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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A112493 Triangle read by rows, T(n, k) = Sum_{j=0..n} C(n-j, n-k)*E2(n, j), where E2 are the second-order Eulerian numbers A201637, for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 11, 25, 15, 1, 26, 130, 210, 105, 1, 57, 546, 1750, 2205, 945, 1, 120, 2037, 11368, 26775, 27720, 10395, 1, 247, 7071, 63805, 247555, 460845, 405405, 135135, 1, 502, 23436, 325930, 1939630, 5735730, 8828820, 6756750, 2027025, 1
Offset: 0

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Author

Wolfdieter Lang, Oct 14 2005

Keywords

Comments

Previous name was: Coefficient triangle of polynomials used for e.g.f.s of Stirling2 diagonals.
For the o.g.f. of diagonal k of the Stirling2 triangle one has a similar result. See A008517 (second-order Eulerian triangle).
A(m,x), the o.g.f. for column m, satisfies the recurrence A(m,x) = x*(x*(d/dx)A(m-1,x) + m*A(m-1,x))/(1-(m+1)*x), for m >= 1 and A(0,x) = 1/(1-x).
The e.g.f. for the sequence in column k+1, k >= 0, of A008278, i.e., for the diagonal k >= 0 of the Stirling2 triangle A048993, is exp(x)*Sum_{m=0..k} a(k,m)*(x^(m+k))/(m+k)!.
It appears that the triangles in this sequence and A124324 have identical columns, except for shifts. - Jörgen Backelin, Jun 20 2022
A refined version of this triangle is given in A356145, which contains a link providing the precise relationship between A124324 and this entry, confirming Jörgen Backelin's observation above. - Tom Copeland, Sep 24 2022

Examples

			Triangle starts:
  [1]
  [1, 1]
  [1, 4,  3]
  [1, 11, 25,  15]
  [1, 26, 130, 210,  105]
  [1, 57, 546, 1750, 2205, 945]
  ...
The e.g.f. of [0,0,1,7,25,65,...], the k=3 column of A008278, but with offset n=0, is exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!).
Third row [1,4,3]: There are three plane increasing trees on 3 vertices. The number of colors are shown to the right of a vertex.
...................................................
....1o.(1+t)...........1o.t*(1+t).....1o.t*(1+t)...
....|................. /.\............/.\..........
....|................ /...\........../...\.........
....2o.(1+t)........2o.....3o......3o....2o........
....|..............................................
....|..............................................
....3o.............................................
...................................................
The total number of trees is (1+t)^2 + t*(1+t) + t*(1+t) = 1+4*t+3*t^2 = R(2,t).
		

Crossrefs

Row sums give A006351(k+1), k>=0.
The column sequences start with A000012 (powers of 1), A000295 (Eulerian numbers), A112495, A112496, A112497.
Antidiagonal sums give A000110.
Cf. A356145.

Programs

  • Maple
    T := (n, k) -> add(combinat:-eulerian2(n, j)*binomial(n-j, n-k), j=0..n):
    seq(seq(T(n, k), k=0..n), n=0..9); # Peter Luschny, Apr 11 2016
  • Mathematica
    max = 11; f[x_, t_] := -1 - (1 + t)/t*ProductLog[-t/(1 + t)*Exp[(x - t)/(1 + t)]]; coes = CoefficientList[ Series[f[x, t], {x, 0, max}, {t, 0, max}], {x, t}]* Range[0, max]!; Table[coes[[n, k]], {n, 0, max}, {k, 1, n - 1}] // Flatten (* Jean-François Alcover, Nov 22 2012, from e.g.f. *)

