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 13 results. Next

A174632 Partial sums of A029768.

Original entry on oeis.org

0, 1, 2, 4, 11, 47, 292, 2368, 23427, 272263, 3628872, 54525252, 911484163, 16775498551, 337021458884, 7338279413680, 172130372061035, 4327036966579151, 116046966039565672, 3307263639537314116
Offset: 0

Views

Author

Jonathan Vos Post, Mar 24 2010

Keywords

Comments

Partial sums of number of increasing mobiles with n elements. In an increasing rooted tree, nodes are numbered and numbers increase as you move away from root. The subsequence of primes in this partial sum begins: 2, 11, 47, 272263.

Examples

			a(x) = 0 + 1 + 1 + 2 + 7 + 36 + 245 + 2076 + 21059 + 248836 = 272263 is prime.
		

Crossrefs

Programs

  • Maple
    S:= rhs(dsolve({diff(a(x), x) = log(1/(1-a(x)))+1, a(0)=0}, a(x), series, order=31)):
    L:= [seq(coeff(S, x, j)*j!, j=0..30)]:
    ListTools:-PartialSums(L); # Robert Israel, Dec 21 2017

Formula

a(n) = SUM[i=o..n] A029768(i).

A008544 Triple factorial numbers: Product_{k=0..n-1} (3*k+2).

Original entry on oeis.org

1, 2, 10, 80, 880, 12320, 209440, 4188800, 96342400, 2504902400, 72642169600, 2324549427200, 81359229952000, 3091650738176000, 126757680265216000, 5577337931669504000, 262134882788466688000
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

a(n-1), n >= 1, enumerates increasing plane (aka ordered) trees with n vertices (one of them a root labeled 1) where each vertex with outdegree r >= 0 comes in r+1 types (like an (r+1)-ary vertex). See the increasing tree comments under A004747. - Wolfdieter Lang, Oct 12 2007
An example for the case of 3 vertices is shown below. For the enumeration of non-plane trees of this type see A029768. - Peter Bala, Aug 30 2011
a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 2. - Peter Luschny, Jun 23 2011
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
Partial products of A016789. - Reinhard Zumkeller, Sep 20 2013
The Mathar conjecture is true. Generally from the factorial form, the last term is the "extra" product beyond the prior term, from k=n-1 and 3k+2 evaluates to 3*(n-1)+2 = 3n-1, yielding a(n) = a(n-1)*(3n-1) (eqn1). Similarly, a(n) = a(n-2)*(3n-1)*(3(n-2)+2) = a(n-2)*(3n-1)*(3n-4) (eqn2) and a(n) = a(n-3)*(3n-1)*(3n-4)*(3*(n-2)+2) = a(n-3)*(3n-1)*(3n-4)*(3n-7) (eqn3). We equate (eqn2) and (eqn3) to get a(n-2)*(3n-1)*(3n-4) = a(n-3)*(3n-1)*(3n-4)*(3n-7) or a(n-2)+(7-3n)*a(n-3) = 0 (eqn4). From (eqn1) we have a(n)+(1-3n)*a(n-1) = 0 (eqn5). Combining (eqn4) and (eqn5) yields a(n)+(1-3n)*a(n-1)+a(n-2)+(7-3n)*a(n-3) = 0. - Bill McEachen, Jan 01 2016
a(n-1), n>=1, is the dimension of the n-th component of the operad encoding the multilinearization of the following identity in nonassociative algebras: s*(a,a,b)-(s+t)*(a,b,a)+t*(b,a,a)=0, for any given pair of scalars (s,t). Here (a,b,c) is the associator (ab)c-a(bc). This is proved in the referenced article on associator dependent algebras by Bremner and me. - Vladimir Dotsenko, Mar 22 2022

Examples

			a(2) = 10 from the described trees with 3 vertices: there are three trees with a root vertex (label 1) with outdegree r=2 (like the three 3-stars each with one different ray missing) and the four trees with a root (r=1 and label 1) a vertex with (r=1) and a leaf (r=0). Assigning labels 2 and 3 yields 2*3+4=10 such trees.
a(2) = 10. The 10 possible plane increasing trees on 3 vertices, where vertices of outdegree 1 come in 2 colors (denoted a or b) and vertices of outdegree 2 come in 3 colors (a, b or c), are:
.
   1a    1b    1a    1b        1a       1b       1c
   |     |     |     |        / \      / \      / \
   2a    2b    2b    2a      2   3    2   3    2   3
   |     |     |     |
   3     3     3     3         1a       1b       1c
                              / \      / \      / \
                             3   2    3   2    3   2
		

Crossrefs

a(n) = A004747(n+1, 1) (first column of triangle). Cf. A051141.
Cf. A225470, A290596 (first columns).
Subsequence of A007661.

