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.

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A143494 Triangle read by rows: 2-Stirling numbers of the second kind.

Original entry on oeis.org

1, 2, 1, 4, 5, 1, 8, 19, 9, 1, 16, 65, 55, 14, 1, 32, 211, 285, 125, 20, 1, 64, 665, 1351, 910, 245, 27, 1, 128, 2059, 6069, 5901, 2380, 434, 35, 1, 256, 6305, 26335, 35574, 20181, 5418, 714, 44, 1, 512, 19171, 111645, 204205, 156660, 58107, 11130, 1110, 54, 1
Offset: 2

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 2 of the r-Stirling numbers of the second kind. The 2-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1 and 2 belong to distinct subsets.
More generally, the r-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the numbers 1, 2, ..., r belong to distinct subsets. The case r = 1 gives the usual Stirling numbers of the second kind A008277; for other cases see A143495 (r = 3) and A143496 (r = 4).
The lower unitriangular array of r-Stirling numbers of the second kind equals the matrix product P^(r-1) * S (with suitable offsets in the row and column indexing), where P is Pascal's triangle, A007318 and S is the array of Stirling numbers of the second kind, A008277.
For the definition of and entries relating to the corresponding r-Stirling numbers of the first kind see A143491. For entries on r-Lah numbers refer to A143497. The theory of r-Stirling numbers of both kinds is developed in [Broder].
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-2)*E^n*x^2 = Sum_{k = 0..n} T(n+2,k+2)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 2..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_2(x) = x^2. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 2-Eulerian numbers E_2(n,j) := A144696(n,j): T(n,k) = 2!/k!*Sum_ {j = n-k..n-2} E_2(n,j)*binomial(j,n-k) for n >= k >= 2. (End)
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,2), n>=k>=2, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column no. k from (A20) with k->2, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(2*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(2*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. (End)

Examples

			Triangle begins
  n\k|...2....3....4....5....6....7
  =================================
  2..|...1
  3..|...2....1
  4..|...4....5....1
  5..|...8...19....9....1
  6..|..16...65...55...14....1
  7..|..32..211..285..125...20....1
  ...
T(4,3) = 5. The set {1,2,3,4} can be partitioned into three subsets such that 1 and 2 belong to different subsets in 5 ways: {{1}{2}{3,4}}, {{1}{3}{2,4}}, {{1}{4}{2,3}}, {{2}{3}{1,4}} and {{2}{4}{1,3}}; the remaining possibility {{1,2}{3}{4}} is not allowed.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    5   1       |  =  | 4   3   1      ||0  2   1       ||0  0  1       | ...
| 8   19   9   1   |     | 8   7   4  1   ||0  4   3  1    ||0  0  2  1    |
|16   65  55  14  1|     |16  15  11  6  1||0  8   7  4  1 ||0  0  4  3  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - x)). See A055248. (End)
		

Crossrefs

A001047 (column 3), A005493 (row sums), A008277, A016269 (column 4), A025211 (column 5), A049444 (matrix inverse), A074051 (alt. row sums).

Programs

  • Maple
    with combinat: T := (n, k) -> (1/(k-2)!)*add ((-1)^(k-i)*binomial(k-2,i)*(i+2)^(n-2),i = 0..k-2): for n from 2 to 11 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - StirlingS2[n-1, k]; Flatten[ Table[ t[n, k], {n, 2, 11}, {k, 2, n}]] (* Jean-François Alcover, Dec 02 2011 *)
  • Sage
    @CachedFunction
    def stirling2r(n, k, r) :
        if n < r: return 0
        if n == r: return 1 if k == r else 0
        return stirling2r(n-1,k-1,r) + k*stirling2r(n-1,k,r)
    A143494 = lambda n,k: stirling2r(n, k, 2)
    for n in (2..6):
        [A143494(n, k) for k in (2..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+2,k+2) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+2)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - Stirling2(n-1,k) for n, k >= 2.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 2, with boundary conditions T(n,1) = T(1,n) = 0 for all n, T(2,2) = 1 and T(2,k) = 0 for k > 2. Special cases: T(n,2) = 2^(n-2); T(n,3) = 3^(n-2) - 2^(n-2).
As a sum of monomial functions of degree m: T(n+m,n) = Sum_{2 <= i_1 <= ... <= i_m <= n} (i_1*i_2*...*i_m). For example, T(6,4) = Sum_{2 <= i <= j <= 4} (i*j) = 2*2 + 2*3 + 2*4 + 3*3 + 3*4 + 4*4 = 55.
E.g.f. column k+2 (with offset 2): 1/k!*exp(2*x)*(exp(x) - 1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-2*x)*(1-3*x)*...*(1-k*x)).
E.g.f.: exp(2*t + x*(exp(t) - 1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+2,k+2) *x^k*t^n/n! = Sum_{n >= 0} B_n(2;x)*t^n/n! = 1 + (2 + x)*t/1! + (4 + 5*x + x^2)*t^2/2! + ..., where the row polynomial B_n(2;x) := Sum_{k = 0..n} T(n+2,k+2)*x^k denotes the 2-Bell polynomial.
Dobinski-type identities: Row polynomial B_n(2;x) = exp(-x)*Sum_{i >= 0} (i + 2)^n*x^i/i!. Sum_{k = 0..n} k!*T(n+2,k+2)*x^k = Sum_{i >= 0} (i + 2)^n*x^i/(1 + x)^(i+1).
The T(n,k) are the connection coefficients between falling factorials and the shifted monomials (x + 2)^(n-2). For example, from row 4 we have 4 + 5*x + x*(x - 1) = (x + 2)^2, while from row 5 we have 8 + 19*x + 9*x*(x - 1) + x*(x - 1)*(x - 2) = (x + 2)^3.
The row sums of the array are the 2-Bell numbers, B_n(2;1), equal to A005493(n-2). The alternating row sums are the complementary 2-Bell numbers, B_n(2;-1), equal to (-1)^n*A074051(n-2).
This array is the matrix product P * S, where P denotes the Pascal triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
Also, this array equals the transpose of the upper triangular array A126351. The inverse array is A049444, the signed 2-Stirling numbers of the first kind. See A143491 for the unsigned version of the inverse.
Let f(x) = exp(exp(x)). Then for n >= 1, the row polynomials R(n,x) are given by R(n+2,exp(x)) = 1/f(x)*(d/dx)^n(exp(2*x)*f(x)). Similar formulas hold for A008277, A039755, A105794, A111577 and A154537. - Peter Bala, Mar 01 2012

