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 15 results. Next

A298013 Differentiate the g.f. underlying A032188 and evaluate it at x=1.

Original entry on oeis.org

0, 1, 8, 91, 1334, 23413, 506652, 12386183
Offset: 0

Views

Author

N. J. A. Sloane, Jan 11 2018

Keywords

Comments

See Riordan [1980] for details.

Crossrefs

Cf. A032188.

A006351 Number of series-parallel networks with n labeled edges. Also called yoke-chains by Cayley and MacMahon.

Original entry on oeis.org

1, 2, 8, 52, 472, 5504, 78416, 1320064, 25637824, 564275648, 13879795712, 377332365568, 11234698041088, 363581406419456, 12707452084972544, 477027941930515456, 19142041172838025216, 817675811320888020992, 37044610820729973813248, 1774189422608238694776832
Offset: 1

Views

Author

Keywords

Comments

For a simple relationship to series-reduced rooted trees, partitions of n, and phylogenetic trees among other combinatoric constructs, see comments in A000311. - Tom Copeland, Jan 06 2021

Examples

			D^3(1) = (12*x^2+56*x+52)/(x-1)^6. Evaluated at x = 0 this gives a(4) = 52.
a(3) = 8: The 8 possible increasing plane trees on 3 vertices with vertices of outdegree k >= 1 coming in 2 colors, B or W, are
.......................................................
.1B..1B..1W..1W.....1B.......1W........1B........1W....
.|...|...|...|...../.\....../..\....../..\....../..\...
.2B..2W..2B..2W...2...3....2....3....3....2....3....2..
.|...|...|...|.........................................
.3...3...3...3.........................................
G.f. = x + 2*x^2 + 8*x^3 + 52*x^4 + 472*x^5 + 5504*x^6 + 78416*x^7 + ...
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, p. 417.
  • P. A. MacMahon, Yoke-trains and multipartite compositions in connexion with the analytical forms called "trees", Proc. London Math. Soc. 22 (1891), 330-346; reprinted in Coll. Papers I, pp. 600-616. Page 333 gives A000084 = 2*A000669.
  • P. A. MacMahon, The combination of resistances, The Electrician, 28 (1892), 601-602; reprinted in Coll. Papers I, pp. 617-619.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 142.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.40(a), S(x).

Crossrefs

Cf. A000311, A000084 (for unlabeled case), A032188. A140945.

Programs

  • Maple
    read transforms; t1 := 2*ln(1+x)-x; t2 := series(t1,x,10); t3 := seriestoseries(t2,'revogf'); t4 := SERIESTOLISTMULT(%);
    # N denotes all series-parallel networks, S = series networks, P = parallel networks;
    spec := [ N, N=Union(Z,S,P),S=Set(Union(Z,P),card>=2), P=Set(Union(Z,S), card>=2)}, labeled ]: A006351 := n->combstruct[count](spec,size=n);
    A006351 := n -> add(combinat[eulerian2](n-1,k)*2^(n-k-1),k=0..n-1):
    seq(A006351(n), n=1..18); # Peter Luschny, Nov 16 2012
  • Mathematica
    max = 18; f[x_] := 2*Log[1+x]-x; Rest[ CoefficientList[ InverseSeries[ Series[ f[x], {x, 0, max}], x], x]]*Range[max]! (* Jean-François Alcover, Nov 25 2011 *)
  • Maxima
    a(n):=if n=1 then 1 else ((n-1)!*sum(binomial(n+k-1,n-1)* sum((-1)^(j)*binomial(k,j)*sum((binomial(j,l)*(j-l)!*2^(j-l)*(-1)^l* stirling1(n-l+j-1,j-l))/(n-l+j-1)!,l,0,j),j,1,k),k,1,n-1)); /* Vladimir Kruchinin, Jan 24 2012 */
    
  • PARI
    x='x+O('x^66); Vec(serlaplace(serreverse( 2*log(1+x) - 1*x ))) \\ Joerg Arndt, May 01 2013
  • Sage
    # uses[eulerian2 from A201637]
    def A006351(n): return add(A201637(n-1, k)*2^(n-k-1) for k in (0..n-1))
    [A006351(n) for n in (1..18)]  # Peter Luschny, Nov 16 2012
    