Programs

  • Haskell
    a008544 n = a008544_list !! n
    a008544_list = scanl (*) 1 a016789_list
    -- Reinhard Zumkeller, Sep 20 2013
    
  • Magma
    [Round((Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6) )/ Sqrt(3)*3^n/4^(n-1)): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015
    
  • Magma
    [Round(3^n*Gamma(n+2/3)/Gamma(2/3)): n in [0..20]]; // G. C. Greubel, Mar 31 2019
  • Maple
    a := n -> mul(3*k-1, k = 1..n);
    A008544 := n -> mul(k, k = select(k-> k mod 3 = 2, [$1 .. 3*n])): seq(A008544(n), n = 0 .. 16); # Peter Luschny, Jun 23 2011
  • Mathematica
    k = 3; b[1]=2; b[n_]:= b[n] = b[n-1]+k; a[0]=1; a[1]=2; a[n_]:= a[n] = a[n-1]*b[n]; Table[a[n], {n,0,20}] (* Roger L. Bagula, Sep 17 2008 *)
    Product[3 k + 2, {k, 0, # - 1}] & /@ Range[0, 16] (* Michael De Vlieger, Jan 02 2016 *)
    Table[3^n*Pochhammer[2/3, n], {n,0,20}] (* G. C. Greubel, Mar 31 2019 *)
  • Maxima
    a(n):=((n)!*sum(binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k),k,floor(n/2),n)); /* Vladimir Kruchinin, Sep 28 2013 */
    
  • PARI
    a(n) = prod(k=0,n-1, 3*k+2 );
    
  • PARI
    vector(20, n, n--; round(3^n*gamma(n+2/3)/gamma(2/3))) \\ G. C. Greubel, Mar 31 2019
    
  • Sage
    @CachedFunction
    def A008544(n): return 1 if n == 0 else (3*n-1)*A008544(n-1)
    [A008544(n) for n in (0..16)]  # Peter Luschny, May 20 2013
    
  • Sage
    [3^n*rising_factorial(2/3, n) for n in (0..20)] # G. C. Greubel, Mar 31 2019
    

Formula

a(n) = Product_{k=0..n-1} (3*k+2) = A007661(3*n-1) (with A007661(-1) = 1).
E.g.f.: (1-3*x)^(-2/3).
a(n) = 2*A034000(n), n >= 1, a(0) = 1.
a(n) ~ 2^(1/2)*Pi^(1/2)*Gamma(2/3)^-1*n^(1/6)*3^n*e^-n*n^n*{1 - 1/36*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = (Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6))/sqrt(3)*3^n/4^(n-1). - Jeremy L. Martin, Mar 31 2002 (typo fixed by Vincenzo Librandi, Feb 21 2015)
From Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003: (Start)
a(n) = A084939(n)/A000142(n)*A000079(n).
a(n) = 3^n*Pochhammer(2/3, n) = 3^n*Gamma(n+2/3)/Gamma(2/3). (End)
Let T = A094638 and c(t) = column vector(1, t, t^2, t^3, t^4, t^5,...), then A008544 = unsigned [ T * c(-3) ] and the list partition transform A133314 of [1,T * c(-3)] gives [1,T * c(3)] with all odd terms negated, which equals a signed version of A007559; i.e., LPT[(1,signed A008544)] = signed A007559. Also LPT[A007559] = (1,-A008544) and e.g.f. [1,T * c(t)] = (1-x*t)^(-1/t) for t = 3 or -3. Analogous results hold for the double factorial, quadruple factorial and so on. - Tom Copeland, Dec 22 2007
G.f.: 1/(1-2x/(1-3x/(1-5x/(1-6x/(1-8x/(1-9x/(1-11x/(1-12x/(1-...))))))))) (continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-1)^n*Sum_{k=0..n} 3^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
G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+2)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k+2)/(x*(3*k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
D-finite with recurrence: a(n) = (9*(n-2)*(n-1)+2)*a(n-2) + 4*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 09 2013
a(n) = n!*Sum_{k=floor(n/2)..n} binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k). - Vladimir Kruchinin, Sep 28 2013
Recurrence equation: a(n) = 3*a(n-1) + (3*n - 4)^2*a(n-2) with a(0) = 1 and a(1) = 2. A024396 satisfies the same recurrence (but with different initial conditions). This observation leads to a continued fraction expansion for the constant A193534 due to Euler. - Peter Bala, Feb 20 2015
a(n) = A225470(n, 0), n >= 0. - Wolfdieter Lang, May 29 2017
G.f.: Hypergeometric2F0(1, 2/3; -; 3*x). - G. C. Greubel, Mar 31 2019
D-finite with recurrence: a(n) + (-3*n+1)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
G.f.: 1/(1-2*x-6*x^2/(1-8*x-30*x^2/(1-14*x-72*x^2/(1-20*x-132*x^2/(1-...))))) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 28 2020
G.f.: 1/G(0), where G(k) = 1 - (6*k+2)*x - 3*(k+1)*(3*k+2)*x^2/G(k+1). - Nikolaos Pantelidis, Feb 28 2020
Sum_{n>=0} 1/a(n) = 1 + (e/3)^(1/3) * (Gamma(2/3) - Gamma(2/3, 1/3)). - Amiram Eldar, Mar 01 2022