A094816 Triangle read by rows: T(n,k) are the coefficients of Charlier polynomials: A046716 transposed, for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 8, 6, 1, 1, 24, 29, 10, 1, 1, 89, 145, 75, 15, 1, 1, 415, 814, 545, 160, 21, 1, 1, 2372, 5243, 4179, 1575, 301, 28, 1, 1, 16072, 38618, 34860, 15659, 3836, 518, 36, 1, 1, 125673, 321690, 318926, 163191, 47775, 8274, 834, 45, 1, 1, 1112083, 2995011
Offset: 0

Views

Author

Philippe Deléham, Jun 12 2004

Keywords

Comments

The a-sequence for this Sheffer matrix is A027641(n)/A027642(n) (Bernoulli numbers) and the z-sequence is A130189(n)/ A130190(n). See the W. Lang link.
Take the lower triangular matrix in A049020 and invert it, then read by rows! - N. J. A. Sloane, Feb 07 2009
Exponential Riordan array [exp(x), log(1/(1-x))]. Equal to A007318*A132393. - Paul Barry, Apr 23 2009
A signed version of the triangle appears in [Gessel]. - Peter Bala, Aug 31 2012
T(n,k) is the number of permutations over all subsets of {1,2,...,n} (Cf. A000522) that have exactly k cycles. T(3,2) = 6: We permute the elements of the subsets {1,2}, {1,3}, {2,3}. Each has one permutation with 2 cycles. We permute the elements of {1,2,3} and there are three permutations that have 2 cycles. 3*1 + 1*3 = 6. - Geoffrey Critzer, Feb 24 2013
From Wolfdieter Lang, Jul 28 2017: (Start)
In Chihara's book the row polynomials (with rising powers) are the Charlier polynomials (-1)^n*C^(a)_n(-x), with a = -1, n >= 0. See p. 170, eq. (1.4).
In Ismail's book the present Charlier polynomials are denoted by C_n(-x;a=1) on p. 177, eq. (6.1.25). (End)
The triangle T(n,k) is a representative of the parametric family of triangles T(m,n,k), whose columns are the coefficients of the standard expansion of the function f(x) = (-log(1-x))^(k)*exp(-m*x)/k! for the case m=-1. See A381082. - Igor Victorovich Statsenko, Feb 14 2025

Examples

			From _Paul Barry_, Apr 23 2009: (Start)
Triangle begins
  1;
  1,     1;
  1,     3,     1;
  1,     8,     6,     1;
  1,    24,    29,    10,     1;
  1,    89,   145,    75,    15,    1;
  1,   415,   814,   545,   160,   21,   1;
  1,  2372,  5243,  4179,  1575,  301,  28,  1;
  1, 16072, 38618, 34860, 15659, 3836, 518, 36, 1;
Production matrix is
  1, 1;
  0, 2, 1;
  0, 1, 3,  1;
  0, 1, 3,  4,  1;
  0, 1, 4,  6,  5,  1;
  0, 1, 5, 10, 10,  6,  1;
  0, 1, 6, 15, 20, 15,  7,  1;
  0, 1, 7, 21, 35, 35, 21,  8, 1;
  0, 1, 8, 28, 56, 70, 56, 28, 9, 1; (End)
		

References

  • T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach, New York, London, Paris, 1978, Ch. VI, 1., pp. 170-172.
  • Classical and Quantum Orthogonal Polynomials in One Variable, Cambridge University Press, 2005, EMA, Vol. 98, p. 177.

Crossrefs

Columns k=0..4 give A000012, A002104, A381021, A381022, A381023.
Diagonals: A000012, A000217.
Row sums A000522, alternating row sums A024000.
KummerU(-n,1-n-x,z): this sequence (z=1), |A137346| (z=2), A327997 (z=3).

Programs

  • Maple
    A094816 := (n,k) -> (-1)^(n-k)*add(binomial(-j-1,-n-1)*Stirling1(j,k), j=0..n):
    seq(seq(A094816(n, k), k=0..n), n=0..9); # Peter Luschny, Apr 10 2016
  • Mathematica
    nn=10;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[ Exp[x]/(1-x)^y,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Feb 24 2013 *)
    Flatten[Table[(-1)^(n-k) Sum[Binomial[-j-1,-n-1] StirlingS1[j,k],{j,0,n}], {n,0,9},{k,0,n}]] (* Peter Luschny, Apr 10 2016 *)
    p[n_] := HypergeometricU[-n, 1 - n - x, 1];
    Table[CoefficientList[p[n], x], {n,0,9}] // Flatten (* Peter Luschny, Oct 27 2019 *)
  • PARI
    {T(n, k)= local(A); if( k<0 || k>n, 0, A = x * O(x^n); polcoeff( n! * polcoeff( exp(x + A) / (1 - x + A)^y, n), k))} /* Michael Somos, Nov 19 2006 */
    
  • Sage
    def a_row(n):
        s = sum(binomial(n,k)*rising_factorial(x,k) for k in (0..n))
        return expand(s).list()
    [a_row(n) for n in (0..9)] # Peter Luschny, Jun 28 2019

