A088368 G.f. satisfies: A(x) = Sum_{n>=0} n!*x^n*A(x)^n.
1, 1, 3, 13, 69, 421, 2867, 21477, 175769, 1567273, 15213955, 160727997, 1846282381, 23013527421, 310284575683, 4506744095141, 70199956070705, 1167389338452753, 20636801363971139, 386304535988493101, 7630926750477398037, 158584458024427667669
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 69*x^4 + 421*x^5 + 2867*x^6 +... where A(x) = 1 + x*A(x) + 2!*x^2*A(x)^2 + 3!*x^3*A(x)^3 + 4!*x^4*A(x)^4 +... Related expansions: A(x)^2 = 1 + 2*x + 7*x^2 + 32*x^3 + 173*x^4 + 1058*x^5 +... A(x)^3 = 1 + 3*x + 12*x^2 + 58*x^3 + 321*x^4 + 1977*x^5 +... A(x)^4 = 1 + 4*x + 18*x^2 + 92*x^3 + 523*x^4 + 3256*x^5 +... A(x)^5 = 1 + 5*x + 25*x^2 + 135*x^3 + 790*x^4 + 4986*x^5 +...
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..200
- David Callan, Sets, Lists and Noncrossing Partitions, arXiv:0711.4841 [math.CO], 2007-2008.
- David Callan and Emeric Deutsch, The Run Transform, arXiv preprint arXiv:1112.3639 [math.CO], 2011.
- Jun Yan, Results on pattern avoidance in parking functions, arXiv:2404.07958 [math.CO], 2024. See p. 11.
- Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 8.
Crossrefs
Cf. A198916.
Programs
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Mathematica
FrequencyDistribution[list_List] := Module[{set = Union[list]}, Table[{set[[i]], Count[list, set[[i]]]}, {i, Length[set]}]]; a[0] = 1; a[n_]/;n>=1 := a[n] = Apply[Plus,Module[{frequencies},Map[(frequencies=Map[Last,FrequencyDistribution[ # ]]; Sum[frequencies]!*Apply[Multinomial,frequencies]* Product[Map[a,# ]])&,Partitions[n]-1 ]]] Table[a[n],{n,0,15}] - David Callan, Jul 25 2008 -
PARI
{a(n)=polcoeff(1/x*serreverse(x/sum(m=0,n,m!*x^m)+x^2*O(x^n)),n)} for(n=0, 30, print1(a(n),", ")) -
PARI
/* Recursive continued fraction: */ {a(n)=local(A=1+x,CF=1+x*O(x^(n+2))); for(i=1,n, for(k=1, n+1, CF=1/(1-((n-k+1)\2+1)*x*A*CF));A=CF); polcoeff(A,n)} for(n=0, 30, print1(a(n),", ")) -
PARI
/* Differential Equation */ {a(n) = my(A=1+x); for(i=0,n, A = 1 + x*A*(x*A^2)'/(x*A+ x^2*O(x^n))' ); polcoeff(A,n)} for(n=0, 30, print1(a(n),", ")) \\ Paul D. Hanna, Apr 01 2018
Formula
G.f.: A(x) = (1/x)*Series_Reversion( x/[Sum_{n>=0} n!*x^n] ).
G.f. satisfies: A(x) = 1/(1 - x*A(x)/(1 - x*A(x)/(1 - 2*x*A(x)/(1 - 2*x*A(x)/(1 - 3*x*A(x)/(1 - 3*x*A(x)/(1 - 4*x*A(x)/(1 - ...)))))))), a recursive continued fraction.
G.f. satisfies: A(x/F(x)) = F(x) where F(x) = Sum_{n>=0} n!*x^n.
G.f. A(x) satisfies: A = 1 + x*A(x) * (x*A(x)^2)' / (x*A(x))'. - Paul D. Hanna, Apr 01 2018
a(n) ~ exp(1) * n!. - Vaclav Kotesovec, Apr 10 2019
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