A053492 REVEGF transform of [1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, ...].

Original entry on oeis.org

1, 2, 15, 184, 3155, 69516, 1871583, 59542064, 2185497819, 90909876100, 4226300379983, 217152013181544, 12219893000227107, 747440554689309404, 49374719534173925055, 3503183373320829575008, 265693897270211120103563, 21451116469521758657525748
Offset: 1

Views

Author

N. J. A. Sloane, Jan 15 2000

Keywords

Comments

Sequence gives the number of total circled partitions of n. This is the number of ways to partition n into at least two blocks, circle one block, then successively partition each non-singleton block into at least two blocks and circle one of the blocks. Stop when only singleton blocks remain. - Brian Drake, Apr 25 2006
a(n) is also the number of Schroeder trees on n vertices. - Brad R. Jones, May 09 2014
Number of pointed trees on pointed sets k[1...k...n] for any point k. - Gus Wiseman, Sep 27 2015

Examples

			E.g.f.: A(x) = x + 2*x^2/2! + 15*x^3/3! + 184*x^4/4! + 3155*x^5/5! + ...
Related expansions from _Paul D. Hanna_, Jul 07 2012: (Start)
A(x) = x + (exp(x)-1)*x + d/dx (exp(x)-1)^2*x^2/2! + d^2/dx^2 (exp(x)-1)^3*x^3/3! + d^3/dx^3 (exp(x)-1)^4*x^4/4! + ...
log(A(x)/x) = (exp(x)-1) + d/dx (exp(x)-1)^2*x/2! + d^2/dx^2 (exp(x)-1)^3*x^2/3! + d^3/dx^3 (exp(x)-1)^4*x^3/4! + ... (End)
The a(3) = 15 pointed trees are 1[1 2[2 3]], 1[1 3[2 3]], 1[1[1 3] 2], 1[1[1 2] 3], 1[1 2 3], 2[1 2[2 3]], 2[1[1 3] 2], 2[2 3[1 3]], 2[2[1 2] 3], 2[1 2 3], 3[1 3[2 3]], 3[2 3[1 3]], 3[1[1 2] 3], 3[2[1 2] 3], 3[1 2 3].
		

Crossrefs

Programs

  • Maple
    A:= series(RootOf(exp(A053492:=%20n-%3E%20n!%20*%20coeff(A,%20x,%20n);%20%23%20_Brian%20Drake">Z)*_Z+x-2*_Z), x, 30): A053492:= n-> n! * coeff(A, x, n); # _Brian Drake, Apr 25 2006
  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[2*x-x*E^x, {x, 0, 20}], x],x] * Range[0, 20]!] (* Vaclav Kotesovec, Oct 27 2014 *)
  • Maxima
    a(n):= if n=1 then 1 else sum(k!*stirling2(n-1,k)*binomial(n+k-1,n-1),k,1,n-1); /* Vladimir Kruchinin, May 10 2011 */
    
  • PARI
    {a(n) = if( n<1, 0, n! * polcoeff( serreverse( 2*x - x * exp(x + x * O(x^n))), n))}; /* Michael Somos, Jun 06 2012 */
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=x); A=x+sum(m=1, n, Dx(m-1, (exp(x+x*O(x^n))-1)^m*x^m/m!)); n!*polcoeff(A, n)} \\ Paul D. Hanna, Jul 07 2012
    for(n=1, 25, print1(a(n), ", "))
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=x+x^2+x*O(x^n)); A=x*exp(sum(m=1, n, Dx(m-1, (exp(x+x*O(x^n))-1)^m*x^(m-1)/m!)+x*O(x^n))); n!*polcoeff(A, n)} \\ Paul D. Hanna, Jul 07 2012
    for(n=1, 25, print1(a(n), ", "))
    