Formula

E.g.f.: exp(t)/(1-t)^x = Sum_{n>=0} C(x,n)*t^n/n!.
Sum_{k = 0..n} T(n, k)*x^k = C(x, n), Charlier polynomials; C(x, n)= A024000(n), A000012(n), A000522(n), A001339(n), A082030(n), A095000(n), A095177(n), A096307(n), A096341(n), A095722(n), A095740(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Feb 27 2013
T(n+1, k) = (n+1)*T(n, k) + T(n, k-1) - n*T(n-1, k) with T(0, 0) = 1, T(0, k) = 0 if k>0, T(n, k) = 0 if k<0.
PS*A008275*PS as infinite lower triangular matrices, where PS is a triangle with PS(n, k) = (-1)^k*A007318(n, k). PS = 1/PS. - Gerald McGarvey, Aug 20 2009
T(n,k) = (-1)^(n-k)*Sum_{j=0..n} C(-j-1, -n-1)*S1(j, k) where S1 are the signed Stirling numbers of the first kind. - Peter Luschny, Apr 10 2016
Absolute values T(n,k) of triangle (-1)^(n+k) T(n,k) where row n gives coefficients of x^k, 0 <= k <= n, in expansion of Sum_{k=0..n} binomial(n,k) (-1)^(n-k) x^{(k)}, where x^{(k)} := Product_{i=0..k-1} (x-i), k >= 1, and x^{(0)} := 1, the falling factorial powers. - Daniel Forgues, Oct 13 2019
From Peter Bala, Oct 23 2019: (Start)
The n-th row polynomial is
R(n, x) = Sum_{k = 0..n} (-1)^k*binomial(n, k)*k! * binomial(-x, k).
These polynomials occur in series acceleration formulas for the constant
1/e = n! * Sum_{k >= 0} (-1)^k/(k!*R(n,k)*R(n,k+1)), n >= 0. (cf. A068985, A094816 and A137346). (End)
R(n, x) = KummerU[-n, 1 - n - x, 1]. - Peter Luschny, Oct 27 2019
Sum_{j=0..m} (-1)^(m-j) * Bell(n+j) * T(m,j) = m! * Sum_{k=0..n} binomial(k,m) * Stirling2(n,k). - Vaclav Kotesovec, Aug 06 2021
From Natalia L. Skirrow, Jun 11 2025: (Start)
G.f.: 2F0(1,y;x/(1-x)) / (1-x).
Polynomial for the n-th row is R(n,y) = 2F0(-n,y;-1).
Falling g.f. for n-th row: Sum_{k=0..n} a(n,k)*(y)_k = [x^0] 2F0(1,-n;-1/x) * (1+log(1/(1-x)))^y = [x^n] e^x * Gamma(n+1,x) * (1+log(1/(1-x)))^y, where (y)_k = y!/(y-k)! denotes the falling factorial. (End)

A187646 (Signless) Central Stirling numbers of the first kind s(2n,n).

Original entry on oeis.org

1, 1, 11, 225, 6769, 269325, 13339535, 790943153, 54631129553, 4308105301929, 381922055502195, 37600535086859745, 4070384057007569521, 480544558742733545125, 61445535102359115635655, 8459574446076318147830625, 1247677142707273537964543265, 196258640868140652967646352465
Offset: 0

Views

Author

Emanuele Munarini, Mar 12 2011

Keywords

Comments

Number of permutations with n cycles on a set of size 2n.

Crossrefs

Programs

  • Maple
    seq(abs(Stirling1(2*n,n)), n=0..20);
  • Mathematica
    Table[Abs[StirlingS1[2n, n]], {n, 0, 12}]
    N[1 + 1/(2 LambertW[-1, -Exp[-1/2]/2]), 50] (* The constant z in the asymptotic - Vladimir Reshetnikov, Oct 08 2016 *)
  • Maxima
    makelist(abs(stirling1(2*n,n)),n,0,12);
    
  • PARI
    for(n=0,50, print1(abs(stirling(2*n, n, 1)), ", ")) \\ G. C. Greubel, Nov 09 2017

Formula

Asymptotic: a(n) ~ (2*n/(e*z*(1-z)))^n*sqrt((1-z)/(2*Pi*n*(2z-1))), where z=0.715331862959... is a root of the equation z = 2*(z-1)*log(1-z). - Vaclav Kotesovec, May 30 2011
Equivalent: a(n) ~ n!*(2*r^2/(r-1))^n/(2*Pi*n*sqrt(r-2)), where r=A226278. - Natalia L. Skirrow, Jul 13 2025
From Seiichi Manyama, May 20 2025: (Start)
a(n) = A132393(2*n,n).
a(n) = (2*n)! * [x^(2*n)] (-log(1 - x))^n / n!. (End)

A045756 Expansion of e.g.f. (1-9*x)^(-1/9), 9-factorial numbers.

Original entry on oeis.org

1, 1, 10, 190, 5320, 196840, 9054640, 498005200, 31872332800, 2326680294400, 190787784140800, 17361688356812800, 1736168835681280000, 189242403089259520000, 22330603564532623360000, 2835986652695643166720000, 385694184766607470673920000, 55925656791158083247718400000
Offset: 0

Views

Author

Keywords

Comments

Nine-fold factorials of numbers 9k+1, k = 0, 1, 2, ... - M. F. Hasler, Feb 14 2020

Crossrefs

Cf. A008542, A048994, A114806 (9-fold factorials), A132393.
Cf. k-fold factorials : A000142 ("1-fold"), A001147 (2-fold), A007559 (3), A007696 (4), A008548 (5), A008542 (6), A045754 (7), A045755 (8), A144773 (10), A256268 (combined table).