Formula

For n >= 2, A006351(n) = 2*A000311(n) = A005640(n)/2^n. Row sums of A140945.
E.g.f. is reversion of 2*log(1+x)-x.
Also exponential transform of A000311, define b by 1+sum b_n x^n / n! = exp ( 1 + sum a_n x^n /n!).
E.g.f.: A(x), B(x)=x*A(x) satisfies the differential equation B'(x)=(1+B(x))/(1-B(x)). - Vladimir Kruchinin, Jan 18 2011
From Peter Bala, Sep 05 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A'(x) = (1+A)/(1-A) with A(0) = 0. Hence the inverse function A^-1(x) = int {t = 0..x} (1-t)/(1+t) = 2*log(1+x)-x, which yields A(x) = -1-2*W(-1/2*exp((x-1)/2)), where W is the Lambert W function.
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-x)*g(x)). Compare with A032188.
Applying [Bergeron et al., Theorem 1] to the result x = int {t = 0..A(x)} 1/phi(t), where phi(t) = (1+t)/(1-t) = 1 + 2*t + 2*t^2 + 2*t^3 + ..., leads to the following combinatorial interpretation for the sequence: a(n) gives the number of plane increasing trees on n vertices where each vertex of outdegree k >=1 can be in one of 2 colors. An example is given below. (End)
A134991 gives (b.+c.)^n = 0^n , for (b_n)=A000311(n+1) and (c_0)=1, (c_1)=-1, and (c_n)=-2* A000311(n) = -A006351(n) otherwise. E.g., umbrally, (b.+c.)^2 = b_2*c_0 + 2 b_1*c_1 + b_0*c_2 =0. - Tom Copeland, Oct 19 2011
G.f.: 1/S(0) where S(k) = 1 - x*(k+1) - x*(k+1)/S(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 18 2011
a(n) = ((n-1)!*sum(k=1..n-1, C(n+k-1,n-1)*sum(j=1..k, (-1)^(j)*C(k,j)* sum(l=0..j, (C(j,l)*(j-l)!*2^(j-l)*(-1)^l*stirling1(n-l+j-1,j-l))/ (n-l+j-1)!)))), n>1, a(1)=1. - Vladimir Kruchinin, Jan 24 2012
E.g.f.: A(x) = exp(B(x))-1 where B(x) is the e.g.f. of A000311. - Vladimir Kruchinin, Sep 25 2012
a(n) = sum_{k=0..n-1} A201637(n-1,k)*2^(n-k-1). - Peter Luschny, Nov 16 2012
G.f.: -1 + 2/Q(0), where Q(k)= 1 - k*x - x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
a(n) ~ sqrt(2)*n^(n-1)/((2*log(2)-1)^(n-1/2)*exp(n)). - Vaclav Kotesovec, Jul 17 2013
G.f.: Q(0)/(1-x), where Q(k) = 1 - x*(k+1)/( x*(k+1) - (1 -x*(k+1))*(1 -x*(k+2))/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 10 2013
a(1) = 1; a(n) = a(n-1) + Sum_{k=1..n-1} binomial(n-1,k) * a(k) * a(n-k). - Ilya Gutkovskiy, Aug 28 2020
Conjecture: a(n) = A379459(n-2,0) = A379460(n-1,0) for n > 1 with a(1) = 1. - Mikhail Kurkov, Jan 16 2025

A134685 Irregular triangle read by rows: coefficients C(j,k) of a partition transform for direct Lagrange inversion.

Original entry on oeis.org

1, -1, 3, -1, -15, 10, -1, 105, -105, 10, 15, -1, -945, 1260, -280, -210, 35, 21, -1, 10395, -17325, 6300, 3150, -280, -1260, -378, 35, 56, 28, -1, -135135, 270270, -138600, -51975, 15400, 34650, 6930, -2100, -1575, -2520, -630, 126, 84, 36, -1
Offset: 1