  • PARI
    \p100 \\ set precision
    {A=Vec(sum(n=0, 400, 1./(2 - n*x +O(x^25))^(n+1)) )}
    for(n=1, #A, print1(round(A[n]), ", ")) \\ Paul D. Hanna, Oct 27 2014

Formula

E.g.f. is the compositional inverse of 2*x - x*exp(x). - Brian Drake, Apr 25 2006
E.g.f.: x + Sum_{n>=1} d^(n-1)/dx^(n-1) (exp(x)-1)^n*x^n / n!. - Paul D. Hanna, Jul 07 2012
E.g.f.: x*exp( Sum_{n>=1} d^(n-1)/dx^(n-1) (exp(x)-1)^n*x^(n-1) / n! ). - Paul D. Hanna, Jul 07 2012
a(n) = Sum_{k=1..n-1} k!*Stirling2(n-1,k)*C(n+k-1,n-1), n > 1, a(1)=1. - Vladimir Kruchinin, May 10 2011
O.g.f.: x*Sum_{n>=0} 1/(2 - n*x)^(n+1). - Paul D. Hanna, Oct 27 2014
a(n) ~ n^(n-1) * (LambertW(2*exp(1)))^n / (sqrt(1+LambertW(2*exp(1))) * 2^n * exp(n) * (LambertW(2*exp(1))-1)^(2*n-1)). - Vaclav Kotesovec, Oct 27 2014

Extensions

Signs removed by Michael Somos, based on Brian Drake's remark, Jun 06 2012

A055356 Triangle of increasing mobiles (circular rooted trees) with n nodes and k leaves.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 4, 2, 0, 1, 11, 18, 6, 0, 1, 26, 98, 96, 24, 0, 1, 57, 424, 874, 600, 120, 0, 1, 120, 1614, 6040, 8244, 4320, 720, 0, 1, 247, 5682, 35458, 83500, 83628, 35280, 5040, 0, 1, 502, 19022, 187288, 701164, 1169768, 915984, 322560, 40320, 0, 1
Offset: 1

Views

Author

Christian G. Bower, May 15 2000

Keywords

Comments

In an increasing rooted tree, nodes are numbered and numbers increase as you move away from root.
Also related to the solution of the equation df/dt=f e^f (see the Maple code). - F. Chapoton, Jul 16 2004

Examples

			Triangle begins
  1;
  1,  0;
  1,  1,   0;
  1,  4,   2,   0;
  1, 11,  18,   6,   0;
  1, 26,  98,  96,  24,   0;
  1, 57, 424, 874, 600, 120, 0;
  ...
		

Crossrefs

Row sums give A029768 (p(n,1)).
Alternating row sums give A089963 (p(n+1,-1)).

Programs

  • Maple
    P[1]:=1;for n from 1 to 8 do P[n+1]:=simplify((1+n*x)*P[n]+x*diff(P[n],x)) end; # F. Chapoton, Jul 16 2004
  • Mathematica
    P[1][_] = 1;
    P[n_][x_] := P[n][x] = (1 + (n-1) x) P[n-1][x] + x P[n-1]'[x] // Expand;
    row[1] = {1};
    row[n_] := Append[CoefficientList[P[n-1][x], x], 0];
    Array[row, 10] // Flatten (* Jean-François Alcover, Nov 17 2018, after F. Chapoton *)
  • PARI
    A(n)={my(v=vector(n)); v[1]=y; for(n=2, #v, v[n]=v[n-1] + sum(k=1, n-2, binomial(n-2, k)*v[k]*v[n-k])); vector(#v, i, Vecrev(v[i]/y, i))}
    { my(T=A(10)); for(i=1, #T, print(T[i])) } \\ Andrew Howroyd, Sep 23 2018

Formula

Let p(n,x) be the polynomial with coefficients equal to the n-th row of the triangle in ascending powers of x, e.g., p(4,x) = 1+4*x+2*x^2; then p(n+1,x) = (1+(n-1)*x)*p(n,x) + x*p'(n,x). - Ben Whitmore, May 12 2021
Recurrence: T(n,k) = (n-2) * T(n-1,k-1) + k * T(n-1,k) for n >= 1, 1 <= k <= n with T(1,1) = 1 and T(n,k) = 0 for n < 1, k < 1 or k > n. - Georg Fischer, Oct 27 2021
Conjecture: row polynomials are R(n-2,0) for n > 1 where R(n,k) = R(n-1,k+1) + x*Sum_{i=0..n-1} Sum_{j=0..k} binomial(n-1, i)*R(n-i-1,j)*R(i,k-j) for n > 0, k >= 0 with R(0,k) = 1 for k >= 0. - Mikhail Kurkov, Apr 11 2025

A032200 Number of rooted compound windmills (mobiles) of n nodes.