Programs

  • GAP
    List([0..20], n-> Product([0..n-1], j-> 9*j+1) ); # G. C. Greubel, Nov 11 2019
  • Magma
    [1] cat [(&*[9*j+1: j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Nov 11 2019
    
  • Maple
    seq( mul(9*j+1, j=0..n-1), n=0..20); # G. C. Greubel, Nov 11 2019
  • Mathematica
    Table[9^n*Pochhammer[1/9, n], {n,0,20}] (* G. C. Greubel, Nov 11 2019 *)
  • PARI
    vector(21, n, prod(j=0,n-2, 9*j+1) ) \\ G. C. Greubel, Nov 11 2019
    
  • Sage
    [product( (9*j+1) for j in (0..n-1)) for n in (0..20)] # G. C. Greubel, Nov 11 2019
    

Formula

a(n+1) = (9*n+1)(!^9) = Product_{k=0..n-1} (9*k+1), n >= 0.
E.g.f. (1-9*x)^(-1/9).
D-finite with recurrence: a(n) +(-9*n+8)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
a(n) = A114806(9n-8). - M. F. Hasler, Feb 14 2020
a(n) = Sum_{k = 0..n} (-9)^(n - k) * A048994(n, k) = Sum_{k = 0..n} 9^(n - k) * A132393(n, k). Philippe Deléham, Sep 20 2008
a(n) = (-8)^n * sum_{k = 0..n} (9/8)^k * s(n + 1, n + 1 - k), where s(n, k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = 9^n * Gamma(n + 1/9) / Gamma(1/9). - Artur Jasinski Aug 23 2016
a(n) ~ sqrt(2 * Pi) * 9^n * n^(n - 7/18)/(Gamma(1/9) * exp(n)). - Ilya Gutkovskiy, Sep 10 2016
Sum_{n>=0} 1/a(n) = 1 + (e/9^8)^(1/9)*(Gamma(1/9) - Gamma(1/9, 1/9)). - Amiram Eldar, Dec 21 2022

Extensions

a(0)=1 inserted; merged with A144772; formulas and programs changed accordingly by Georg Fischer, Feb 15 2020

A143495 Triangle read by rows: 3-Stirling numbers of the second kind.

Original entry on oeis.org

1, 3, 1, 9, 7, 1, 27, 37, 12, 1, 81, 175, 97, 18, 1, 243, 781, 660, 205, 25, 1, 729, 3367, 4081, 1890, 380, 33, 1, 2187, 14197, 23772, 15421, 4550, 644, 42, 1, 6561, 58975, 133057, 116298, 47271, 9702, 1022, 52, 1, 19683, 242461, 724260, 830845, 447195
Offset: 3

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 3 of the r-Stirling numbers of the second kind. The 3-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2 and 3 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 3-Stirling numbers of the first kind is A143492. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 3-Lah numbers refer to A143498.
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-3)*E^n*x^3 = Sum_{k = 0..n} T(n+3,k+3)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 3..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_3(x) = x^3. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 3-Eulerian numbers E_3(n,j) := A144697(n,j): T(n,k) = (3!/k!)*Sum_{j = n-k..n-3} E_3(n,j)*binomial(j,n-k) for n >= k >= 3.
(End)
T(n,k) = S(n,k,3), n>=k>=3, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->3, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(3*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(3*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. - Wolfdieter Lang, Sep 29 2011

Examples

			Triangle begins
  n\k|....3....4....5....6....7....8
  ==================================
  3..|....1
  4..|....3....1
  5..|....9....7....1
  6..|...27...37...12....1
  7..|...81..175...97...18....1
  8..|..243..781..660..205...25....1
  ...
T(5,4) = 7. The set {1,2,3,4,5} can be partitioned into four subsets such that 1, 2 and 3 belong to different subsets in 7 ways: {{1}{2}{3}{4,5}}, {{1}{2}{5}{3,4}}, {{1}{2}{4}{3,5}}, {{1}{3}{4}{2,5}}, {{1}{3}{5}{2,4}}, {{2}{3}{4}{1,5}} and {{2}{3}{5}{1,4}}.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 3    1           |     | 3   1          ||0  1           ||0  1          |
| 9    7   1       |  =  | 9   4   1      ||0  3   1       ||0  0  1       | ...
|27   37  12   1   |     |27  13   5  1   ||0  9   4  1    ||0  0  3  1    |
|81  175  97  18  1|     |81  40  18  6  1||0 27  13  5  1 ||0  0  9  4  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 3*x), x/(1 - x)). See A106516. (End)
		

Crossrefs

Cf. A005061 (column 4), A005494 (row sums), A008277, A016753 (column 5), A028025 (column 6), A049458 (matrix inverse), A106516, A143492, A143494, A143496, A143498.

Programs

  • Maple
    A143495 := (n, k) -> (1/(k-3)!)*add((-1)^(k-i-1)*binomial(k-3,i)*(i+3)^(n-3), i = 0..k-3): for n from 3 to 12 do seq(A143495(n, k), k = 3..n) end do;
  • Mathematica
    nmax = 12; t[n_, k_] := 1/(k-3)!* Sum[ (-1)^(k-j-1)*Binomial[k-3, j]*(j+3)^(n-3), {j, 0, k-3}]; Flatten[ Table[ t[n, k], {n, 3, nmax}, {k, 3, n}]] (* Jean-François Alcover, Dec 07 2011, after Maple *)
  • Sage
    @CachedFunction
    def stirling2r(n, k, r) :
        if n < r: return 0
        if n == r: return 1 if k == r else 0
        return stirling2r(n-1, k-1, r) + k*stirling2r(n-1, k, r)
    A143495 = lambda n, k: stirling2r(n, k, 3)
    for n in (3..8): [A143495(n, k) for k in (3..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+3,k+3) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+3)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 3*Stirling2(n-1,k) + 2*Stirling2(n-2,k), n,k >= 3.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 3, with boundary conditions: T(n,2) = T(2,n) = 0 for all n; T(3,3) = 1; T(3,k) = 0 for k > 3.
Special cases: T(n,3) = 3^(n-3); T(n,4) = 4^(n-3) - 3^(n-3).
E.g.f. (k+3) column (with offset 3): (1/k!)*exp(3x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-3*x)*(1-4*x)*...*(1-k*x)).
E.g.f.: exp(3*t + x*(exp(t)-1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+3,k+3)*x^k*t^n/n! = Sum_{n >= 0} B_n(3;x)*t^n/n! = 1 + (3+x)*t/1! + (9+7*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(3;x) := Sum_{k = 0..n} T(n+3,k+3)*x^k, may be called the 3-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(3;x) = exp(-x)*Sum_{i >= 0} (i+3)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+3,k+3)*x^k = Sum_{i >= 0} (i+3)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+3)^(n-3). For example, 9 + 7*x + x*(x-1) = (x+3)^2 and 27 + 37*x + 12x*(x-1) + x*(x-1)*(x-2) = (x+3)^3.
This array is the matrix product P^2 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]). The inverse array is A049458, the signed 3-Stirling numbers of the first kind.