Formula

a(k, m) = 0 if k < m, a(k, -1):=0, a(0, 0)=1, a(k, m)=(m+1)*a(k-1, m) + (k+m-1)*a(k-1, m-1) else.
From Peter Bala, Sep 30 2011: (Start)
E.g.f.: A(x,t) = -1-((1+t)/t)*LambertW(-(t/(1+t))*exp((x-t)/(1+t))) = x + (1+t)*x^2/2! + (1+4*t+3*t^2)*x^3/3! + .... A(x,t) is the inverse function of (1+t)*log(1+x)-t*x.
A(x,t) satisfies the partial differential equation (1-x*t)*dA/dx = 1 + A + t*(1+t)*dA/dt. It follows that the row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) =(n*t+1)*R(n,t) + t*(1+t)*dR(n,t)/dt. Cf. A054589 and A075856. The polynomials t/(1+t)*R(n,t) are the row polynomials of A134991.
The generating function A(x,t) satisfies the autonomous differential equation dA/dx = (1+A)/(1-t*A). Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees on n+1 vertices where the non-leaf vertices of outdegree k come in t^(k-1)*(1+t) colors. An example is given below. Cf. A006351, which corresponds to the case t = 1. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x) = (1+x)/(1-x*t). Then R(n,t) = (f(x)*d/dx)^n(f(x)) evaluated at x = 0. (End)
Sum_{j=0..n} T(n-j,j) = A000110(n). - Alois P. Heinz, Jun 20 2022
From Mikhail Kurkov, Apr 01 2025: (Start)
E.g.f.: B(y) = -w/(x*(1+w)) where w = LambertW(-x/(1+x)*exp((y-x)/(1+x))) satisfies the first-order ordinary differential equation (1+x)*B'(y) = B(y)*(1+x*B(y))^2, hence row polynomials are P(n,x) = P(n-1,x) + x*Sum_{j=0..n-1} binomial(n, j)*P(j,x)*P(n-j-1,x) for n > 0 with P(0,x) = 1 (see MathOverflow link).
Conjecture: row polynomials are P(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling1(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(j+k)/((n+j-k)!*(i-j)!*k!). (End)
Conjecture: g.f. satisfies 1/(1 - x - x*y/(1 - 2*x - 2*x*y/(1 - 3*x - 3*x*y/(1 - 4*x - 4*x*y/(1 - 5*x - 5*x*y/(1 - ...)))))) (see A383019 for conjectures about combinatorial interpretation and algorithm for efficient computing). - Mikhail Kurkov, Apr 21 2025

Extensions

New name from Peter Luschny, Apr 11 2016

A001298 Stirling numbers of the second kind S(n+4, n).

Original entry on oeis.org

0, 1, 31, 301, 1701, 6951, 22827, 63987, 159027, 359502, 752752, 1479478, 2757118, 4910178, 8408778, 13916778, 22350954, 34952799, 53374629, 79781779, 116972779, 168519505, 238929405, 333832005, 460192005, 626551380, 843303006, 1122998436, 1480692556
Offset: 0

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Author

Keywords

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [n*(n+1)*(n+2)*(n+3)*(n+4)*(15*n^3 + 30*n^2 + 5*n - 2)/5760: n in [0..50]]; // G. C. Greubel, Oct 22 2017
  • Maple
    A001298:=-(1+22*z+58*z**2+24*z**3)/(z-1)**9; # Simon Plouffe in his 1992 dissertation, without the leading 0
  • Mathematica
    Table[StirlingS2[n+4, n], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)
    a[ n_] := n (n + 1) (n + 2) (n + 3) (n + 4) (15 n^3 + 30 n^2 + 5 n - 2) / 5760; (* Michael Somos, Sep 04 2017 *)
  • PARI
    {a(n) = n * (n+1) * (n+2) * (n+3) * (n+4) * (15*n^3 + 30*n^2 + 5*n - 2) / 5760}; /* Michael Somos, Sep 04 2017 */
    
  • Sage
    [stirling_number2(n+4,n) for n in range(0, 24)] # Zerinvary Lajos, May 16 2009
    

Formula

G.f.: x(1 + 22x + 58x^2 + 24x^3)/(1 - x)^9. - Paul Barry, Aug 05 2004
a(n) = Stirling2(n+4, n) = Sum_{L=1..n} (Sum_{k=1..L} (Sum_{j=1..k} (Sum_{i=1..j} i*j*k*L))) = (n+4)*(n+3)*(n+2)*(n+1)*n *(15*n^3 + 30*n^2 + 5*n - 2)/5760 = (15*n^3 + 30*n^2 + 5*n - 2)*binomial(n+4, 5)/48. - Vladeta Jovovic, Jan 31 2005
E.g.f. with offset -3: exp(x)*(1*(x^4)/4! + 26*(x^5)/5! + 130*(x^6)/6! + 210*(x^7)/7! +105*(x^8)/8!). For the coefficients [1, 26, 130, 210, 105] see triangle A112493. E.g.f.: x*exp(x)*(15*x^7 + 600*x^6 + 8600*x^5 + 55248*x^4 + 162960*x^3 + 202560*x^2 + 83520*x + 5760)/5760. Above given e.g.f. differentiated three times.
O.g.f. is D^4(x/(1-x)), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
a(n) = A000915(-4-n) for all n in Z. - Michael Somos, Sep 04 2017

Extensions

Name edited and initial zero added by Nathaniel Johnston, Apr 30 2011

A111999 T(n, k) = [x^k] (-1)^n*Sum_{k=0..n} E2(n, n-k)*(1+x)^(n-k) where E2(n, k) are the second-order Eulerian numbers. Triangle read by rows, T(n, k) for n >= 1 and 0 <= k <= n.