Original entry on oeis.org

1, 1, 2, 4, 9, 20, 51, 128, 345, 940, 2632, 7450, 21434, 62174, 182146, 537369, 1596133, 4767379, 14312919, 43162856, 130695821, 397184252, 1211057426, 3703794849, 11358759346, 34923477315, 107627138308, 332404636811
Offset: 1

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Author

Keywords

Comments

Also the number of locally necklace plane trees with n nodes, where a plane tree is locally necklace if the sequence of branches directly under any given node is lexicographically minimal among its cyclic permutations. - Gus Wiseman, Sep 05 2018

Examples

			From _Gus Wiseman_, Sep 05 2018: (Start)
The a(5) = 9 locally necklace plane trees:
  ((((o))))
  (((oo)))
  ((o(o)))
  (o((o)))
  ((o)(o))
  ((ooo))
  (o(oo))
  (oo(o))
  (oooo)
(End)
		

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 241 (3.3.84).

Crossrefs

Programs

  • Mathematica
    neckQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And];
    neckplane[n_]:=If[n==1,{{}},Join@@Table[Select[Tuples[neckplane/@c],neckQ],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[neckplane[n]],{n,10}] (* Gus Wiseman, Sep 05 2018 *)
  • PARI
    CIK(p,n)={sum(d=1, n, eulerphi(d)/d*log(subst(1/(1+O(x*x^(n\d))-p), x, x^d)))}
    seq(n)={my(p=O(1));for(i=1, n, p=1+CIK(x*p, i)); Vec(p)} \\ Andrew Howroyd, Jun 20 2018

Formula

Shifts left under "CIK" (necklace, indistinct, unlabeled) transform.

A038037 Number of labeled rooted compound windmills (mobiles) with n nodes.

Original entry on oeis.org

1, 2, 9, 68, 730, 10164, 173838, 3524688, 82627200, 2198295360, 65431163160, 2154106470240, 77714083773456, 3048821300491680, 129221979665461200, 5884296038166954240, 286492923374605966080, 14851359950834255500800
Offset: 1

Views

Author

Christian G. Bower, Sep 15 1998

Keywords

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 241 (3.3.83).

Crossrefs

Programs

  • Maple
    logtr:= proc(p) local b; b:=proc(n) option remember; local k; if n=0 then 1 else p(n)- add(k *binomial(n,k) *p(n-k) *b(k), k=1..n-1)/n fi end end: b:= logtr(-a): a:= n-> `if`(n<=1,1, -n*b(n-1)): seq(a(n), n=1..25); # Alois P. Heinz, Sep 14 2008
  • Mathematica
    a[n_] = Sum[Binomial[n, j]*Abs[StirlingS1[n-1, j]]*j!, {j, 0, n}]; Array[a, 18]
    (* Jean-François Alcover, Jun 22 2011, after Vladimir Kruchinin *)
  • PARI
    Vec(serlaplace(serreverse(x/(1 - log(1-x + O(x^20)))))) \\ Andrew Howroyd, Sep 19 2018

Formula

Divides by n and shifts left under "CIJ" (necklace, indistinct, labeled) transform.
E.g.f. A(x) satisfies A(x) = x-x*log(1-A(x)). [Corrected by Andrey Zabolotskiy, Sep 16 2022]
a(n) = Sum_{j=0..n} binomial(n,j)*abs(Stirling1(n-1,j))*j!, n > 0. - Vladimir Kruchinin, Feb 03 2011
a(n) ~ sqrt(-1-LambertW(-1,-exp(-2))) * (-LambertW(-1,-exp(-2)))^(n-1) * n^(n-1) / exp(n). - Vaclav Kotesovec, Dec 27 2013
E.g.f.: series reversion of x/(1 - log(1-x)). - Andrew Howroyd, Sep 19 2018

A007549 Number of increasing rooted connected graphs where every block is a complete graph.