A032188 Number of labeled series-reduced mobiles (circular rooted trees) with n leaves (root has degree 0 or >= 2).

Original entry on oeis.org

1, 1, 5, 41, 469, 6889, 123605, 2620169, 64074901, 1775623081, 54989743445, 1882140936521, 70552399533589, 2874543652787689, 126484802362553045, 5977683917752887689, 301983995802099667861, 16239818347465293071401, 926248570498763547197525, 55847464116157184894240201
Offset: 1

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Author

Keywords

Comments

With offset 0, a(n) = number of partitions of the multiset {1,1,2,2,...,n,n} into lists of strictly decreasing lists, called blocks, such that the concatenation of all blocks in a list has the Stirling property: all entries between the two occurrences of i exceed i, 1<=i<=n. For example, with slashes separating blocks, a(2) = 5 counts 1/1/2/2; 1/2/2/1; 2/2/1/1; 1/2/2 1; 2/2 1/1, but not, for instance, 2 1/2/1 because it fails the Stirling test for i=2. - David Callan, Nov 21 2011

Examples

			D^3(1) = (24*x^2-64*x+41)/(2*x-1)^6. Evaluated at x = 0 this gives a(4) = 41.
a(3) = 5: Denote the colors of the vertices by the letters a,b,c, .... The 5 possible increasing plane trees on 3 vertices with vertices of outdegree k coming in 2^(k-1) colors are
.
   1a       1a        1b        1a        1b
   |       /  \      /  \      /  \      /  \
   2a     2    3    2    3    3    2    3    2
   |
   3
		

Crossrefs

Programs

  • Maple
    Order := 20; t1 := solve(series((ln(1-A)+2*A),A)=x,A); A000311 := n->n!*coeff(t1,x,n);
    # With offset 0:
    a := n -> add(combinat:-eulerian2(n,k)*2^k,k=0..n):
    seq(a(n),n=0..19); # Peter Luschny, Jul 09 2015
  • Mathematica
    For[y=x+O[x]^21; oldy=0, y=!=oldy, oldy=y; y=((1-y)Log[1-y]+x*y+y-x)/(2y-1), Null]; Table[n!Coefficient[y, x, n], {n, 1, 20}]
    Rest[CoefficientList[InverseSeries[Series[2*x + Log[1-x],{x,0,20}],x],x] * Range[0,20]!] (* Vaclav Kotesovec, Jan 08 2014 *)
  • Maxima
    a(n):=sum((n+k-1)!*sum(1/(k-j)!*sum((2^l*(-1)^(n+l+1)*stirling1(n-l+j-1,j-l))/(l!*(n-l+j-1)!),l,0,j),j,0,k),k,0,n-1); /* Vladimir Kruchinin, Feb 06 2012 */
    
  • PARI
    N = 66; x = 'x + O('x^N);
    Q(k) = if(k>N, 1,  1 + (k+1)*x - 2*x*(k+1)/Q(k+1) );
    gf = 1/Q(0); Vec(gf) \\ Joerg Arndt, May 01 2013
    
  • PARI
    {my(n=20); Vec(serlaplace(serreverse(2*x+log(1-x + O(x*x^n)))))} \\ Andrew Howroyd, Jan 16 2018

Formula

Doubles (index 2+) under "CIJ" (necklace, indistinct, labeled) transform.
E.g.f. A(x) satisfies log(1-A(x))+2*A(x)-x = 0. - Vladeta Jovovic, Dec 06 2002
With offset 0, second Eulerian transform of the powers of 2 [A000079]. See A001147 for definition of SET. - Ross La Haye, Feb 14 2005
From Peter Bala, Sep 05 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A'(x) = (1-A)/(1-2*A) with A(0) = 0. Hence the inverse function A^-1(x) = int {t = 0..x} (1-2*t)/(1-t) = 2*x+log(1-x). The expansion of A(x) can be found by inverting the above integral using the method of [Dominici, Theorem 4.1] to arrive at the result a(n) = D^(n-1)(1) evaluated at x = 0, where D denotes the operator g(x) -> d/dx((1-x)/(1-2*x)*g(x)). Compare with A006351.
Applying [Bergeron et al., Theorem 1] to the result x = int {t = 0..A(x)} 1/phi(t), where phi(t) = (1-t)/(1-2*t) = 1+t+2*t^2+4*t^3+8*t^4+... leads to the following combinatorial interpretation for this sequence: a(n) gives the number of plane increasing trees on n vertices where each vertex of outdegree k >= 1 can be colored in 2^(k-1) ways. An example is given below. (End)
The integral from 0 to infinity w.r.t. w of exp(-2w)(1-z*w)^(-1/z) gives an o.g.f. for the series with offset 0. Consequently, a(n)= sum(j=1 to infinity): St1d(n,j)/(2^(n+j-1)) where St1d(n,j) is the j-th element of the n-th diagonal of A132393 with offset=1; e.g., a(3)= 5 = 0/2^3 + 2/2^4 + 11/2^5 + 35/2^6 + 85/2^7 + ... . - Tom Copeland, Sep 15 2011
A signed o.g.f., with Γ(v,x) the incomplete gamma function (see A111999 with u=1), is g(z) = (2/z)^(-(1/z)-1) exp(2/z) Γ((1/z)+1,2/z)/z. - Tom Copeland, Sep 16 2011
With offset 0, a(n) = Sum[T(n+k,k), k=1..n] where T(n,k) are the associated Stirling numbers of the first kind (A008306). For example, a(3) = 41 = 6 + 20 + 15. - David Callan, Nov 21 2011
a(n) = sum(k=0..n-1, (n+k-1)!*sum(j=0..k, 1/(k-j)!*sum(l=0..j, (2^l*(-1)^(n+l+1)*stirling1(n-l+j-1,j-l))/(l!*(n-l+j-1)!)))), n>0. - Vladimir Kruchinin, Feb 06 2012
G.f.: 1/Q(0), where Q(k)= 1 + (k+1)*x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
a(n) ~ n^(n-1) / (2 * exp(n) * (1-log(2))^(n-1/2)). - Vaclav Kotesovec, Jan 08 2014
a(n) = A032034(n)/2. - Alois P. Heinz, Jul 04 2018
E.g.f: series reversion of 2*x + log(1-x). - Andrew Howroyd, Sep 19 2018