Original entry on oeis.org

-1, 3, 2, -15, -20, -6, 105, 210, 130, 24, -945, -2520, -2380, -924, -120, 10395, 34650, 44100, 26432, 7308, 720, -135135, -540540, -866250, -705320, -303660, -64224, -5040, 2027025, 9459450, 18288270, 18858840, 11098780, 3678840, 623376, 40320, -34459425, -183783600, -416215800
Offset: 1

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

Previous name was: A triangle that converts certain binomials into triangle A008276 (diagonals of signed Stirling1 triangle A008275).
Stirling1(n,n-m) = A008275(n,n-m) = Sum_{k=0..m-1}a(m,k)*binomial(n,2*m-k).
The unsigned column sequences start with A001147, A000906 = 2*A000457, 2*|A112000|, 4*|A112001|.
The general results on the convolution of the refined partition polynomials of A133932, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these unsigned polynomials. - Tom Copeland, Sep 20 2016

Examples

			Triangle starts:
  [1]      -1;
  [2]       3,       2;
  [3]     -15,     -20,       -6;
  [4]     105,     210,      130,       24;
  [5]    -945,   -2520,    -2380,     -924,     -120;
  [6]   10395,   34650,    44100,    26432,     7308,     720;
  [7] -135135, -540540,  -866250,  -705320,  -303660,  -64224,  -5040;
  [8] 2027025, 9459450, 18288270, 18858840, 11098780, 3678840, 623376, 40320.
		

References

  • Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 152. Table C_{m, nu}.

Crossrefs

Row sums give A032188(m+1)*(-1)^m, m>=1. Unsigned row sums give A032188(m+1), m>=1.
Cf. A008517 (second-order Eulerian triangle) for a similar formula for |Stirling1(n, n-m)|.

Programs

  • Maple
    CoeffList := p -> op(PolynomialTools:-CoefficientList(p, x)):
    E2 := (n, k) -> combinat[eulerian2](n, k):
    poly := n -> (-1)^n*add(E2(n, n-k)*(1+x)^(n-k), k = 0..n):
    seq(CoeffList(poly(n)), n = 1..8); # Peter Luschny, Feb 05 2021
  • Mathematica
    a[m_, k_] := a[m, k] = Which[m < k + 1, 0, And[m == 1, k == 0], -1, k == -1, 0, True, -(2 m - k - 1)*(a[m - 1, k] + a[m - 1, k - 1])]; Table[a[m, k], {m, 9}, {k, 0, m - 1}] // Flatten (* Michael De Vlieger, Sep 23 2016 *)

Formula

a(m, k)=0 if m
From Tom Copeland, May 05 2010 (updated Sep 12 2011): (Start)
The integral from 0 to infinity w.r.t. w of
exp[-w(u+1)] (1+u*z*w)^(1/z) gives a power series, f(u,z), in z for reversed row polynomials in u of A111999, related to an Euler transform of diagonals of A008275.
Let g(u,x) be obtained from f(u,z) by replacing z^n with x^(n+1)/(n+1)!;
g(u,x)= x - u^2 x^2/2! + (2 u^3 + 3 u^4) x^3/3! - (6 u^4 + 20 u^5 + 15 u^6) x^4/4! + ... , an e.g.f. associated to f(u,z).
Then g^(-1)(u,x)=(1+u)*x - log(1+u*x) is the comp. inverse of g(u,x) in x, and, consequently, A133932 is a refinement of A111999.
With h(u,x)= 1/(dg^(-1)/dx)= (1+u*x)/(1+(1+u)*u*x),
g(u,x)=exp[x*h(u,t)d/dt] t, evaluated at t=0. Also, dg(u,x)/dx = h(u,g(u,x)). (End)
From Tom Copeland, May 06 2010: (Start)
For m,k>0, a(m,k) = Sum(j=2 to 2m-k+1): (-1)^(2m-k+1+j) C(2m-k+1,j) St1d(j,m),
where C(n,j) is the binomial coefficient and St1d(j,m) is the (j-m)-th element of the m-th subdiagonal of A008275 for (j-m)>0 and is 0 otherwise,
e.g., St1d(1,1) = 0, St1d(2,1) = -1, St1d(3,1) = -3, St1d(4,1) = -6. (End)
From Tom Copeland, Sep 03 2011 (updated Sep 12 2011): (Start)
The integral from 0 to infinity w.r.t. w of
exp[-w*(u+1)/u] (1+u*z*w)^(1/(u^2*z)) gives a power series, F(u,z), in z for the row polynomials in u of A111999.
Let G(u,x) be obtained from F(u,z) by replacing z^n with x^(n+1)/(n+1)!;
G(u,x) = x - x^2/2! + (3 + 2 u) x^3/3! - (15 + 20 u + 6 u^2) x^4/4! + ... , an e.g.f. for A111999 associated to F(u,z).
G^(-1)(u,x) = ((1+u)*u*x - log(1+u*x))/u^2 is the comp. inverse of G(u,x) in x.
With H(u,x) = 1/(dG^(-1)/dx) = (1+u*x)/(1+(1+u)*x),
G(u,x) = exp[x*H(u,t)d/dt] t, evaluated at t=0. Also, dG(u,x)/dx = H(u,G(u,x)). (End)
From Tom Copeland, Sep 16 2011: (Start)
f(u,z) and F(u,z) are expressible in terms of the incomplete gamma function Γ(v,p)(see Laplace Transforms for Power-law Functions at EqWorld):
With K(p,s) = p^(-s-1) exp(p) Γ(s+1,p),
f(u,z) = K(p,s)/(u*z) with p=(u+1)/(u*z) and s=1/z , and
F(u,z) = K(p,s)/(u*z) with p=(u+1)/(u^2*z) and s=1/(u^2*z). (End)
Diagonals of A008306 are reversed rows of A111999 (see P. Bala). - Tom Copeland, May 08 2012