Original entry on oeis.org

1, 1, 3, 14, 89, 716, 6967, 79524, 1041541, 15393100, 253377811, 4596600004, 91112351537, 1959073928124, 45414287553455, 1129046241331316, 29965290866974493, 845605519848379436, 25282324544244718411, 798348403914242674980, 26549922456617388029641
Offset: 1

Views

Author

Keywords

Comments

In an increasing rooted graph, nodes are numbered and the numbers increase as you move away from the root.
(a(n+1)/a(n))/n tends to 1/A073003 = 1.676875... (same limit as A029768). - Vaclav Kotesovec, Jul 26 2014

References

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

Crossrefs

Cf. A029768.
Row sums of A078341. Column k=1 of A264436.

Programs

  • Maple
    exptr:= proc(p) local g; g:= proc(n) option remember; p(n) +add(binomial(n-1, k-1) *p(k) *g(n-k), k=1..n-1) end: end: b:= exptr(exptr(a)): a:= n-> `if`(n=0, 1, b(n-1)): seq(a(n), n=1..30); # Alois P. Heinz, Oct 07 2008
  • Mathematica
    exptr[p_] := Module[{g}, g[n_] := g[n] = p[n] + Sum[ Binomial[n-1, k-1]*p[k]*g[n-k], {k, 1, n-1}]; g]; b = exptr[ exptr[a] ]; a[n_] := If[n == 0, 1, b[n-1]]; Table[ a[n], {n, 1, 19}] (* Jean-François Alcover, May 10 2012, after Alois P. Heinz *)

Formula

Shifts left when exponentiated twice.
Conjecture: a(n) = Sum_{i=0..2^(n-2) - 1} b(i) for n > 1 with a(1) = 1 where b(n) = (L(n) + 2)*b(f(n)) + Sum_{k=0..L(n) - 1} (1 - R(n,k))*b(f(n) + 2^k*(1 - R(n,k))) for n > 0 with b(0) = 1, L(n) = A000523(n), f(n) = A053645(n) and where R(n,k) = floor(n/2^k) mod 2. Here R(n,k) is the (k+1)-th bit from the right side in the binary expansion of n. - Mikhail Kurkov, Jul 21 2024
Conjecture: a(n) = D^(n-1)(exp(x)) evaluated at x = 0, where D denotes the operator exp(x)*(1 + x)*d/dx. - Peter Bala, Feb 24 2025

Extensions

New description from Christian G. Bower, Oct 15 1998

A131178 Non-plane increasing unary binary (0-1-2) trees where the nodes of outdegree 1 come in 2 colors.

Original entry on oeis.org

1, 2, 5, 16, 64, 308, 1730, 11104, 80176, 643232, 5676560, 54650176, 569980384, 6401959328, 77042282000, 988949446144, 13488013248256, 194780492544512, 2969094574403840, 47640794742439936, 802644553810683904, 14166772337295285248, 261410917571703825920
Offset: 1

Views

Author

Wenjin Woan, Oct 31 2007

Keywords

Comments

A labeled tree of size n is a rooted tree on n nodes that are labeled by distinct integers from the set {1,...,n}. An increasing tree is a labeled tree such that the sequence of labels along any branch starting at the root is increasing. Thus the root of an increasing tree will be labeled 1. In unary binary trees (sometimes called 0-1-2 trees) the outdegree of a node is either 0, 1 or 2. Here we are counting non-plane (where the subtrees stemming from a node are not ordered between themselves) increasing unary binary trees where the nodes of outdegree 1 come in two colors. An example is given below. - Peter Bala, Sep 01 2011
The number of plane increasing 0-1-2 trees on n nodes, where the nodes of outdegree 1 come in two colors, is equal to n!. Other examples of sequences counting increasing trees include A000111, A000670, A008544, A008545, A029768 and A080635. - Peter Bala, Sep 01 2011
Number of plane increasing 0-1-2 trees, where the nodes of outdegree 1 come in 2 colors, avoiding pattern T213. See A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 16*x^4 + 64*x^5 + 308*x^6 + 1730*x^7 + 11104*x^8 + ...
a(3) = 5: Denoting the two types of node of outdegree 1 by the letters a or b, the 5 possible trees are
.
.  1a    1b    1a    1b      1
.  |     |     |     |      / \
.  2a    2b    2b    2a    2   3
.  |     |     |     |
.  3     3     3     3
- _Peter Bala_, Sep 01 2011
		