A265609 Array read by ascending antidiagonals: A(n,k) the rising factorial, also known as Pochhammer symbol, for n >= 0 and k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 6, 6, 0, 1, 4, 12, 24, 24, 0, 1, 5, 20, 60, 120, 120, 0, 1, 6, 30, 120, 360, 720, 720, 0, 1, 7, 42, 210, 840, 2520, 5040, 5040, 0, 1, 8, 56, 336, 1680, 6720, 20160, 40320, 40320, 0
Offset: 0

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Author

Peter Luschny, Dec 19 2015

Keywords

Comments

The Pochhammer function is defined P(x,n) = x*(x+1)*...*(x+n-1). By convention P(0,0) = 1.
From Antti Karttunen, Dec 19 2015: (Start)
Apart from the initial row of zeros, if we discard the leftmost column and divide the rest of terms A(n,k) with (n+k) [where k is now the once-decremented column index of the new, shifted position] we get the same array back. See the given recursive formula.
When the numbers in array are viewed in factorial base (A007623), certain repeating patterns can be discerned, at least in a few of the topmost rows. See comment in A001710 and arrays A265890, A265892. (End)
A(n,k) is the k-th moment (about 0) of a gamma (Erlang) distribution with shape parameter n and rate parameter 1. - Geoffrey Critzer, Dec 24 2018

Examples

			Square array A(n,k) [where n=row, k=column] is read by ascending antidiagonals as:
A(0,0), A(1,0), A(0,1), A(2,0), A(1,1), A(0,2), A(3,0), A(2,1), A(1,2), A(0,3), ...
Array starts:
n\k [0  1   2    3     4      5        6         7          8]
--------------------------------------------------------------
[0] [1, 0,  0,   0,    0,     0,       0,        0,         0]
[1] [1, 1,  2,   6,   24,   120,     720,     5040,     40320]
[2] [1, 2,  6,  24,  120,   720,    5040,    40320,    362880]
[3] [1, 3, 12,  60,  360,  2520,   20160,   181440,   1814400]
[4] [1, 4, 20, 120,  840,  6720,   60480,   604800,   6652800]
[5] [1, 5, 30, 210, 1680, 15120,  151200,  1663200,  19958400]
[6] [1, 6, 42, 336, 3024, 30240,  332640,  3991680,  51891840]
[7] [1, 7, 56, 504, 5040, 55440,  665280,  8648640, 121080960]
[8] [1, 8, 72, 720, 7920, 95040, 1235520, 17297280, 259459200]
.
Seen as a triangle, T(n, k) = Pochhammer(n - k, k), the first few rows are:
   [0] 1;
   [1] 1, 0;
   [2] 1, 1,  0;
   [3] 1, 2,  2,   0;
   [4] 1, 3,  6,   6,    0;
   [5] 1, 4, 12,  24,   24,    0;
   [6] 1, 5, 20,  60,  120,  120,     0;
   [7] 1, 6, 30, 120,  360,  720,   720,     0;
   [8] 1, 7, 42, 210,  840, 2520,  5040,  5040,     0;
   [9] 1, 8, 56, 336, 1680, 6720, 20160, 40320, 40320, 0.
		

References

  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1994.
  • H. S. Wall, Analytic Theory of Continued Fractions, Chelsea 1973, p. 355.

Crossrefs

Triangle giving terms only up to column k=n: A124320.
Row 0: A000007, row 1: A000142, row 3: A001710 (from k=1 onward, shifted two terms left).
Column 0: A000012, column 1: A001477, column 2: A002378, columns 3-7: A007531, A052762, A052787, A053625, A159083 (shifted 2 .. 6 terms left respectively, i.e. without the extra initial zeros), column 8: A239035.
Row sums of the triangle: A000522.
A(n, n) = A000407(n-1) for n>0.
2^n*A(1/2,n) = A001147(n).
Cf. also A007623, A008279 (falling factorial), A173333, A257505, A265890, A265892.