Extensions

New name from Peter Luschny, Feb 05 2021

A001662 Coefficients of Airey's converging factor.

Original entry on oeis.org

0, 1, 1, -1, -1, 13, -47, -73, 2447, -16811, -15551, 1726511, -18994849, 10979677, 2983409137, -48421103257, 135002366063, 10125320047141, -232033147779359, 1305952009204319, 58740282660173759, -1862057132555380307, 16905219421196907793, 527257187244811805207
Offset: 0

Keywords

Comments

A051711 times the coefficient in expansion of W(exp(x)) about x=1, where W is the Lambert function. - Paolo Bonzini, Jun 22 2016
The polynomials with coefficients in triangle A008517, evaluated at -1.

Examples

			G.f. = x + x^2 - x^3 - x^4 + 13*x^5 - 47*x^6 - 73*x^7 + 2447*x^8 + ... - _Michael Somos_, Jun 23 2019
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    with(combinat); A001662 := proc(n) add((-1)^k*eulerian2(n-1,k),k=0..n-1) end:
    seq(A001662(i),i=0..23); # Peter Luschny, Nov 13 2012
  • Mathematica
    a[0] = 0; a[n_] := Sum[ (n+k-1)! * Sum[ (-1)^j/(k-j)! * Sum[ 1/i! * StirlingS1[n-i+j-1, j-i] / (n-i+j-1)!, {i, 0, j}] * 2^(n-j-1), {j, 0, k}], {k, 0, n-1}]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jul 26 2013, after Vladimir Kruchinin *)
    a[ n_] := If[ n < 1, 0, 2^(n - 1) Sum[ (-2)^-j StirlingS1[n - i + j - 1, j - i] Binomial[n + k - 1, n + j - 1] Binomial[n + j - 1, i], {k, 0, n - 1}, {j, 0, k}, {i, 0, j}]]; (* Michael Somos, Jun 23 2019 *)
    len := 12; gf := (1/2) (LambertW[Exp[x + 1]] - 1);
    ser := Series[gf, { x, 0, len}]; norm := Table[n! 4^n, {n, 0, len}];
    CoefficientList[ser, x] * norm (* Peter Luschny, Jun 24 2019 *)
  • Maxima
    a(n):= if n=0 then 1 else (sum((n+k-1)!*sum(((-1)^(j)/(k-j)!*sum((1/i! *stirling1(n-i+j-1, j-i))/(n-i+j-1)!, i, 0, j))*2^(n-j-1), j, 0, k), k, 0, n-1)); /* Vladimir Kruchinin, Nov 11 2012 */
  • SageMath
    @CachedFunction
    def eulerian2(n, k):
        if k==0: return 1
        elif k==n: return 0
        return eulerian2(n-1, k)*(k+1)+eulerian2(n-1, k-1)*(2*n-k-1)
    def A001662(n): return add((-1)^k*eulerian2(n-1,k) for k in (0..n-1))
    [A001662(m) for m in (0..23)] # Peter Luschny, Nov 13 2012
    