Views

Author

Tom Copeland, Jan 26 2008, Sep 13 2008

Keywords

Comments

Let f(t) = u(t) - u(0) = Ev[exp(u.* t) - u(0)] = log{Ev[(exp(z.* t))/z_0]} = Ev[-log(1- a.* t)], where the operator Ev denotes umbral evaluation of the umbral variables u., z. or a., e.g., Ev[a.^n + a.^m] = a_n + a_m . The relation between z_n and u_n is given in reference in A127671 and u_n = (n-1)! * a_n .
If u_1 is not equal to 0, then the compositional inverse for these expressions is given by g(t) = Sum_{j>=1} P(j,t) where, with u_n denoted by (n') for brevity,
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -(2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 10 (1')(2')(3') - (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 105 (1') (2')^2 (3') + 10 (1')^2 (3')^2 + 15 (1')^2 (2') (4') - (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 1260 (1') (2')^3 (3') - 280 (1')^2 (2') (3')^2 - 210 (1')^2 (2')^2 (4') + 35 (1')^3 (3')(4') + 21 (1')^3 (2')(5') - (1')^4 (6') ] * t^6 / 6!
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 17325 (1') (2')^4 (3') + (1')^2 [ 6300 (2')^2 (3')^2 + 3150 (2')^3 (4')] - (1')^3 [280 (3')^3 + 1260 (2')(3')(4') + 378 (2')^2(5')] + (1')^4 [35 (4')^2 + 56 (3')(5') + 28 (2')(6')] - (1')^5 (7') ] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 270270 (1') (2')^5 (3') - (1')^2 [ 138600 (2')^3 (3')^2 + 51975 (2')^4 (4')] + (1')^3 [15400 (2')(3')^3 + 34650 (2')^2(3')(4') + 6930 (2')^3(5')] - (1')^4 [2100 (3')^2(4') + 1575 (2')(4')^2 + 2520 (2')(3')(5') + 630 (2')^2(6') ] + (1')^5 [126 (4')(5') + 84 (3')(6') + 36 (2')(7')] - (1')^6 (8') ] * t^8 / 8!
...
Substituting ((m-1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n-1 listed in Abramowitz and Stegun on page 831 (in the reversed order) and the number of partitions in P(n,t) is given by A000041(n-1).
Combinatorial interpretations are given in the link.
From Tom Copeland, Jul 10 2018: (Start)
Coefficients occurring in prolongation for the special Euclidean group SE(2) and special affine group SA(2) in the Olver presentation on moving frames (MFP) in slides 33 and 42. These are a result of applying an iterated derivative of the form h(x)d/dx = d/dy as in this entry (more generally as g(x) d/dx as discussed in A145271). See also p. 6 of Olver's paper on contact forms, but note that the 12 should be a 15 in the formula for the compositional inverse of S(t).
Change variables in the MFP to obtain connections to the partition polynomials Prt_n = n! * P(n,1) above. Let delta and beta in the formulas for the equi-affine curves in MFP be L and B, respectively, and D_y = (1/(L-B*u_x)) d/dx = (1/w_x) d/dx. Then v_(yy) = (1/B) [-w_(xx)/(w_x)^3] in MFP (there is an overall sign error in MFP for v_(yy) and higher derivatives w.r.t. y), and (d/dy)^n v = v_n = (1/B)* [(1/w_1)*(d/dx)]^(n-2) [-w_2/(w_1)^3] for n > 1, with w_n = (d/dx)^n w. Consequently, in the partition polynomials Prt_n for n > 1 here substitute (n') = -B*u_n = w_n for n > 1 and (1') = L-B*u_1 = w_1, where u_n = (d/dx)^n u, and then divide by B. For example, v_4 = (1/B)*Prt_4 = (1/B)*4!*P(4,1) = (1/B) (L-B*u_n)^(-7) [-15*(-B*u_2)^3 + 10 (L-B*u_1)(-B*u_2)(-B*u_3) - (L-B*u_1)^2 (-B*u_4)], agreeing with v_4 in MFP except for the overall sign.
For the SE(2) transformation formulas in MFP, let w_x = cos(phi) + sin(phi)*u_x, and then the same transformations apply as above with cos(phi) and sin(phi) substituted for L and -B, respectively. (End)

Examples

			Examples and checks:
1) Let u_1 = -1 and u_n = 1 for n>1,
then f(t) = exp(u.*t) - u(0) = exp(t)-2t-1
and g(t) = [e.g.f. of signed A000311];
therefore, the row sums of unsigned [C(j,k)] are A000311 =
(0,1,1,4,26,236,2752,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),4!*P(4,1),...).
2) Let u_1 = -1 and u_n = (n-1)! for n>1,
then f(t) = -log(1-t)-2t
and g(t) = [e.g.f. of signed (0,A032188)]
with (0,A032188) = (0,1,1,5,41,469,6889,...) = (0,-P(1,1),2!*P(2,1),-3!P(3,1),...).
3) Let u_1 = -1 and u_n = (-1)^n (n-2)! for n>1, then
f(t) = (1+t) log(1+t) - 2t
and g(t) = [e.g.f. of signed (0,A074059)]
with (0,A074059) = (0,1,1,2,7,34,213,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),...).
4) Let u_1 = 1, u_2 = -1 and u_n = 0 for n>2,
then f(t) = t(1-t/2)
and g(t) = [e.g.f. of (0,A001147)] = 1 - (1-2t)^(1/2)
with (0,A001147) = (0,1,1,3,15,105,945...) =(0,P(1,1),2!*P(2,1),3!*P(3,1),...).
5) Let u_1 = 1, u_2 = -2 and u_n = 0 for n>2,
then f(t)= t(1-t)
and g(t) = t * [o.g.f. of A000108] = [1 - (1-4t)^(1/2)] / 2
with (0,A000108) = (0,1,1,2,5,14,42,...) = (0,P(1,1),P(2,1),P(3,1),...).
.
From _Peter Luschny_, Feb 19 2021: (Start)
Triangle starts:
 [1]  1;
 [2] -1;
 [3]  3,     -1;
 [4] -15,     10,    -1;
 [5]  105,   -105,   [10, 15],  -1;
 [6] -945,    1260,  [-280, -210], [35, 21],  -1;
 [7]  10395, -17325, [6300, 3150], [-280, -1260, -378], [35, 56, 28], -1;
 [8] -135135, 270270, [-138600, -51975], [15400, 34650, 6930], [-2100, -1575, -2520, -630], [126, 84, 36], -1
The coefficients can be seen as a refinement of the Ward numbers: Let R(n, k) = Sum T(n, k), where the sum collects adjacent terms with equal sign, as indicated by the square brackets in the table, then R(n+1, k+1) = (-1)^(n-k)*W(n, k), where W(n, k) are the Ward numbers A181996, for n >= 0 and 0 <= k <= n-1.  (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 831.
  • D. S. Alexander, A History of Complex Dynamics: From Schröder to Fatou to Julia, Friedrich Vieweg & Sohn, 1994, p. 10.
  • J. Riordan, Combinatorial Identities, Robert E. Krieger Pub. Co., 1979, (unsigned partition polynomials in Table 5.2 on p. 181, but may have errors).