Programs

  • Maple
    for n from 0 to 8 do seq(pochhammer(n,k), k=0..8) od;
  • Mathematica
    Table[Pochhammer[n, k], {n, 0, 8}, {k, 0, 8}]
  • Sage
    for n in (0..8): print([rising_factorial(n,k) for k in (0..8)])
    
  • Scheme
    (define (A265609 n) (A265609bi (A025581 n) (A002262 n)))
    (define (A265609bi row col) (if (zero? col) 1 (* (+ row col -1) (A265609bi row (- col 1)))))
    ;; Antti Karttunen, Dec 19 2015

Formula

A(n,k) = Gamma(n+k)/Gamma(n) for n > 0 and n^k for n=0.
A(n,k) = Sum_{j=0..k} n^j*S1(k,j), S1(n,k) the Stirling cycle numbers A132393(n,k).
A(n,k) = (k-1)!/(Sum_{j=0..k-1} (-1)^j*binomial(k-1, j)/(j+n)) for n >= 1, k >= 1.
A(n,k) = (n+k-1)*A(n,k-1) for k >= 1, A(n,0) = 1. - Antti Karttunen, Dec 19 2015
E.g.f. for row k: 1/(1-x)^k. - Geoffrey Critzer, Dec 24 2018
A(n, k) = FallingFactorial(n + k - 1, k). - Peter Luschny, Mar 22 2022
G.f. for row n as a continued fraction of Stieltjes type: 1/(1 - n*x/(1 - x/(1 - (n+1)*x/(1 - 2*x/(1 - (n+2)*x/(1 - 3*x/(1 - ... ))))))). See Wall, Chapter XVIII, equation 92.5. Cf. A226513. - Peter Bala, Aug 27 2023

A143496 4-Stirling numbers of the second kind.

Original entry on oeis.org

1, 4, 1, 16, 9, 1, 64, 61, 15, 1, 256, 369, 151, 22, 1, 1024, 2101, 1275, 305, 30, 1, 4096, 11529, 9751, 3410, 545, 39, 1, 16384, 61741, 70035, 33621, 7770, 896, 49, 1, 65536, 325089, 481951, 305382, 95781, 15834, 1386, 60, 1, 262144, 1690981, 3216795
Offset: 4

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 4 of the r-Stirling numbers of the second kind. The 4-Stirling numbers of the second kind count the ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2, 3 and 4 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 4-Stirling numbers of the first kind is A143493. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 4-Lah numbers refer to A143499.
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,4), n >= k >= 4, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->4, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(4*x),exp(x)-1) with e.g.f. of column no. m >= 0: exp(4*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393.
(End)

Examples

			Triangle begins
n\k|.....4.....5.....6.....7.....8.....9
========================================
4..|.....1
5..|.....4.....1
6..|....16.....9.....1
7..|....64....61....15.....1
8..|...256...369...151....22.....1
9..|..1024..2101..1275...305....30.....1
...
T(6,5) = 9. The set {1,2,3,4,5,6} can be partitioned into five subsets such that 1, 2, 3 and 4 belong to different subsets in 9 ways: {{1,5}{2}{3}{4}{6}}, {{1,6}{2}{3}{4}{5}}, {{2,5}{1}{3}{4}{6}}, {{2,6}{1}{3}{4}{5}}, {{3,5}{1}{2}{4}{6}}, {{3,6}{1}{2}{4}{5}}, {{4,5}{1}{2}{3}{6}}, {{4,6}{1}{2}{3}{5}} and {{5,6}{1}{2}{3}{4}}.
		

Crossrefs

Cf. A003468 (column 7), A005060 (column 5), A008277, A016103 (column 6), A045379 (row sums), A049459 (matrix inverse), A143493, A143494, A143495, A143499.

Programs

  • Maple
    with combinat: T := (n, k) -> 1/(k-4)!*add ((-1)^(k-i)*binomial(k-4,i)*(i+4)^(n-4),i = 0..k-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - 6*StirlingS2[n-1, k] + 11*StirlingS2[n-2, k] - 6*StirlingS2[n-3, k]; Flatten[ Table[ t[n, k], {n, 4, 13}, {k, 4, n}]] (* Jean-François Alcover, Dec 02 2011 *)

Formula

T(n+4,k+4) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+4)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 6*Stirling2(n-1,k) + 11*Stirling2(n-2,k) - 6*Stirling2(n-3,k) for n,k >= 4.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 4 with boundary conditions: T(n,3) = T(3,n) = 0 for all n; T(4,4) = 1; T(4,k) = 0 for k > 4. Special cases: T(n,4) = 4^(n-4); T(n,5) = 5^(n-4) - 4^(n-4).
E.g.f. (k+4)-th column (with offset 4): (1/k!)*exp(4*x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n>=k} T(n,k)*x^n = x^k/((1-4*x)*(1-5*x)*...*(1-k*x)).
E.g.f.: exp(4*t + x*(exp(t)-1)) = Sum_{n = 0..infinity} Sum_(k = 0..n) T(n+4,k+4)*x^k*t^n/n! = Sum_{n = 0..infinity} B_n(4;x)*t^n/n! = 1 + (4+x)*t/1! + (16+9*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(4;x) := Sum_{k = 0..n} T(n+4,k+4)*x^k, may be called the 4-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(4;x) = exp(-x)*Sum_{i = 0..infinity} (i+4)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+4,k+4)*x^k = Sum_{i = 0..infinity} (i+4)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+4)^(n-4). For example, 16 + 9*x + x*(x-1) = (x+4)^2; 64 + 61*x + 15*x*(x-1) + x*(x-1)*(x-2) = (x+4)^3.
This array is the matrix product P^3 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
The inverse array is A049459, the signed 4-Stirling numbers of the first kind.
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-4)*E^n*x^4 = Sum_{k = 0..n} T(n+4,k+4)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k=4..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_4(x) = x^4. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 4-Eulerian numbers E_4(n,j) := A144698(n,j): T(n,k) = 4!/k!*Sum_{j = n-k..n-4} E_4(n,j)*binomial(j,n-k) for n >= k >= 4.
(End)

A144828 Partial products of successive terms of A017113; a(0)=1.