Crossrefs

Programs

  • Maple
    E:=  (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)):
    S:= map(simplify,series(E,x,101)):
    seq(coeff(S,x,j)*j!, j=1..100); # Robert Israel, Nov 23 2016
  • Mathematica
    max = 25; f[x_] := (2*(Exp[Sqrt[2]*x] - 1))/((2 + Sqrt[2]) - (2 - Sqrt[2])*Exp[Sqrt[2]*x]); Drop[ Simplify[ CoefficientList[ Series[f[x], {x, 0, max}], x]*Range[0, max]!], 1] (* Jean-François Alcover, Oct 05 2011 *)
  • PARI
    x='x+O('x^66); /* that many terms */
    default(realprecision,1000); /* working with floats here */
    egf=(2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x));
    round(Vec(serlaplace(egf))) /* show terms */
    /* Joerg Arndt, Sep 01 2011 */
    
  • PARI
    /* the following program should be preferred. */
    Vec( serlaplace( serreverse( intformal( 1/(1+2*x+1/2*x^2) + O(x^66) ) ) ) )
    \\ Joerg Arndt, Mar 01 2014
    
  • PARI
    {a(n) = if( n<1, 0, n! * polcoeff( 2 / (-2 + quadgen(8) * (-1 + 2 / (1 - exp(-quadgen(8) * x + x * O(x^n))))), n))};

Formula

E.g.f.: A(x) = (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)) = x+2*x^2/2!+5*x^3/3!+16*x^4/4!+64*x^5/5!+....
From Peter Bala, Sep 01 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A' = 1+2*A+1/2*A^2 with A(0) = 0. It follows that the inverse function A(x)^-1 may be expressed as an integral A(x)^-1 = int {t = 0..x} 1/(1+2*t+1/2*t^2).
Applying [Dominici, Theorem 4.1] to invert the integral gives the following method for calculating the terms of the sequence: let f(x) = 1+2*x+1/2*x^2. Let D be the operator f(x)*d/dx. Then a(n) = D^n(f(x)) evaluated at x = 0. Compare with A000111(n+1) = D^n(1+x+x^2/2!) evaluated at x = 0.
(End)
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ) and m=1; (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - 1/2*x^2*(k+1)*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2013
a(n) ~ n! * 2^((n+3)/2) / log(3+2*sqrt(2))^(n+1). - Vaclav Kotesovec, Oct 08 2013
G.f.: conjecture: T(0)/(1-2*x) -1, where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-2*x*(k+1))*(1-2*x*(k+2))/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2013
E.g.f.: x/(T(0)-x), where T(k) = 4*k + 1 + x^2/(8*k+6 + x^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2013

Extensions

Terms >= 80176 from Peter Bala, Sep 01 2011
Changed offset to 1 to agree with name and example. - Michael Somos, Nov 23 2016

A108528 Number of increasing mobiles (cycle rooted trees) with n generators.

Original entry on oeis.org

1, 2, 10, 92, 1216, 20792, 435520, 10793792, 308874016, 10021509632, 363509706880, 14576530558592, 640275236943616, 30573223563625472, 1576805482203235840, 87353392124392020992, 5173324070004374358016, 326160898887563325581312, 21810458629345555407462400
Offset: 1

Views

Author

Christian G. Bower, Jun 07 2005

Keywords

Comments

In an increasing rooted tree, nodes are numbered and numbers increase as you move away from root.

Crossrefs

Programs

  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[Log[(1+x)*Sqrt[1-x^2]], {x, 0, 20}], x],x] * Range[0, 20]!] (* Vaclav Kotesovec, Jan 08 2014 *)
  • PARI
    {a(n)=n!*polcoeff(serreverse(log((1+x)*sqrt(1-x^2+O(x^(n+2))))),n)} \\ Paul D. Hanna, Sep 11 2010

Formula

E.g.f. satisfies 2*A(x) = x - 1 + A'(x) - log(1-A(x)).
From Paul D. Hanna, Sep 11 2010: (Start)
E.g.f. satisfies: (1+A(x))*sqrt(1-A(x)^2) = exp(x).
E.g.f.: A(x) = Series_Reversion[ log((1+x)*sqrt(1-x^2)) ]. (End)
a(n) ~ 2^(n-2) * sqrt(3) * n^(n-1) / (exp(n) * (log(27/16))^(n-1/2)). - Vaclav Kotesovec, Jan 08 2014

A190015 Triangle T(n,k) for solving differential equation A'(x)=G(A(x)), G(0)!=0.