Crossrefs

Cf. A145271, (A134991, A019538) = (reduced array, associated g(x)).
Cf. A181996 (Ward numbers).

Programs

  • Mathematica
    rows[n_] := {{1}}~Join~Module[{h = 1/(1 + Sum[u[k] y^k/k!, {k, n-1}] + O[y]^n), g = y, r}, r = Reap[Do[g = h D[g, y]; Sow[Expand[Normal@g /. {y -> 0}]], {k, n}]][[2, 1, ;;]]; Table[Coefficient[r[[k]], Product[u[t], {t, p}]], {k, 2, n}, {p, Reverse@Sort[Sort /@ IntegerPartitions[k-1]]}]];
    rows[8] // Flatten (* Andrei Zabolotskii, Feb 19 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [2!^e(2)*e(2)!*3!^e(3)*e(3)! ... n!^e(n)*e(n)! ].
From Tom Copeland, Sep 05 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u.*exp(u.*t)]
= 1/(u_1+(u_2)*t+(u_3)*t^2/2!+(u_4)*t^3/3!+...),
an e.g.f. for the partition polynomials of A133314
(signed A049019) with an index shift.
Then for the partition polynomials of A134685,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (Cf. A000311 and A134991)(End)
From Tom Copeland, Oct 30 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*h(y)d/dy] exp(t*y) eval. at y=0, the raising/creation and lowering/annihilation operators
defined by R PS(n,t)=PS(n+1,t) and L PS(n,t)= n*PS(n-1,t) are
R = t*h(d/dt) = t * 1/[u_1+(u_2)*d/dt+(u_3)*(d/dt)^2/2!+...], and
L = f(d/dt)=(u_1)*d/dt+(u_2)*(d/dt)^2/2!+(u_3)*(d/dt)^3/3!+....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2! + u_3 * x^2/3! + ... + u_n * x^(n-1)/n!]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix by u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) u_{n+1-k}. Then P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314. - Tom Copeland, Jul 15 2016
Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. - Tom Copeland, Aug 27 2016
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -u5, b2 = 15 u2 u4 + 10 u3^2, b3 = -105 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers' equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2,..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1,u_1=1,u_2,...,u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 105 u2^2 u3 + 1 * 15 u2 u4 + 1 * 10 u3^2 - 0 * u5 = 315 u2^4 - 210 u2^2 u3 + 15 u2 u4 + 10 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 10 u2 u3 - u4) + 6!/(3!*3!) * (3 u2^2 - u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the Bell polynomials of A036040 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016
Terms in rows 5-8 reordered by Andrei Zabolotskii, Feb 19 2024

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

A005172 Number of labeled rooted trees of subsets of an n-set.

Original entry on oeis.org

1, 4, 32, 416, 7552, 176128, 5018624, 168968192, 6563282944, 288909131776, 14212910809088, 772776684683264, 46017323176296448, 2978458881388183552, 208198894960190160896, 15631251601179130462208, 1254492810303112820555776, 107174403941451434687463424, 9711022458989438255300083712
Offset: 1

Keywords

Comments

Each node is a subset of the labeled set {1,...,n}. If the subset node is empty, it must have at least two children.
John W. Layman observes that this is the Stirling transform of A005264.

Examples

			x + 4*x^2 + 32*x^3 + 416*x^4 + 7552*x^5 + 176128*x^6 + 5018624*x^7 + ...
D^3(1) = 32*(12*x^2+28*x+13)/(2*x-1)^6. Evaluated at x = 0 this gives a(4) = 416.
From _Peter Bala_, Sep 06 2011: (Start)
a(3) = 32: The 32 increasing plane trees on 3 vertices with vertices of outdegree k coming in 2^(k+1) colors are
.
           1(x4 colors)      1(x8 colors)      1(x8 colors)
           |                / \               / \
           2(x4 colors)    2   3             3   2
           |
           3
.
  Totals: 16        +        8        +        8     = 32.   (End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.26.