Formula

Let b(n) = 0, 1, -1, 1, 1, -13,.. be the sequence with all signs but one reversed: b(1)=a(1), b(n)=-a(n) for n<>1. Define the e.g.f. B(x) = 2*Sum_{n>=0} b(n)*(x/2)^n/n!. B(x) satisfies exp(B(x)) = 1 + 2*x - B(x). [Bernstein/Sloane S52]
Similarly, c(0)=1, c(n)=-a(n+1) are the alternating row sums of the second-order Eulerian numbers A340556, or c(n) = E2poly(n,-1). - Peter Luschny, Feb 13 2021
a(n) = Sum_{k=0..n-1} (n+k-1)!*Sum_{j=0..k} ((-1)^j/(k-j)!)*Sum_{i=0..j} ((1/i!)*Stirling1(n-i+j-1,j-i)/(n-i+j-1)!)*2^(n-j-1), n > 0, a(0)=1. - Vladimir Kruchinin, Nov 11 2012
From Sergei N. Gladkovskii, Nov 24 2012, Aug 22 2013: (Start)
Continued fractions:
G.f.: 2*x - x/G(0) where G(k) = 1 - 2*x*k + x*(k+1)/G(k+1).
G.f.: 2*x - 2*x/U(0) where U(k) = 1 + 1/(1 - 4*x*(k+1)/U(k+1)).
G.f.: A(x) = x/G(0) where G(k) = 1 - 2*x*(k+1) + x*(k+1)/G(k+1).
G.f.: 2*x - x*W(0) where W(k) = 1 + x*(2*k+1)/( x*(2*k+1) + 1/(1 + x*(2*k+2)/( x*(2*k+2) + 1/W(k+1)))). (End)
a(n) = 4^n * Sum_{i=1..n} Stirling2(n,i)*A013703(i)/2^(2*i+1). - Paolo Bonzini, Jun 23 2016
E.g.f.: 1/2*(LambertW(exp(4*x+1))-1). - Vladimir Kruchinin, Feb 18 2018
a(0) = 0; a(1) = 1; a(n) = 2 * a(n-1) - Sum_{k=1..n-1} binomial(n-1,k) * a(k) * a(n-k). - Ilya Gutkovskiy, Aug 28 2020

Extensions

More terms from James Sellers, Dec 07 1999
Reverted to converging factors definition by Paolo Bonzini, Jun 23 2016

A008789 a(n) = n^(n+3).

Original entry on oeis.org

0, 1, 32, 729, 16384, 390625, 10077696, 282475249, 8589934592, 282429536481, 10000000000000, 379749833583241, 15407021574586368, 665416609183179841, 30491346729331195904, 1477891880035400390625, 75557863725914323419136
Offset: 0

Keywords

Programs

  • GAP
    List([0..20], n-> n^(n+3)); # G. C. Greubel, Sep 11 2019
  • Magma
    [n^(n+3): n in [0..20]]; // Vincenzo Librandi, Jun 11 2013
    
  • Maple
    printlevel := -1; a := [0]; T := x->-LambertW(-x); f := series((T(x)*(1+8*T(x)+6*(T(x))^2)/(1-T(x))^7),x,24); for m from 1 to 23 do a := [op(a),op(2*m-1,f)*m! ] od; print(a); # Len Smiley, Nov 19 2001
  • Mathematica
    Table[n^(n+3),{n,0,20}](* Vladimir Joseph Stephan Orlovsky, Dec 26 2010 *)
  • PARI
    vector(20, n, (n-1)^(n+2)) \\ G. C. Greubel, Sep 11 2019
    
  • Sage
    [n^(n+3) for n in (0..20)] # G. C. Greubel, Sep 11 2019
    

Formula

E.g.f.(x): T*(1 +8*T +6*T^2)*(1-T)^(-7); where T=T(x) is Euler's tree function (see A000169). - Len Smiley, Nov 19 2001
See A008517 and A134991 for similar e.g.f.s and diagonals of A048993. - Tom Copeland, Oct 03 2011
E.g.f.: d^3/dx^3 {x^3/(T(x)^3*(1-T(x)))}, where T(x) = Sum_{n>=1} n^(n-1)*x^n/n! is the tree function of A000169. - Peter Bala, Aug 05 2012
a(n) = n*A008788(n). - R. J. Mathar, Oct 31 2015

A008790 a(n) = n^(n+4).

Original entry on oeis.org

0, 1, 64, 2187, 65536, 1953125, 60466176, 1977326743, 68719476736, 2541865828329, 100000000000000, 4177248169415651, 184884258895036416, 8650415919381337933, 426878854210636742656, 22168378200531005859375
Offset: 0

Keywords

Programs

Formula

E.g.f.: T*(1 +22*T +58*T^2 +24*T^3)*(1-T)^(-9); where T is Euler's tree function (see A000169). - Len Smiley, Nov 17 2001
See A008517 and A134991 for similar e.g.f.s and diagonals of A048993. - Tom Copeland, Oct 03 2011
E.g.f.: d^4/dx^4 {x^4/(T(x)^4*(1-T(x)))}, where T(x) = Sum_{n>=1} n^(n-1)*x^n/n! is the tree function of A000169. - Peter Bala, Aug 05 2012

A002539 Eulerian numbers of the second kind: <>.

