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 1-6 of 6 results.

A281118 a(1)=1, a(n>1) = number of tree-factorizations of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 4, 2, 2, 1, 6, 1, 2, 2, 12, 1, 6, 1, 6, 2, 2, 1, 20, 2, 2, 4, 6, 1, 8, 1, 32, 2, 2, 2, 28, 1, 2, 2, 20, 1, 8, 1, 6, 6, 2, 1, 76, 2, 6, 2, 6, 1, 20, 2, 20, 2, 2, 1, 38, 1, 2, 6, 112, 2, 8, 1, 6, 2, 8, 1, 116, 1, 2, 6, 6, 2, 8, 1, 76, 12, 2, 1
Offset: 1

Views

Author

Gus Wiseman, Jan 15 2017

Keywords

Comments

A tree-factorization of n>=2 is either (case 1) the number n or (case 2) a sequence of two or more tree-factorizations, one of each part of a weakly increasing factorization of n. These are rooted plane trees and the ordering of branches is important. For example, {{2,2},9}, {2,{2,9}}, {{2,2},{3,3}}, {6,{2,3}}, and {{2,3},6} are distinct tree-factorizations of 36, but {9,{2,2}}, {{2,9},2}, and {{3,3},{2,2}} are not.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

Examples

			The a(30)=8 tree-factorizations are 30, 2*15, 2*(3*5), 3*10, 3*(2*5), 5*6, 5*(2*3), 2*3*5.
		

Crossrefs

Programs

  • Mathematica
    postfacs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[postfacs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    treefacs[n_]:=If[n<=1,{{}},Prepend[Join@@Function[q,Tuples[treefacs/@q]]/@DeleteCases[postfacs[n],{n}],n]];
    Table[Length[treefacs[n]],{n,2,83}]
  • PARI
    seq(n)={my(v=vector(n), w=vector(n)); w[1]=v[1]=1; for(k=2, n, w[k]=v[k]+1; forstep(j=n\k*k, k, -k, my(i=j, e=0); while(i%k==0, i/=k; e++; v[j] += w[k]^e*v[i]))); w} \\ Andrew Howroyd, Nov 18 2018

Formula

a(p^n) = A289501(n) for prime p. - Andrew Howroyd, Nov 18 2018

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

A052894 a(n) is the number of Schröder trees on n vertices with a prescribed root.

Original entry on oeis.org

1, 1, 5, 46, 631, 11586, 267369, 7442758, 242833091, 9090987610, 384209125453, 18096001098462, 939991769248239, 53388611049236386, 3291647968944928337, 218948960832551848438, 15629052780600654123739, 1191728692751208814306986, 96675526164560545405689141
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Previous name was: A simple grammar.
Number of pointed trees on pointed sets k[1...k...n] with a prescribed point k. - Gus Wiseman, Sep 27 2015

Examples

			The a(4) = 46 pointed trees of the form rootpoint[pointedbranch ... pointedbranch] on 1[1 2 3 4] are 1[1 2[2 3[3 4]]], 1[1 2[2 4[3 4]]], 1[1 2[2[2 4] 3]], 1[1 2[2[2 3] 4]], 1[1 2[2 3 4]], 1[1 3[2 3[3 4]]], 1[1 3[2[2 4] 3]], 1[1 3[3 4[2 4]]], 1[1 3[3[2 3] 4]], 1[1 3[2 3 4]], 1[1 4[2 4[3 4]]], 1[1 4[3 4[2 4]]], 1[1 4[2[2 3] 4]], 1[1 4[3[2 3] 4]], 1[1 4[2 3 4]], 1[1[1 3[3 4]] 2], 1[1[1 4[3 4]] 2], 1[1[1[1 4] 3] 2], 1[1[1[1 3] 4] 2], 1[1[1 3 4] 2], 1[1[1 2[2 4]] 3], 1[1[1 4[2 4]] 3], 1[1[1[1 4] 2] 3], 1[1[1[1 2] 4] 3], 1[1[1 2 4] 3], 1[1[1 2[2 3]] 4], 1[1[1 3[2 3]] 4], 1[1[1[1 3] 2] 4], 1[1[1[1 2] 3] 4], 1[1[1 2 3] 4], 1[1[1 2] 3[3 4]], 1[1[1 2] 4[3 4]], 1[1[1 3] 2[2 4]], 1[1[1 3] 4[2 4]], 1[1[1 4] 2[2 3]], 1[1[1 4] 3[2 3]], 1[1 2 3[3 4]], 1[1 2 4[3 4]], 1[1 2[2 4] 3], 1[1 3 4[2 4]], 1[1 2[2 3] 4], 1[1 3[2 3] 4], 1[1[1 4] 2 3], 1[1[1 3] 2 4], 1[1[1 2] 3 4], 1[1 2 3 4]. - _Gus Wiseman_, Sep 27 2015
		

Crossrefs

Programs

  • Maple
    spec := [S,{C=Set(B,1 <= card),B=Prod(Z,S),S=Sequence(C)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    m = 20; (* number of terms *)
    Rest@CoefficientList[InverseSeries[Series[2*x - x*E^x, {x, 0, m}], x], x]*Range[0, m-1]! (* Jean-François Alcover, Oct 11 2022 *)
  • PARI
    {a(n) = local(A=1); A = (1/x)*serreverse(2*x - x*exp(x +x^2*O(x^n) )); n!*polcoeff(A, n)}
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 19 2015
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=1); A = 1 + (1/x)*sum(m=1, n+1, Dx(m-1, (exp(x +x*O(x^n)) - 1)^m * x^m/m!)); n!*polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 19 2015
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=1+x+x*O(x^n)); A = exp(sum(m=1, n+1, Dx(m-1, (exp(x +x*O(x^n)) - 1)^m * x^(m-1)/m!)+x*O(x^n))); n!*polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 19 2015

