis.org

1, 2, 3, 3, 10, 13, 4, 21, 74, 75, 5, 36, 219, 730, 541, 6, 55, 484, 3045, 9002, 4683, 7, 78, 905, 8676, 52923, 133210, 47293, 8, 105, 1518, 19855, 194404, 1103781, 2299754, 545835, 9, 136, 2359, 39390, 544505, 5227236, 26857659, 45375130, 7087261
Offset: 1

Views

Author

Ralf Stephan, May 02 2004

Keywords

Comments

Also, r times the number of (r+1)-level labeled linear rooted trees with n leaves.
"AIJ" (ordered, indistinct, labeled) transform of {r,r,r,...}.
Stirling transform of r^n*n!, i.e. of e.g.f. 1/(1-r*x).
Also, Bo(r,s) is ((x*d/dx)^n)(1/(1+r-r*x)) evaluated at x=1.
r-th ordered Bell polynomial (A019538) evaluated at n.
Bo(r,n) is the n-th moment of a geometric distribution with probability parameter = 1/(r+1). Here, geometric distribution is the number of failures prior to the first success. - Geoffrey Critzer, Jan 01 2019
Row r (starting at r=0), Bo(r+1, n), is the Akiyama-Tanigawa algorithm applied to the powers of r+1. See Python program below. - Shel Kaphan, May 03 2024

Examples

			Array begins as:
  1,  3,   13,    75,     541,     4683,      47293, ...
  2, 10,   74,   730,    9002,   133210,    2299754, ...
  3, 21,  219,  3045,   52923,  1103781,   26857659, ...
  4, 36,  484,  8676,  194404,  5227236,  163978084, ...
  5, 55,  905, 19855,  544505, 17919055,  687978905, ...
  6, 78, 1518, 39390, 1277646, 49729758, 2258233998, ...
		

Crossrefs

Columns include A014105, A094421.
Main diagonal is A094420.
Antidiagonal sums are A094422.

Programs

  • Magma
    A094416:= func< n,k | (&+[Factorial(j)*n^j*StirlingSecond(k,j): j in [0..k]]) >;
    [A094416(n-k+1,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Jan 12 2024
    
  • Mathematica
    Bo[, 0]=1; Bo[r, n_]:= Bo[r, n]= r*Sum[Binomial[n,k] Bo[r,n-k], {k, n}];
    Table[Bo[r-n+1, n], {r, 10}, {n, r}] // Flatten (* Jean-François Alcover, Nov 03 2018 *)
  • Python
    # The Akiyama-Tanigawa algorithm applied to the powers of r + 1
    # generates the rows. Adds one row (r=0) and one column (n=0).
    # Adapted from Peter Luschny on A371568.
    def f(n, r): return (r + 1)**n
    def ATtransform(r, len, f):
      A = [0] * len
      R = [0] * len
      for n in range(len):
          R[n] = f(n, r)
          for j in range(n, 0, -1):
              R[j - 1] = j * (R[j] - R[j - 1])
          A[n] = R[0]
      return A
    for r in range(8): print([r], ATtransform(r, 8, f)) # Shel Kaphan, May 03 2024
  • SageMath
    def A094416(n,k): return sum(factorial(j)*n^j*stirling_number2(k,j) for j in range(k+1)) # array
    flatten([[A094416(n-k+1,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Jan 12 2024
    

Formula

E.g.f.: 1/(1 + r*(1 - exp(x))).
Bo(r, n) = Sum_{k=0..n} k!*r^k*Stirling2(n, k) = 1/(r+1) * Sum_{k>=1} k^n * (r/(r+1))^k, for r>0, n>0.
Recurrence: Bo(r, n) = r * Sum_{k=1..n} C(n, k)*Bo(r, n-k), with Bo(r, 0) = 1.
Bo(r,0) = 1, Bo(r,n) = r*Bo(r,n-1) - (r+1)*Sum_{j=1..n-1} (-1)^j * binomial(n-1,j) * Bo(r,n-j). - Seiichi Manyama, Nov 17 2023

Extensions

Offset corrected by Geoffrey Critzer, Jan 01 2019

A255927 a(n) = (3/4) * Sum_{k>=0} (3*k)^n/4^k.

