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EXAMPLE
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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 +...
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MATHEMATICA
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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 *)
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