Crossrefs

See A225170 for another version.

Programs

  • Maple
    with(combinat); A005172 := n -> add(eulerian2(n-1,k)*2^(2*n-k-2),k=0..n-1): seq(A005172(n), n=1..16); # Peter Luschny, Nov 10 2012
    A005172_list := proc(len) local A, n; A[1] := 1; for n from 2 to len do
    A[n] := 2*A[n-1] + add(binomial(n,j)*A[j]*A[n-j], j=1..n-1) od:
    convert(A,list) end: A005172_list(19); # Peter Luschny, May 24 2017
  • Mathematica
    max = 16; g[x_] := -1/2 - ProductLog[-E^(-1/2 + x)/2]; Drop[ CoefficientList[ Series[ g[x], {x, 0, max}], x]*Range[0, max]!, 1](* Jean-François Alcover, Nov 17 2011, after 1st e.g.f. *)
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ -1/2 - ProductLog[ -Exp[ -1/2 + z] / 2], {z, 0, n}]] (* Michael Somos, Jun 07 2012 *)
    a[1] = 1; a[n_] := (Sum[(n + k - 1)!*Sum[(-1)^j/(k - j)!*Sum[(-1)^i*2^(n - i + j - 1)*StirlingS1[n - i + j - 1, j - i]/((n - i + j - 1)!*i!), {i, 0, j}], {j, 1, k}], {k, 1, n - 1}]);
    Array[a, 20] (* Jean-François Alcover, Jun 24 2018, after Vladimir Kruchinin *)
    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] (2n - k - 1)]]; A005172[n_] := Sum[Eulerian2[n - 1, k] 2^(2 n - k - 2), {k, 0, n - 1}]; Table[A005172[n], {n, 19}] (* Peter Luschny, Jun 24 2018 *)
  • Maxima
    a(n):=if n=1 then 1 else (sum((n+k-1)!*sum((-1)^j/(k-j)!*sum((-1)^i*2^(n-i+j-1)*stirling1(n-i+j-1,j-i)/((n-i+j-1)!*i!),i,0,j),j,1,k),k,1,n-1)); /* Vladimir Kruchinin, Jan 30 2012 */
    
  • PARI
    {a(n)=local(A=1+x);for(i=0,n,A=1+intformal(A*(1+A+x*O(x^n))^2));n!*polcoeff(A,n)} \\ Paul D. Hanna, Sep 06 2008
    
  • PARI
    N=66; x='x+O('x^N);
    Q(k)=if(k>N,1,1-2*(k+1)*x-2*x*(k+1)/Q(k+1));
    gf=1/Q(0);  Vec(gf) \\ Joerg Arndt, May 01 2013
  • Sage
    @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)
    A005172 = lambda n: add(eulerian2(n-1,k)*2^(2*n-k-2) for k in (0..n-1))
    [A005172(n) for n in (1..16)]  # Peter Luschny, Nov 10 2012
    

Formula

E.g.f.: -1/2 - LambertW ( - exp( -1/2 + x) / 2 ).
E.g.f.: A(x) = 1 + Integral A(x)*(1 + A(x))^2 dx. - Paul D. Hanna, Sep 06 2008
From Peter Bala, Sep 06 2011: (Start)
The generating function A(x) = x+4*x^2/2!+32*x^3/3!+... satisfies the autonomous differential equation A'(x) = (1+2*A)/(1-2*A) with A(0) = 0. Hence the inverse function A^-1(x) = Integral_{t = 0..x} (1-2*t)/(1+2*t) dt = log(1+2*x)-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+2*x)/(1-2*x)*g(x)). Compare with A032188.
Applying [Bergeron et al., Theorem 1] to the result x = Integral_{t = 0..A(x)} 1/phi(t) dt, where phi(t) = (1+2*t)/(1-2*t) = 1+4*t+8*t^2+16*t^3+32*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)
a(n) = Sum_{k=1..n-1} (n+k-1)!*Sum_{j=1..k} (-1)^j/(k-j)!*Sum_{i=0..j} (-1)^i* 2^(n-i+j-1)*Stirling1(n-i+j-1,j-i)/((n-i+j-1)!*i!), n>1, a(1)=1. - Vladimir Kruchinin, Jan 30 2012
Let p(n,w) = w*Sum_{k=0..n-1}((-1)^k*E2(n-1,k)*w^k)/(1+w)^(2*n-1), E2 the second-order Eulerian numbers as defined by Knuth, then a(n) = -p(n,-1/2). - Peter Luschny, Nov 10 2012
a(n) ~ (2/(2*log(2)-1))^(n-1/2)*n^(n-1)/exp(n). - Vaclav Kotesovec, Jan 05 2013
G.f.: 1/Q(0), where Q(k)= 1 - 2*(k+1)*x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
a(n) = 2*a(n-1) + Sum_{j=1..n-1} binomial(n,j)*a(j)*a(n-j) for n>1, a(1) = 1. - Peter Luschny, May 24 2017
a(1) = 1; a(n) = n! * [x^n] exp(x + Sum_{k=1..n-1} a(k)*x^k/k!). - Ilya Gutkovskiy, Oct 18 2017

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

A112487 a(n) = Sum_{k=0..n} E2(n, k)*2^k, where E2(n, k) are the second-order Eulerian numbers A340556.