Original entry on oeis.org

1, 22, 328, 4400, 58140, 785304, 11026296, 162186912, 2507481216, 40788301824, 697929436800, 12550904017920, 236908271543040, 4687098165573120, 97049168010017280, 2099830209402931200, 47405948832458496000, 1115089078488795648000, 27290469545695931904000, 694002594415741341696000
Offset: 0

Keywords

Comments

Eulerian permutations of the multiset {1,1,2,2,...,n+3,n+3} with n ascents.Eulerian permutations have the restriction that for all m, all integers between the two copies of m are less than m. In particular, the two 1s are always next to each other.
The sequence gives the Eulerian numbers <<3,0>>, <<4,1>>, <<5,2>>, <<6,3>>, ... (and in particular the offset is 0).

Examples

			For instance, a(1) = 22 because among the 7!! = 105 permutations of {1,1,2,2,3,3,4,4} selected according to the definition of Eulerian numbers of the second kind, only 22 contain n = 1 descent, namely : 11223443, 11224433, 11233244, 11233442, 11244233, 11332244, 11334422, 11442233, 12213344, 12233144, 12233441, 12244133, 13312244, 13344122, 14412233, 22113344, 22331144, 22334411, 22441133, 33112244, 33441122, 44112233. - _Jean-François Alcover_, Mar 28 2011
		

References

  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2nd edition; Addison-Wesley, 1994, pp. 270-271.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

3rd diagonal of A008517, third column of A112007.

Programs

  • Mathematica
    b[1]=1; b[2]=22; b[n_] := b[n] = ((n-1)*(n-1)!*n^3 - (n+2)*(n+3)*b[n-2]*n + (n*(2*n+5)-4)*b[n-1]) / (n-1); a[n_] := b[n+1]; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Mar 23 2011, updated Oct 12 2015 *)
  • PARI
    {a=vector(30,n,1);a[2]=22;for(n=3,#a,a[n]=(n-1)!*n^3+((n*(2*n+5)-4)*a[n-1] - n*(n+2)*(n+3)*a[n-2])/(n-1));a} \\ Uses offet 1 for technical reasons. - M. F. Hasler, Sep 19 2015

Formula

a(n)= (n+5)*a(n-1) + (n+1)*A002538(n+1), n>=1, a(0)=1.
Recurrence: (n-1)*n^2*a(n) = (n-1)*(3*n^3 + 12*n^2 + 6*n + 1)*a(n-1) - (n+1)*(3*n^4 + 15*n^3 + 13*n^2 - 15*n - 4)*a(n-2) + n*(n+1)^3*(n+2)*(n+3)*a(n-3). - Vaclav Kotesovec, May 24 2014
a(n) ~ n! * n^5 * log(n) * (log(n)*(1/2+181/(24*n)) + gamma*(1+181/(12*n)) - 2 - 65/(3*n)), where gamma is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, May 24 2014

Extensions

Formulas adapted for offset 0 by Vaclav Kotesovec, May 24 2014
More terms from M. F. Hasler, Sep 19 2015

A106828 Triangle T(n,k) read by rows: associated Stirling numbers of first kind (n >= 0 and 0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 0, 0, 1, 0, 2, 0, 6, 3, 0, 24, 20, 0, 120, 130, 15, 0, 720, 924, 210, 0, 5040, 7308, 2380, 105, 0, 40320, 64224, 26432, 2520, 0, 362880, 623376, 303660, 44100, 945, 0, 3628800, 6636960, 3678840, 705320, 34650, 0, 39916800, 76998240, 47324376, 11098780, 866250, 10395
Offset: 0

Author

N. J. A. Sloane, May 22 2005

Keywords

Comments

Another version of the triangle is in A008306.
A signed version of this triangle is given by the exponential Riordan array [1, log(1+t)-t]. Its row sums are (-1)^n*(1-n). Another version is [1, log(1-t)+t], whose row sums are 1-n. - Paul Barry, May 10 2008

Examples

			Rows 0 though 7 are:
  1;
  0,
  0,   1;
  0,   2,
  0,   6,   3;
  0,  24,  20,
  0, 120, 130,  15;
  0, 720, 924, 210;
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 256.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 75.

Crossrefs

See A008306 for more information.
Cf. A008619 (row lengths), A000166 (row sums).

Programs

  • Haskell
    a106828 n k = a106828_tabf !! n !! k
    a106828_row n = a106828_tabf !! n
    a106828_tabf = map (fst . fst) $ iterate f (([1], [0]), 1) where
       f ((us, vs), x) =
         ((vs, map (* x) $ zipWith (+) ([0] ++ us) (vs ++ [0])), x + 1)
    -- Reinhard Zumkeller, Aug 05 2013
    