Original entry on oeis.org

1, 1, 5, 33, 285, 3081, 40005, 606033, 10491885, 204343641, 4422082005, 105265315233, 2733583519485, 76902684021801, 2329889536156005, 75629701786875633, 2618654297178083085, 96336948993312237561, 3752590641305604502005, 154294551397830418471233, 6677999524135208461382685
Offset: 0

Views

Author

Karol A. Penson, Sep 03 2015

Keywords

Examples

			a(5) = 729*hypergeom([2,2,2,2,2],[1,1,1,1],1/4)/16 = 3081.
		

Crossrefs

Programs

  • Maple
    S:= series(3/(4-exp(3*x)), x, 51):
    seq(coeff(S,x,n)*n!, n=0..50); # Robert Israel, Sep 03 2015
    seq(add(combinat:-eulerian1(n,k)*4^k, k=0..n), n=0..20); # Peter Luschny, Jun 27 2019
  • Mathematica
    a[n_] := 3^(n+1)/4 HurwitzLerchPhi[1/4, -n, 0];
    Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Sep 18 2018 *)
    Eulerian1[0, 0] = 1; Eulerian1[n_, k_] := Sum[(-1)^j (k-j+1)^n Binomial[n+1, j], {j, 0, k+1}]; Table[Sum[Eulerian1[n, k] 4^k, {k, 0, n}], {n, 0, 20}] (* Jean-François Alcover, Jul 13 2019, after Peter Luschny *)
  • PARI
    a(n) = sum(k=0, n, stirling(n,k,2)*k!*3^(n-k)); \\ Michel Marcus, Sep 03 2015

Formula

a(n) = Sum_{k>=0} Stirling2(n,k)*k!*3^(n-k).
E.g.f.: 3/(4-exp(3*x)).
Special values of the generalized hypergeometric function n_F_(n-1):
a(n) = (3^(n+1)/16) * hypergeom([2,2,..2],[1,1,..1],1/4), where the sequence in the first square bracket ("upper" parameters) has n elements all equal to 2 whereas the sequence in the second square bracket ("lower" parameters) has n-1 elements all equal to 1.
Example: a(5) = 729 * hypergeom([2,2,2,2,2],[1,1,1,1],1/4)/16 = 3081.
a(n) is the n-th moment of the discrete weight function W(x) = (3/4)*sum(k>=0, Dirac(x-3*k)/4^k), n>=1.
a(n) ~ n! * 3^(n+1) / ((log(2))^(n+1) * 2^(n+3)). - Vaclav Kotesovec, Jul 09 2018
G.f.: Sum_{j>=0} j!*x^j / Product_{k=1..j} (1 - 3*k*x). - Ilya Gutkovskiy, Apr 04 2019
a(n) = A_{4}(n) where A_{n}(x) are the Eulerian polynomials as defined in A326323. - Peter Luschny, Jun 27 2019

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 18 2018

A090358 G.f. satisfies A^6 = BINOMIAL(A^5).

Original entry on oeis.org

1, 1, 6, 66, 1071, 23151, 627236, 20452976, 779947641, 34050858041, 1674497370602, 91575747294582, 5512402585832847, 362148111801511407, 25783279860096503952, 1977349647140061768364, 162508269041154881377519
Offset: 0

Views

Author

Paul D. Hanna, Nov 26 2003

Keywords

Comments

In general, if A^n = BINOMIAL(A^(n-1)), then for all integer m>0 there exists an integer sequence B such that B^d = BINOMIAL(A^m) where d=gcd(m+1,n). Also, coefficients of A(k*x)^n = k-th binomial transform of coefficients in A(k*x)^(n-1) for all k>0.
In general, if g.f. satisfies A(x)^(k+1) = A(x/(1-x))^k / (1-x), k>0, then a(n) ~ (n-1)! / (k*(k+1) * (log((k+1)/k))^(n+1)). - Vaclav Kotesovec, Nov 19 2014

Examples

			A^6 = BINOMIAL(A090362), since A090362=A^5. Also,
BINOMIAL(A) = A090359^2 since 2=gcd(1+1,6),
BINOMIAL(A^2) = A090360^3 since 3=gcd(2+1,6) and
BINOMIAL(A^3) = A090361^2 since 2=gcd(3+1,6).
		