Original entry on oeis.org

1, 2, 10, 82, 938, 13778, 247210, 5240338, 128149802, 3551246162, 109979486890, 3764281873042, 141104799067178, 5749087305575378, 252969604725106090, 11955367835505775378, 603967991604199335722, 32479636694930586142802, 1852497140997527094395050
Offset: 0

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

Previous name: Row sums of triangle A112486.

Programs

  • Maple
    A112487 := proc(n)
        add(A112486(n,k),k=0..n) ;
    end proc: # R. J. Mathar, Dec 19 2013
    seq(op(k, convert(asympt(GAMMA(n, 2*n)*exp(2*n)/(2*n)^n, n, 20), polynom))*(-1)^(k+1)*n^k, k = 1..19); # Maple 2017, Vaclav Kotesovec, Aug 14 2017
    E2 := (n, k) -> `if`(k=0, k^n, combinat:-eulerian2(n, k-1));
    a := n -> add(E2(n, k)*2^k, k=0..n):
    seq(a(n), n=0..17); # Peter Luschny, Feb 13 2021
  • Mathematica
    a[n_] := (n-1)!*(Sum[ Binomial[n+k-1, n-1]* Sum[(-1)^(n+j-1)*Binomial[k, j]* Sum[(Binomial[j, l]*(j-l)!*2^(j-l)*(-1)^l*StirlingS2[n-l+j-1, j-l])/(n-l+j-1)!, {l, 0, j}], {j, 0, k}], {k, 0, n-1}]); Table[a[n], {n, 1, 18}] (* Jean-François Alcover, Feb 26 2013, after Vladimir Kruchinin *)
    T[n_, k_] := T[n, k] = If[k == 0, Boole[n == 0], If[n < 0, 0, k T[n - 1, k] + (2 n - k) T[n - 1, k - 1]]]; a[n_] := Sum[T[n, k] 2^k, {k, 0, n}];
    Table[a[n], {n, 0, 17}] (* Peter Luschny, Feb 13 2021 *)
  • Maxima
    a(n):=n!*(sum(binomial(n+k, n)*sum((-1)^(n+j)*binomial(k, j)*sum((binomial(j, l)*(j-l)!*2^(j-l)*(-1)^l*stirling2(n-l+j, j-l))/(n-l+j)!, l, 0, j), j, 0, k), k, 0, n)); /* Vladimir Kruchinin, Feb 14 2012 */
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=exp(intformal(A+A^2)+x*O(x^n)));n!*polcoeff(A,n)} \\ Paul D. Hanna, Jun 30 2009
    

Formula

a(n) = Sum_{m=0..n} A112486(n, m), n >= 0.
a(n) = 2*A032188(n+1), n > 0. - Vladeta Jovovic, Jul 11 2007
From Paul D. Hanna, Jun 30 2009: (Start)
E.g.f. A(x) satisfies: A'(x) = A(x)^2 + A(x)^3.
E.g.f. A(x) satisfies: A(x) = exp( Integral[A(x) + A(x)^2]dx ) with A(0)=1. (End)
E.g.f. A(x) satisfies: A(x) = 2*exp(A(x)) - (2+x), where A(x) = Sum_{n>=0} a(n)*x^(n+1)/(n+1)! (the e.g.f. when offset=1). - Paul D. Hanna, Sep 23 2011
From Tom Copeland, Oct 05 2011: (Start)
With c(0)= 0 and c(n+1)= (-1)^n a(n) for n>=0, c(n)=(-1)^(n+1) PW(n,-2) with PW the Ward polynomials A134991. E.g.f. for the c(n) is A(x) = -(x+2)-LW{-2 exp[-(x+2)]}, where LW(x) is a suitable branch of the Lambert W Fct. (see A135338).
The compositional inverse is B(x) = x + 2(exp(x) - x - 1). These results are a special case of A134685 with u(x)=B(x), i.e., u_1=1 and (u_n)=2 for n>0.
Let h(x) = 1/(dB(x)/dx) = 1/[1+2(exp(x)-1)], then c(n) is given by (h(x)*d/dx)^n x, evaluated at x=0, i.e., A(x) = exp(x*h(u)*d/du) u, evaluated at u=0. Also, dA(x)/dx = h(A(x)).
The e.g.f. A(x) = -v * Sum_(j>=1) D(j-1,u) (-z)^j/ j! where u=-(x+2), v=1+u, z=(1+v)/(v^2) and D(j-1,u) are the polynomials of A042977. (End)
a(n) = n!*Sum_{k=0..n} binomial(n+k, n)*Sum_{j=0..k} (-1)^(n+j)*binomial(k, j)*Sum_{l=0..j} binomial(j, l)*(j-l)!*2^(j-l)*(-1)^l*Stirling2(n-l+j, j-l)/(n-l+j)!. - Vladimir Kruchinin, Feb 14 2012
G.f.: 1/Q(0), where Q(k)= 1 + k*x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
a(n) ~ n^n / (exp(n) * (1-log(2))^(n+1/2)). - Vaclav Kotesovec, Aug 14 2017
a(0) = 1; a(n) = n * a(n-1) + Sum_{k=0..n-1} binomial(n,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Jul 02 2020