  • Maple
    A106828 := (n,k) -> add(binomial(j,n-2*k)*combinat:-eulerian2(n-k,j),j=0..n-k):
    seq(print(seq(A106828(n,k),k=0..iquo(n,2))),n=0..9); # Peter Luschny, Apr 20 2011 (revised Jan 13 2016)
    # Second program, after David Callan in A008306:
    A106828 := proc(n, k) option remember; if k = 0 then k^n elif k = 1 then (n-1)! elif n <= 2*k-1 then 0 else (n-1)*(procname(n-1, k) + procname(n-2, k-1)) fi end: seq((seq(A106828(n, k), k = 0..iquo(n, 2))), n=0..12); # Peter Luschny, Aug 24 2021
  • Mathematica
    Eulerian2[n_, k_] := Eulerian2[n, k] = If[k == 0, 1, If[k == n, 0, Eulerian2[n - 1, k] (k + 1) + Eulerian2[n - 1, k - 1] (2 n - k - 1)]];
    T[n_, k_] := Sum[Binomial[j, n - 2 k] Eulerian2[n - k, j], {j, 0, n - k}];
    Table[T[n, k], {n, 0, 12}, {k, 0, n/2}] (* Jean-François Alcover, Jun 13 2019 *)
  • PARI
    E2(n, m) = sum(k=0, n-m, (-1)^(n+k)*binomial(2*n+1, k)*stirling(2*n-m-k+1, n-m-k+1, 1)); \\ A008517
    T(n, k) = if ((n==0) && (k==0), 1, sum(j=0, n-k, binomial(j, n-2*k)*E2(n-k, j+1))); \\ Michel Marcus, Dec 07 2021
    
  • Python
    from math import factorial
    def A106828(n, k):
        return k**n if k == 0 else factorial(n-1) if k == 1 else 0 if n <= 2*k - 1 else (n - 1)*(A106828(n-1, k) + A106828(n-2, k-1))
    for n in range(14): print([A106828(n, k) for k in range(n//2 + 1)])
    # Mélika Tebni, Dec 07 2021, after second Maple script.
  • Sage
    # uses[bell_transform from A264428]
    # Computes the full triangle 0<=k<=n.
    def A106828_row(n):
        g = lambda k: factorial(k) if k>0 else 0
        s = [g(k) for k in (0..n)]
        return bell_transform(n, s)
    [A106828_row(n) for n in (0..8)] # Peter Luschny, Jan 13 2016
    

Formula

T(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, j), where E2 are the second-order Eulerian numbers (A008517). - Peter Luschny, Jan 13 2016
Also the Bell transform of the sequence g(k) = k! if k>0 else 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 13 2016

Extensions

Removed extra 0 in row 1 from Michael Somos, Jan 19 2011

A214406 Triangle of second-order Eulerian numbers of type B.

Original entry on oeis.org

1, 1, 1, 1, 8, 3, 1, 33, 71, 15, 1, 112, 718, 744, 105, 1, 353, 5270, 14542, 9129, 945, 1, 1080, 33057, 191384, 300291, 129072, 10395, 1, 3265, 190125, 2033885, 6338915, 6524739, 2071215, 135135, 1, 9824, 1038780, 18990320, 103829590, 204889344, 150895836, 37237680, 2027025
Offset: 0

Author

Peter Bala, Jul 17 2012

Keywords

Comments

The second-order Eulerian numbers A008517 count Stirling permutations by ascents. A Stirling permutation of order n is a permutation of the multiset {1,1,2,2,...,n,n} such that for each i, 1 <= i <= n, the elements lying between the two occurrences of i are larger than i.
We define a signed Stirling permutation of order n to be a vector (x_0, x_1, ..., x_(2*n)) such that x_0 = 0 and (|x_1|, ... ,|x_(2*n)|) is a Stirling permutation of order n. We say that a signed Stirling permutation (x_0, x_1, ... , x_(2*n)) has an ascent at position j, 0 <= j <= 2*n-1, if |x_j| < |x_(j+1)|. We define T(n,k), the second-order Eulerian numbers of type B, as the number of signed Stirling permutations of order n having k ascents. An example is given below.

Examples

			Row 2: [1,8,3]:
Signed Stirling permutations of order 2
= = = = = = = = = = = = = = = = = = = =
..............ascents...................ascents
(0 2 2 1 1)......1.......(0 -2 -2 1 1).....1
(0 1 2 2 1)......2.......(0 1 -2 -2 1).....2
(0 1 1 2 2)......2.......(0 1 1 -2 -2).....1
(0 2 2 -1 -1)....1.......(0 -2 -2 -1 -1)...1
(0 -1 2 2 -1)....1.......(0 -1 -2 -2 -1)...1
(0 -1 -1 2 2)....1.......(0 -1 -1 -2 -2)...0
............................................
Triangle begins
.n\k.|..0.....1......2.......3......4........5......6
= = = = = = = = = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1.....1
..2..|..1.....8......3
..3..|..1....33.....71......15
..4..|..1...112....718.....744....105
..5..|..1...353...5270...14542...9129......945
..6..|..1..1080..33057..191384..300291..129072..10395
...
Recurrence example: T(4,2) = 11*T(3,1) + 5*T(3,2) = 11*33 + 5*71 = 718.
		

