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 1-2 of 2 results.

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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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

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

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Author

Wolfdieter Lang, Sep 12 2005

Keywords

Comments

Previous name: Row sums of triangle A112486.

Crossrefs

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
Showing 1-2 of 2 results.