Extensions

New name from Peter Luschny, Feb 13 2021

A269940 Triangle read by rows, T(n, k) = Sum_{m=0..k} (-1)^(m + k)*binomial(n + k, n + m) * |Stirling1(n + m, m)|, for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 0, 6, 20, 15, 0, 24, 130, 210, 105, 0, 120, 924, 2380, 2520, 945, 0, 720, 7308, 26432, 44100, 34650, 10395, 0, 5040, 64224, 303660, 705320, 866250, 540540, 135135, 0, 40320, 623376, 3678840, 11098780, 18858840, 18288270, 9459450, 2027025
Offset: 0

Author

Peter Luschny, Mar 27 2016

Keywords

Comments

We propose to call this sequence the 'Ward cycle numbers' and sequence A269939 the 'Ward set numbers'. - Peter Luschny, Nov 25 2022

Examples

			Triangle T(n,k) starts:
  [1]
  [0,   1]
  [0,   2,      3]
  [0,   6,     20,     15]
  [0,  24,    130,    210,    105]
  [0,  120,   924,   2380,   2520,    945]
  [0,  720,  7308,  26432,  44100,  34650,  10395]
  [0, 5040, 64224, 303660, 705320, 866250, 540540, 135135]
		

Crossrefs

Variants: A111999, A259456.
Cf. A269939 (Stirling2 counterpart), A268438, A032188 (row sums).

Programs

  • Maple
    T := (n, k) -> add((-1)^(m+k)*binomial(n+k,n+m)*abs(Stirling1(n+m, m)), m=0..k):
    seq(print(seq(T(n, k), k=0..n)), n=0..6);
    # Alternatively:
    T := proc(n, k) option remember;
        `if`(k=0, k^n,
        `if`(k<=0 or k>n, 0,
        (n+k-1)*(T(n-1, k)+T(n-1, k-1)))) end:
    for n from 0 to 6 do seq(T(n, k), k=0..n) od;
  • Mathematica
    T[n_, k_] := Sum[(-1)^(m+k)*Binomial[n+k, n+m]*Abs[StirlingS1[n+m, m]], {m, 0, k}];
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 12 2022 *)
  • Sage
    T = lambda n, k: sum((-1)^(m+k)*binomial(n+k, n+m)*stirling_number1(n+m, m) for m in (0..k))
    for n in (0..7): print([T(n, k) for k in (0..n)])
    
  • Sage
    # uses[PtransMatrix from A269941]
    PtransMatrix(8, lambda n: n/(n+1), lambda n, k: (-1)^k*falling_factorial(n+k,n))

Formula

T(n,k) = (-1)^k*FF(n+k,n)*P[n,k](n/(n+1)) where P is the P-transform and FF the falling factorial function. For the definition of the P-transform see the link.
T(n,k) = A268438(n,k)*FF(n+k,n)/(2*n)!.

Extensions

Name corrected after notice from Ed Veling by Peter Luschny, Jun 14 2022

A118788 Triangle where T(n,k) = n!/(n-k)!*[x^k] ( x/(2*x + log(1-x)) )^(n+1), for n>=k>=0, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 1, 6, 23, 41, 1, 10, 65, 255, 469, 1, 15, 145, 930, 3679, 6889, 1, 21, 280, 2590, 16429, 65247, 123605, 1, 28, 490, 6090, 54789, 344694, 1371887, 2620169, 1, 36, 798, 12726, 151599, 1338330, 8367785, 33347535, 64074901, 1, 45, 1230, 24360
Offset: 0

Author

Paul D. Hanna, Apr 29 2006

Keywords

Comments

Row sums are A118789, where Sum_{n>=0} A118789(n)*x^n/n! = exp( Sum_{n>=1} A032188(n)*x^n/n! ).
Main diagonal is A032188(n) = number of labeled series-reduced mobiles (circular rooted trees) with n leaves.
Secondary diagonal is A118790.