Crossrefs

Cf. A001813 (row sums), A008517, A039755, A185896, A034940.

Programs

  • Mathematica
    T[n_, k_] /; 0 < k <= n := T[n, k] = (4n-2k-1) T[n-1, k-1] + (2k+1) T[n-1, k]; T[, 0] = 1; T[, _] = 0;
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)

Formula

T(n,k) = Sum_{i = 0..k} (-1)^(i-k)*binomial(2*n+1,k-i)*S(n+i,i), where S(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n = A039755(n,k).
It appears that Sum_{k = 0..n} (-1)^(k+1)*T(n,k)/((2*n-k)*binomial(2*n,k)) = (-1)^n *(2^n-2)*Bernoulli(n)/n.
Recurrence equation: T(n,k) = (4*n-2*k-1)*T(n-1,k-1) + (2*k+1)*T(n-1,k), for n,k >= 0.
The row polynomials R(n,x) may be calculated by means of the recurrence equation R(0,x) = 1 and for n >= 0, R(n,x^2) = (1 - x^2)^(2*n)*d/dx( x/(1-x^2)^(2*n-1)*R(n-1,x^2) ). Equivalently, x*R(n,x^2)/(1 - x^2)^(2*n+1)) = D^n(x), where D is the differential operator x/(1 - x^2)*d/dx.
Another recurrence is R(n+1,x) = 2*x*(1 - x)*d/dx(R(n,x)) + (1 + (4*n+1)*x)*R(n,x). It follows that the row polynomials R(n,x) have only real zeros (apply Liu and Wang, Corollary 1.2 with f(x) = R(n,x) and g(x) = R'(n,x)).
For n >= 0, the rational functions Q(n,x) := R(n,x)/(1 - x)^(2*n+1) are the o.g.f.'s for the diagonals of the type B Stirling numbers of the second kind A039755. They appear to satisfy the semi-orthogonality property Integral_{x = 0..oo} (1 - x)*Q(n,x)*Q(m,x) dx = (-1)^n*(2^(n+m) - 2)*Bernoulli(n+m)/(n+m), for n, m >= 0 but excluding the case (n,m) = (0,0). A similar result holds for the row polynomials of A185896.
Row sums are A001813.
Define functions F(n,z) := Sum_{k >= 0} (2*k+1)^(k+n)*z^k/k!, n = 0,1,2,.... Then exp(-x/2)*F(n,x/2*exp(-x)) = R(n,x)/(1 - x)^(2*n+1). - Peter Bala, Jul 26 2012

Extensions

Missing 1 in data inserted by Jean-François Alcover, Nov 11 2019

A004301 Second-order Eulerian numbers <>.

Original entry on oeis.org

0, 6, 58, 328, 1452, 5610, 19950, 67260, 218848, 695038, 2170626, 6699696, 20507988, 62407890, 189123286, 571432036, 1722945672, 5187185766, 15600353130, 46882846680, 140820504700, 422822222266, 1269221639358, 3809241974028, 11431014253872, 34299887862990
Offset: 2

Keywords

Comments

See A008517 for the definition of second-order Eulerian numbers.

Examples

			G.f. = 6*x^3 + 58*x^4 + 328*x^5 + 1452*x^6 + 5610*x^7 + 19950*x^8 + ...
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd edition. Addison-Wesley, Reading, MA, 1994, p. 270.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

3rd column of A008517.
2nd column of A201637.
Equals the fourth right hand column of triangle A163936. - Johannes W. Meijer, Oct 16 2009

Programs

  • Mathematica
    LinearRecurrence[{10, -40, 82, -91, 52, -12}, {0, 6, 58, 328, 1452, 5610}, 26] (* Jean-François Alcover, Feb 27 2019 *)
  • PARI
    {a(n) = if( n<0, 0, (9*3^n - (12 + 8*n)*2^n + (3 + 6*n + 4*n^2))/2)}; /* Michael Somos, Oct 13 2002 */

Formula

From Michael Somos, Oct 13 2002: (Start)
G.f.: x^3(6-2x-12x^2)/((1-x)^3(1-2x)^2(1-3x)).
a(n) = A008517(n, 3) = (9*3^n - (12+8*n)*2^n + (3+6*n+4*n^2))/2. (End)
a(n) = Sum_{k=0..n-3} (-1)^(n+k)*binomial(2*n+1, k)*Stirling1(2*n-k-2, n-k-2). - Johannes W. Meijer, Oct 16 2009

Extensions

Edited by Olivier Gérard, Mar 28 2011
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