Crossrefs

Programs

  • Magma
    m:=40;
    f:= func< n,x | Exp((&+[(&+[5^(j-1)*Factorial(j)*StirlingSecond(k,j) *x^k/k: j in [1..k]]): k in [1..n+2]])) >;
    R:=PowerSeriesRing(Rationals(), m+1); // A090358
    Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 08 2023
    
  • Mathematica
    nmax = 16; sol = {a[0] -> 1};
    Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^6 - A[x/(1 - x)]^5/(1 - x) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
    sol /. Rule -> Set;
    a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)
    With[{m=40}, CoefficientList[Series[Exp[Sum[Sum[5^(j-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Jun 08 2023 *)
  • PARI
    {a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^5,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^6+B);polcoeff(A,n,x))}
    
  • SageMath
    m=50
    def f(n, x): return exp(sum(sum( 5^(j-1)*factorial(j)* stirling_number2(k,j)*x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
    def A090358_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( f(m,x) ).list()
    A090358_list(m-5) # G. C. Greubel, Jun 08 2023

Formula

G.f. satisfies: A(x)^6 = A(x/(1-x))^5/(1-x).
a(n) ~ (n-1)! / (30 * (log(6/5))^(n+1)). - Vaclav Kotesovec, Nov 19 2014
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*5^(k-1) = 1/5*A094418(n) for n >= 1. - Peter Bala, May 26 2015
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/30) * (5/6)^k). - Seiichi Manyama, May 26 2025

A346984 Expansion of e.g.f. 1 / (6 - 5 * exp(x))^(1/5).

Original entry on oeis.org

1, 1, 7, 85, 1495, 34477, 983983, 33476437, 1322441575, 59492222077, 3002578396255, 168005805229285, 10321907081030167, 690761732852321677, 50015387402165694607, 3895721046926471861365, 324805103526730206129607, 28861947117644330678207389, 2722944810091827410698112959
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 09 2021

Keywords

Comments

Stirling transform of A008548.

Crossrefs

Programs

  • Maple
    g:= proc(n) option remember; `if`(n<2, 1, (5*n-4)*g(n-1)) end:
    b:= proc(n, m) option remember;
         `if`(n=0, g(m), m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..18);  # Alois P. Heinz, Aug 09 2021
  • Mathematica
    nmax = 18; CoefficientList[Series[1/(6 - 5 Exp[x])^(1/5), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[StirlingS2[n, k] 5^k Pochhammer[1/5, k], {k, 0, n}], {n, 0, 18}]

Formula

a(n) = Sum_{k=0..n} Stirling2(n,k) * A008548(k).
a(n) ~ n! / (Gamma(1/5) * 6^(1/5) * n^(4/5) * log(6/5)^(n + 1/5)). - Vaclav Kotesovec, Aug 14 2021
O.g.f. (conjectural): 1/(1 - x/(1 - 6*x/(1 - 6*x/(1 - 12*x/(1 - 11*x/(1 - 18*x/(1 - ... - (5*n-4)*x/(1 - 6*n*x/(1 - ... ))))))))) - a continued fraction of Stieltjes-type. - Peter Bala, Aug 22 2023
a(0) = 1; a(n) = Sum_{k=1..n} (5 - 4*k/n) * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 2023
a(0) = 1; a(n) = a(n-1) - 6*Sum_{k=1..n-1} (-1)^k * binomial(n-1,k) * a(n-k). - Seiichi Manyama, Nov 16 2023

A090362 G.f. satisfies A^6 = BINOMIAL(A)^5 and also equals A090358^5.

Original entry on oeis.org

1, 5, 40, 460, 7220, 148276, 3831760, 120333680, 4460572870, 190679906990, 9230084185456, 498734395394840, 29740372199558420, 1939241402832412180, 137222625361036807760, 10470376552560151801616, 856818090423771231257245
Offset: 0

Views

Author

Paul D. Hanna, Nov 26 2003

Keywords

Comments

See comments in A090358.

Crossrefs

Programs

  • Mathematica
    nmax = 16; sol = {a[0] -> 1};
    Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^6 - A[x/(1 - x)]^5/(1 - x)^5 + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
    sol /. Rule -> Set;
    a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)
  • PARI
    {a(n)=local(A); if(n<0,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^6+B^5);polcoeff(A,n,x))}

Formula

G.f.: A(x)^6 = A(x/(1-x))^5/(1-x)^5.
a(n) ~ (n-1)! / (6 * (log(6/5))^(n+1)). - Vaclav Kotesovec, Nov 19 2014
From Peter Bala, May 26 2015: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*5^k = A094418(n).
BINOMIAL(A(x)) = exp( Sum_{n >= 1} c(n)*x^n/n ) where c(n) = (-1)^n*Sum_{k = 1..n} k!*Stirling2(n,k)*(-6)^k. A(x) = B(x)^5 and BINOMIAL(A(x)) = B(x)^6 where B(x) = 1 + x + 6*x^2 + 66*x^3 + 1071*x^4 + ... is the o.g.f. for A090358. See also A019538. (End)
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/6) * (5/6)^k). - Seiichi Manyama, May 26 2025

A278075 Coefficients of the signed Fubini polynomials in ascending order, F_n(x) = Sum_{k=0..n} (-1)^n*Stirling2(n,k)*k!*(-x)^k.