Original entry on oeis.org

1, 4, 48, 960, 26880, 967680, 42577920, 2214051840, 132843110400, 9033331507200, 686533194547200, 57668788341964800, 5305528527460761600, 530552852746076160000, 57299708096576225280000, 6646766139202842132480000, 824199001261152424427520000, 108794268166472120024432640000
Offset: 0

Views

Author

Philippe Deléham, Sep 21 2008

Keywords

Comments

a(n) is the number of signed permutations of length 4n that are equal to their reverse-inverses. Note that the reverse-inverse of a permutation is equivalent to a 90-degree rotation of the permutation's diagram (see the Hardt and Troyka link). - Justin M. Troyka, Aug 11 2011
Define the bar operation as an operation on signed permutation that flips the sign of each entry. Then a(n) is the number of signed permutations of length 2n that are equal to the bar of their inverses and equal to their reverse-complements (see the Hardt and Troyka link). - Justin M. Troyka, Aug 11 2011

Examples

			a(0)=1, a(1)=4, a(2)=4*12=48, a(3)=4*12*20=960, a(4)=4*12*20*28=26880, ...
Since a(1) = 4, there are 4 signed permutations of 4 that are equal to their reverse-inverses.  These are: (+2,+4,+1,+3), (+3,+1,+4,+2), (-2,-4,-1,-3), (-3,-1,-4,-2). - _Justin M. Troyka_, Aug 11 2011
G.f. = 1 + 4*x + 48*x^2 + 960*x^3 + 26880*x^4 + 967680*x^5 + 42577920*x^6 + ...
		

Crossrefs

Essentially the same as A052714. - N. J. A. Sloane, Feb 03 2013
Sequences of the form m^(n-1)*n!*Catalan(n-1): A001813 (m=1), A052714 (or this sequence) (m=2), A221954 (m=3), A052734 (m=4), A221953 (m=5), A221955 (m=6).

Programs

  • Magma
    [2^k *Factorial(2*k) / Factorial(k): k in [0..20]]; // Vincenzo Librandi, Aug 11 2011
    
  • Maple
    A144828:= n-> 2^n*n!*binomial(2*n,n); seq(A144828(n), n=0..30); # G. C. Greubel, Apr 02 2021
  • Mathematica
    Table[4^n (2 n - 1)!!, {n, 0, 15}] (* Vincenzo Librandi, May 14 2015 *)
    Join[{1},FoldList[Times,(8*Range[0,20]+4)]] (* Harvey P. Dale, Dec 01 2015 *)
  • PARI
    a(n)=binomial(2*n,n)*n!<Charles R Greathouse IV, Jan 17 2012
    
  • PARI
    {a(n) = if( n<0, (-1)^n / a(-n), 2^n *(2*n)! / n!)}; /* Michael Somos, Jan 06 2017 */
    
  • Sage
    [2^n*factorial(n+1)*catalan_number(n) for n in (0..30)] # G. C. Greubel, Apr 02 2021

Formula

a(n) = Sum_{k=0..n} A132393(n,k)*4^k*8^(n-k).
a(n) = A052714(n+1). - R. J. Mathar, Oct 01 2008
a(n) = 2^n *(2*n)! / n!. - Justin M. Troyka, Aug 11 2011
G.f.: 1/(1-4x/(1-8x/(1-12x/(1-16x/(1-20x/(1-24x/(1-28x/(1-32x/(1-... (continued fraction). - Philippe Deléham, Jan 07 2012
a(n) = (-4)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
E.g.f.: 1/sqrt(1-8*x). - Philippe Deléham, May 14 2015
a(n) = 4^n * A001147(n). - Philippe Deléham, May 14 2015
a(n) = 8^n * Gamma(n + 1/2) / sqrt(Pi). - Daniel Suteu, Jan 06 2017
0 = a(n)*(8*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) and a(n) = (-1)^n / a(-n) for all n in Z. - Michael Somos, Jan 06 2017
a(n) = 2^n * (n+1)! * Catalan(n). - G. C. Greubel, Apr 02 2021
Sum_{n>=0} 1/a(n) = 1 + e^(1/8)*sqrt(Pi)*erf(1/(2*sqrt(2)))/(2*sqrt(2)), where erf is the error function. - Amiram Eldar, Dec 20 2022

A154955 a(1) = 1, a(2) = -1, followed by 0, 0, 0, ... .

Original entry on oeis.org

1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Mats Granvik, Jan 18 2009

Keywords

Comments

Matrix inverse of A000012.
Moebius transform of the sequence A000035. Dirichlet inverse of A209229. Partial sums of a(n) is characteristic function of 1 (A063524). a(n)=(-1)^(n+1)*A019590(n). a(n) for n >= 1 is Dirichlet convolution of following functions b(n), c(n), a(n) = Sum_{d|n} b(d)*c(n/d): a(n) = A000012(n) * A092673(n). Examples of Dirichlet convolutions with function a(n), i.e. b(n) = Sum_{d|n} a(d)*c(n/d): a(n) * A000012(n) = A000035(n), a(n) * A000027(n) = A026741(n), a(n) * A008683(n) = A092673(n), a(n) * A036987(n-1) = A063524(n), a(n) * A000005(n) = A001227(n). - Jaroslav Krizek, Mar 21 2009
The Kn21 sums, see A180662, of triangle A108299 equal the terms of this sequence. - Johannes W. Meijer, Aug 14 2011
{a(n-1)}A132393.%20-%20_Wolfdieter%20Lang">{n>=1}, gives the alternating row sums of A132393. - _Wolfdieter Lang, May 09 2017
With offset 0 the alternating row sums of A097805. - Peter Luschny, Sep 07 2017

Crossrefs

Programs

Formula

G.f.: A(x) = x - x^2 = x / (1 + x / (1 - x)). - Michael Somos, Jan 03 2013
a(n) = (2/sqrt(3))*sin((2*Pi/3)*n!). - Lorenzo Pinlac, Jan 16 2022
a(n) = [n = 1] - [n = 2], where [] is the Iverson bracket. - Wesley Ivan Hurt, Jun 22 2024
Multiplicative with a(2) = -1, a(2^e) = 0 if e > 1, a(p^e) = 0 if p > 2. - Antti Karttunen, Dec 17 2024

Extensions

Keyword:mult added by Antti Karttunen, Dec 17 2024
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