Examples

			Row sums e.g.f. equals the exponential of the diagonal e.g.f.:
1 + x + 2*x^2/2! + 9*x^3/3! + 71*x^4/4! +...+ A118789(n)*x^n/n! +...
= exp(x + x^2/2! + 5*x^3/3! + 41*x^4/4! +...+ A032188(n)*x^n/n! +...).
Triangle begins:
  1;
  1, 1;
  1, 3, 5;
  1, 6, 23, 41;
  1, 10, 65, 255, 469;
  1, 15, 145, 930, 3679, 6889;
  1, 21, 280, 2590, 16429, 65247, 123605;
  1, 28, 490, 6090, 54789, 344694, 1371887, 2620169;
  1, 36, 798, 12726, 151599, 1338330, 8367785, 33347535, 64074901;
  ...
Triangle is formed from powers of F(x) = x/(2*x + log(1-x)):
  F(x)^1 = (1) + 1/2*x + 7/12*x^2 + 17/24*x^3 + 629/720*x^4 +...
  F(x)^2 = (1 + x) + 17/12*x^2 + 2*x^3 + 671/240*x^4 +...
  F(x)^3 = (1 + 3/2*x + 5/2*x^2) + 4*x^3 + 1489/240*x^4 +...
  F(x)^4 = (1 + 6/3*x + 23/6*x^2 + 41/6*x^3) + 8351/720*x^4 +...
  F(x)^5 = (1 + 10/4*x + 65/12*x^2 + 255/24*x^3 + 469/24*x^4) +...
		

Crossrefs

Third column is A241765.

Programs

  • PARI
    {T(n,k)=local(x=X+X^2*O(X^(k+2)));n!/(n-k)!*polcoeff((x/(2*x+log(1-x)))^(n+1),k,X)}

Formula

Main diagonal has e.g.f.: series_reversion[2*x+log(1-x)].
Conjecture: T(n,k) = Sum_{j=0..k} binomial(n+j, n-k)*A269940(k, j) for 0 <= k <= n. - Mikhail Kurkov, Feb 17 2025

A032034 Shifts left under "AIJ" (ordered, indistinct, labeled) transform.

Original entry on oeis.org

2, 2, 10, 82, 938, 13778, 247210, 5240338, 128149802, 3551246162, 109979486890, 3764281873042, 141104799067178, 5749087305575378, 252969604725106090, 11955367835505775378, 603967991604199335722, 32479636694930586142802, 1852497140997527094395050
Offset: 1

Keywords

Crossrefs

Programs

  • Maple
    with(combinat): A032034 := n -> add(eulerian2(n-1,k)*2^(k+1), k=0..n-1):
    seq(A032034(n), n=1..17); # Peter Luschny, Nov 10 2012
  • 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] (2n-k-1)]];
    a[n_] := Sum[Eulerian2[n-1, k] 2^(k+1), {k, 0, n-1}];
    Array[a, 20] (* Jean-François Alcover, Jun 01 2019, after Peter Luschny *)
  • Maxima
    a(n):=if n=1 then 2 else ((n-1)!*sum(binomial(n+k-1,n-1)*sum((-1)^(j+n+1)*binomial(k,j)*sum((binomial(j,l)*(j-l)!*2^(j-l)*(-1)^l*stirling2(n-l+j-1,j-l))/(n-l+j-1)!,l,0,j),j,1,k),k,1,n-1)); /* Vladimir Kruchinin, Jan 24 2012 */
    
  • PARI
    seq(n)={my(p=O(x)); for(i=1, n, p=intformal(1 + 1/(1-p))); Vec(serlaplace(p))} \\ Andrew Howroyd, Sep 19 2018
  • Sage
    @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)
    A032034 = lambda n: add(eulerian2(n-1,k)*2^(k+1) for k in (0..n-1))
    [A032034(n) for n in (1..17)]  # Peter Luschny, Nov 10 2012
    

Formula

a(n) = ((n-1)!*sum(k=1..n-1, binomial(n+k-1,n-1)*sum(j=1..k, (-1)^(j+n+1)*binomial(k,j)*sum(l=0..j, (binomial(j,l)*(j-l)!*2^(j-l)*(-1)^l*stirling2(n-l+j-1,j-l))/(n-l+j-1)!)))), n>1, a(1)=2. - Vladimir Kruchinin, Jan 24 2012
Let p(n,w) = w*Sum_{k=0..n-1} ((-1)^k*E2(n-1,k)*w^k)/(1+w)^(2*n-1),
E2 the second-order Eulerian numbers as defined by Knuth, then a(n) = p(n,-2). - Peter Luschny, Nov 10 2012
G.f.: 1 + 1/Q(0), where Q(k)= 1 + k*x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
a(n) = 2 * A032188(n). - Alois P. Heinz, Jul 04 2018
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