Original entry on oeis.org

1, 0, 1, 0, -1, 2, 0, 1, -6, 6, 0, -1, 14, -36, 24, 0, 1, -30, 150, -240, 120, 0, -1, 62, -540, 1560, -1800, 720, 0, 1, -126, 1806, -8400, 16800, -15120, 5040, 0, -1, 254, -5796, 40824, -126000, 191520, -141120, 40320, 0, 1, -510, 18150, -186480, 834120, -1905120, 2328480, -1451520, 362880
Offset: 0

Views

Author

Peter Luschny, Jan 09 2017

Keywords

Comments

Signed version of A131689.
Integral_{x=0..1} F_n(x) = B_n(1) where B_n(x) are the Bernoulli polynomials.

Examples

			Triangle of coefficients starts:
[1]
[0,  1]
[0, -1,    2]
[0,  1,   -6,    6]
[0, -1,   14,  -36,    24]
[0,  1,  -30,  150,  -240,   120]
[0, -1,   62, -540,  1560, -1800,    720]
[0,  1, -126, 1806, -8400, 16800, -15120, 5040]
		

Crossrefs

Row sums are A000012, diagonal is A000142.
Cf. A131689 (unsigned), A019538 (n>0, k>0), A090582.
Let F(n, x) = Sum_{k=0..n} T(n,k)*x^k then, apart from possible differences in the sign or the offset, we have: F(n, -5) = A094418(n), F(n, -4) = A094417(n), F(n, -3) = A032033(n), F(n, -2) = A004123(n), F(n, -1) = A000670(n), F(n, 0) = A000007(n), F(n, 1) = A000012(n), F(n, 2) = A000629(n), F(n, 3) = A201339(n), F(n, 4) = A201354(n), F(n, 5) = A201365(n).

Programs

  • Julia
    function T(n, k)
        if k < 0 || k > n return 0 end
        if n == 0 && k == 0 return 1 end
        k*(T(n-1, k-1) - T(n-1, k))
    end
    for n in 0:7
        println([T(n,k) for k in 0:n])
    end
    # Peter Luschny, Mar 26 2020
  • Maple
    F := (n,x) -> add((-1)^n*Stirling2(n,k)*k!*(-x)^k, k=0..n):
    for n from 0 to 10 do PolynomialTools:-CoefficientList(F(n,x), x) od;
  • Mathematica
    T[ n_, k_] := If[ n < 0 || k < 0, 0, (-1)^(n - k) k! StirlingS2[n, k]]; (* Michael Somos, Jul 08 2018 *)
  • PARI
    {T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(n + i) * binomial(k, i) * i^n))};
    /* Michael Somos, Jul 08 2018 */
    

Formula

T(n, k) = (-1)^(n-k) * Stirling2(n, k) * k!.
E.g.f.: 1/(1-x*(1-exp(-t))) = Sum_{n>=0} F_n(x) t^n/n!.
T(n, k) = k*(T(n-1, k-1) - T(n-1, k)) for 0 <= k <= n, T(0, 0) = 1, otherwise 0.
Bernoulli numbers are given by B(n) = Sum_{k = 0..n} T(n, k) / (k+1) with B(1) = 1/2. - Michael Somos, Jul 08 2018
Let F_n(x) be the row polynomials of this sequence and W_n(x) the row polynomials of A163626. Then F_n(1 - x) = W_n(x) and Integral_{x=0..1} U(n, x) = Bernoulli(n, 1) for U in {W, F}. - Peter Luschny, Aug 10 2021
T(n, k) = [z^k] Sum_{k=0..n} Eulerian(n, k)*z^(k+1)*(z-1)^(n-k-1) for n >= 1, where Eulerian(n, k) = A173018(n, k). - Peter Luschny, Aug 15 2022
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