A136653 G.f.: A(x) satisfies: coefficient of x^n in A(x)^(n+1)/(n+1) = 2^(n*(n-1)/2).

1, 1, 1, 4, 39, 748, 27162, 1880872, 252273611, 66358216668, 34506398937158, 35644762692112792, 73356520492898454022, 301274559225693420690360, 2471654510727312089903896948, 40527708183358718551543295827536, 1328579216048284168977214446788083699

Offset

0,4

Comments

a(n) is the number of graphs on vertices 1,...,n such that, when these vertices are arranged counterclockwise around a circle and edges are drawn as straight line segments, the resulting diagram is connected. - Jonathan Novak (j2novak(AT)math.uwaterloo.ca), Apr 30 2010

In this interpretation, both intersecting (set theoretically) and crossing (topologically) edges are considered connected. - Gus Wiseman, Feb 23 2019

Links

Andrew Howroyd, Table of n, a(n) for n = 0..50

Gus Wiseman, The a(4) = 39 graphs connected by overlaps or crossings.

Formula

G.f.: A(x) = x/Series_Reversion( x*Sum_{k=0..n} 2^(k(k-1)/2)*x^k ).

Equals the free cumulant sequence corresponding to A006125. - Jonathan Novak (j2novak(AT)math.uwaterloo.ca), Apr 30 2010

Example

G.f.: A(x) = 1 + x + x^2 + 4*x^3 + 39*x^4 + 748*x^5 + 27162*x^6 +...
Let F(x) = 1 + x + 2*x^2 + 8*x^3 + 64*x^4 + 1024*x^5 +...+ 2^(n*(n-1)/2)*x^n +..
then A(x) = F(x/A(x)), A(x*F(x)) = F(x).
Coefficient of x^n in A(x)^(n+1)/(n+1) = 2^(n*(n-1)/2),
as can be seen by the main diagonal in the array of
coefficients in the initial powers of A(x):
A^1: [(1), 1, 1, 4, 39, 748, 27162, 1880872, 252273611,...;
A^2: [1, (2), 3, 10, 87, 1582, 55914, 3817876, 508370795,...;
A^3: [1, 3, (6), 19, 147, 2517, 86398, 5813550, 768378627,...;
A^4: [1, 4, 10, (32), 223, 3572, 118778, 7870640, 1032387787,...;
A^5: [1, 5, 15, 50, (320), 4771, 153245, 9992130, 1300492845,...;
A^6: [1, 6, 21, 74, 444, (6144), 190023, 12181278, 1572792585,...;
A^7: [1, 7, 28, 105, 602, 7728, (229376), 14441659, 1849390375,...;
A^8: [1, 8, 36, 144, 802, 9568, 271616, (16777216), 2130394591,...;
A^9: [1, 9, 45, 192, 1053, 11718, 317112, 19192320, (2415919104),...;
dividing each diagonal term in row n by (n+1) gives 2^(n*(n-1)/2).
The diagonal above the main diagonal gives coefficients of l.g.f.:
log(F(x)) = x + 3*x^2/2 + 19*x^3/3 + 223*x^4/4 + 4771*x^5/5 +...

Mathematica

max = 15; s = x*Sum[2^(k*(k-1)/2)*x^k, {k, 0, max}] + O[x]^(max+2); x/InverseSeries[s] + O[x]^(max+1) // CoefficientList[#, x]& (* Jean-François Alcover, Sep 03 2017 *)
croXQ[stn_]:=MatchQ[stn, {___, {___, x_, ___, y_, ___}, ___, {___, z_, ___, t_, ___}, ___}/; x<z<y<t||z<x<t<y];
csm[s_]:=With[{c=Select[Tuples[Range[Length[s]], 2], And[OrderedQ[#], UnsameQ@@#, Length[Intersection@@s[[#]]]>0]&]}, If[c=={}, s, csm[Sort[Append[Delete[s, List/@c[[1]]], Union@@s[[c[[1]]]]]]]]];
bicmpts[stn_]:=csm[Union[Subsets[stn, {1}], Select[Subsets[stn, {2}], Intersection@@#!={}&], Select[Subsets[stn, {2}], croXQ]]];
Table[Length[Select[Subsets[Subsets[Range[n], {2}]], And[Union@@#==Range[n], Length[bicmpts[#]]<=1]&]], {n, 0, 5}] (* Gus Wiseman, Feb 23 2019 *)

Prog

(PARI) a(n)=polcoeff(x/serreverse(x*sum(k=0, n, 2^(k*(k-1)/2)*x^k +x*O(x^n))), n)

Crossrefs

Cf. A136652 (log(A(x))), A136654.

Cf. A000699, A002662, A006125, A007297, A016098, A054726, A099947, A306438.

Cf. A324166, A324169, A324172, A324173, A324327, A324328.

Sequence in context: A129463 A299426 A188418 * A165434 A348118 A341473

Adjacent sequences: A136650 A136651 A136652 * A136654 A136655 A136656

Keyword

nonn,changed

Author

Paul D. Hanna, Jan 15 2008

Extensions

Name changed and part of prior name moved to formula section by Paul D. Hanna, Sep 19 2013

Status

approved