Formula

E.g.f.: RootOf(-2*_Z + _Z*exp(x*_Z) + 1).
a(n) = A053492(n)/n.
From Paul D. Hanna, Jun 19 2015: (Start)
E.g.f. A(x) satisfies:
(1) A(x) = (1/x) * Series_Reversion( 2*x - x*exp(x) ).
(2) A(x) = 1 + (1/x) * Sum_{n>=1} d^(n-1)/dx^(n-1) (exp(x)-1)^n * x^n / n!.
(3) A(x) = exp( Sum_{n>=1} d^(n-1)/dx^(n-1) (exp(x)-1)^n * x^(n-1) / n! ).
(End)
(4) A(x) = Sum_{n>=0} exp(n*x*A(x)) / 2^(n+1). - Paul D. Hanna, Apr 07 2018
a(n) = (1/(n+1)!) * Sum_{k=0..n} (n+k)! * Stirling2(n,k). - Seiichi Manyama, Nov 06 2023

Extensions

New name from Vaclav Kotesovec, Feb 16 2015

A281119 Number of complete tree-factorizations of n >= 2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 9, 1, 1, 2, 3, 1, 4, 1, 12, 1, 1, 1, 12, 1, 1, 1, 9, 1, 4, 1, 3, 3, 1, 1, 29, 1, 3, 1, 3, 1, 9, 1, 9, 1, 1, 1, 17, 1, 1, 3, 34, 1, 4, 1, 3, 1, 4, 1, 44, 1, 1, 3, 3, 1, 4, 1, 29, 5, 1, 1
Offset: 2

Views

Author

Gus Wiseman, Jan 15 2017

Keywords

Comments

A tree-factorization of n>=2 is either (case 1) the number n or (case 2) a sequence of two or more tree-factorizations, one of each part of a weakly increasing factorization of n into factors greater than 1. A complete (or total) tree-factorization is a tree-factorization whose leaves are all prime numbers.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

Examples

			The a(36)=12 complete tree-factorizations of 36 are:
(2(2(33))), (2(3(23))), (2(233)),   (3(2(23))),
(3(3(22))), (3(223)),   ((22)(33)), ((23)(23)),
(22(33)),   (23(23)),   (33(22)),   (2233).
		

Crossrefs

Programs

  • Mathematica
    postfacs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[postfacs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    treefacs[n_]:=If[n<=1,{{}},Prepend[Join@@Function[q,Tuples[treefacs/@q]]/@DeleteCases[postfacs[n],{n}],n]];
    Table[Length[Select[treefacs[n],FreeQ[#,_Integer?(!PrimeQ[#]&)]&]],{n,2,83}]
  • PARI
    seq(n)={my(v=vector(n), w=vector(n)); v[1]=1; for(k=2, n, w[k]=v[k]+isprime(k); forstep(j=n\k*k, k, -k, my(i=j, e=0); while(i%k==0, i/=k; e++; v[j]+=w[k]^e*v[i]))); w[2..n]} \\ Andrew Howroyd, Nov 18 2018

Formula

a(p^n) = A196545(n) for prime p. - Andrew Howroyd, Nov 18 2018

A301935 Number of positive subset-sum trees whose composite a positive subset-sum of the integer partition with Heinz number n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 10, 2, 3, 1, 21, 1, 3, 3, 58, 1, 21, 1, 21, 3, 3, 1, 164, 2, 3, 10, 21, 1, 34, 1, 373, 3, 3, 3, 218, 1, 3, 3, 161, 1, 7, 1, 5, 5, 3, 1, 1320, 2, 5, 3, 5, 1, 7, 3, 7, 3, 3, 1, 7, 1, 3, 4, 2558, 3, 7, 1, 5, 3, 6, 1, 7
Offset: 1

Views

Author

Gus Wiseman, Mar 28 2018

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). A positive subset-sum tree with root x is either the symbol x itself, or is obtained by first choosing a positive subset-sum x <= (y_1,...,y_k) with k > 1 and then choosing a positive subset-sum tree with root y_i for each i = 1...k. The composite of a positive subset-sum tree is the positive subset-sum x <= g where x is the root sum and g is the multiset of leaves. We write positive subset-sum trees in the form rootsum(branch,...,branch). For example, 4(1(1,3),2,2(1,1)) is a positive subset-sum tree with composite 4(1,1,1,2,3) and weight 8.

Crossrefs

A301934 Number of positive subset-sum trees of weight n.

Original entry on oeis.org

1, 3, 14, 85, 586, 4331, 33545, 268521, 2204249
Offset: 1

Views

Author

Gus Wiseman, Mar 28 2018

Keywords

Comments

A positive subset-sum tree with root x is either the symbol x itself, or is obtained by first choosing a positive subset-sum x <= (y_1,...,y_k) with k > 1 and then choosing a positive subset-sum tree with root y_i for each i = 1...k. The weight is the sum of the leaves. We write positive subset-sum trees in the form rootsum(branch,...,branch). For example, 4(1(1,3),2,2(1,1)) is a positive subset-sum tree with composite 4(1,1,1,2,3) and weight 8.

Examples

			The a(3) = 14 positive subset-sum trees:
3           3(1,2)       3(1,1,1)     3(1,2(1,1))
2(1,2)      2(1,1,1)     2(1,1(1,1))  2(1(1,1),1)  2(1,2(1,1))
1(1,2)      1(1,1,1)     1(1,1(1,1))  1(1(1,1),1)  1(1,2(1,1))
		

Crossrefs

Showing 1-6 of 6 results.