Original entry on oeis.org

1, 1, 2, 1, 6, 8, 1, 24, 42, 16, 22, 1, 120, 264, 180, 192, 136, 52, 1, 720, 1920, 1248, 540, 1824, 2304, 272, 732, 720, 114, 1, 5040, 15840, 10080, 8064, 18720, 22752, 9612, 7056, 10224, 17928, 3968, 2538, 3072, 240, 1, 40320, 146160, 92160, 70560, 32256, 207360, 249120, 193536, 73728, 61560, 144720, 246816, 101844, 142704, 7936, 51048, 110448, 34304, 8334, 11616, 494, 1
Offset: 0

Views

Author

Vladimir Kruchinin, May 04 2011

Keywords

Comments

For solving the differential equation A'(x)=G(A(x)), where G(0)!=0,
a(n) = 1/n!*sum(pi(i) in P(2*n-1,n), T(n,i)*prod(j=1..n, g(k_j-1))),
where pi(i) is the partition of 2*n-1 into n parts in lexicographic order P(2*n-1,n).
G(x) = g(0)+g(1)*x+g(2)*x^2+...
Examples
A003422 A'(x)=A(x)+1/(1-x)
A000108 A'(x)=1/(1-2*A(x)),
A001147 A'(x)=1/(1-A(x))
A007489 A'(x)=A(x)+x/(1-x)^2+1.
A006351 B'(x)=(1+B(x))/(1-B(x))
A029768 A'(x)=log(1/(1-A(x)))+1.
A001662 B'(x)=1/(1+B(x))
A180254 A'(x)=(1-sqrt(1-4*A(x)))/2
Compare with A145271. There (j')^k = [(d/dx)^j g(x)]^k evaluated at x=0 gives formulas expressed in terms of the coefficients of the Taylor series g(x). If, instead, we express the formulas in terms of the coefficients of the power series of g(x), we obtain a row reversed array for A190015 since the partitions there are in reverse order to the ones here. Simply exchange (j!)^k * (j")^k for (j')^k, where (j")^k = [(d/dx)^j g(x) / j!]^k, to transform from one array to the other. E.g., R^3 g(x) = 1 (0')^1 (1')^3 + 4 (0')^2 (1')^1 (2')^1 + 1 (0')^3 (3')^1 = 1 (O")^1 (1")^3 + 4 (0")^2 (1")^1 2*(2")^1 + 1 (0")^1 3!*(3")^1 = 1 (O")^1 (1")^3 + 8 (0")^2 (1")^1 (2")^1 + 6 (0")^1 (3")^1, the fourth partition polynomial here. - Tom Copeland, Oct 17 2014

Examples

			Triangle begins:
1;
1;
2,1;
6,8,1;
24,42,16,22,1;
120,264,180,192,136,52,1;
720,1920,1248,540,1824,2304,272,732,720,114,1;
5040,15840,10080,8064,18720,22752,9612,7056,10224,17928,3968,2538,3072,240,1;
40320,146160,92160,70560,32256,207360,249120,193536,73728,61560,144720,246816, 101844,142704,7936,51048,110448,34304,8334,11616,494,1;
Example for n=5:
partitions of number 9 into  5 parts in lexicographic order:
[1,1,1,1,5]
[1,1,1,2,4]
[1,1,1,3,3]
[1,1,2,2,3]
[1,2,2,2,2]
a(5) = (24*g(0)^4*g(4) +42*g(0)^3*g(1)*g(3) +16*g(0)^3*g(2)^2 +22*g(0)^2*g(1)^2*g(2) +g(0)*g(1)^4)/5!.
		

Programs

  • Maxima
    /* array of triangle */
    M:[1,1,2,1,6,8,1,24,42,16,22,1,120,264,180,192,136,52,1,720,1920,1248,540,1824,2304,272,732,720,114,1,5040,15840,10080,8064,18720,22752,9612,7056,10224,17928,3968,2538,3072,240,1,40320,146160,92160,70560,32256,207360,249120,193536,73728,61560,144720,246816,101844,142704,7936,51048,110448,34304,8334,11616,494,1];
    /* function of triangle */
    T(n,k):=M[sum(num_partitions(i),i,0,n-1)+k+1];
    /* count number of partitions of n into m parts */
    b(n,m):=if n
    				
  • Maxima
    /* Find triangle */
    Co(n,k):=if k=1  then a(n) else sum(a(i+1)*Co(n-i-1,k-1),i,0,n-k);
    a(n):=if n=1 then 1 else 1/n*sum(Co(n-1,k)*x(k),k,1,n-1);
    makelist(ratsimp(n!*a(n)),n,1,5);
    /* Vladimir Kruchinin, Jun 15 2012 */
    
  • PARI
    serlaplace( serreverse( intformal( 1 / sum(n=0, 9, eval(Str("g"n)) * x^n, x * O(x^9))))) /* Michael Somos, Oct 22 2014 */
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