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A032035 Number of increasing rooted 2,3 cacti (triangular cacti with bridges) with n-1 nodes. 2

%I #38 Aug 28 2020 11:06:20

%S 1,1,1,3,13,77,573,5143,54025,650121,8817001,133049339,2210979381,

%T 40118485237,789221836741,16730904387183,380227386482641,

%U 9221550336940241,237724953543108753,6491255423787076915,187156557809878784797,5681772224922980536413

%N Number of increasing rooted 2,3 cacti (triangular cacti with bridges) with n-1 nodes.

%C Also increasing involution rooted trees with n-1 nodes.

%H Alois P. Heinz, <a href="/A032035/b032035.txt">Table of n, a(n) for n = 1..200</a>

%H O. Bodini, M. Dien, X. Fontaine, A. Genitrini, H. K. Hwang, <a href="https://doi.org/10.1007/978-3-662-49529-2_16">Increasing Diamonds</a>, in LATIN 2016: 12th Latin American Symposium, Ensenada, Mexico, April 11-15, 2016, Proceedings Pages pp 207-219 2016, Lecture Notes in Computer Science Series Volume 9644.

%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>

%H <a href="/index/Ca#cacti">Index entries for sequences related to cacti</a>

%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>

%F E.g.f. of a(n+1) satisfies A'(x) = exp(A(x)+A(x)^2/2).

%F E.g.f. satisfies A''(x) = 1/(1-A(x)).

%F Shifts left 2 places under "AIJ" (ordered, indistinct, labeled) transform.

%F a(1) = a(2) = 1; a(n) = Sum_{k=1..n-2} binomial(n-2,k) * a(k) * a(n-k). - _Ilya Gutkovskiy_, Aug 28 2020

%p A:= proc(n) option remember; if n=0 then x else convert(series(Int(exp(A(n-1)+ A(n-1)^2/2), x), x=0, n+1), polynom) fi end; a:= n-> if n=1 then 1 else coeff(A(n-1), x, n-1)*(n-1)! fi: seq(a(n), n=1..20); # _Alois P. Heinz_, Aug 22 2008

%t CoefficientList[Series[Sqrt[2]*InverseErf[Sqrt[2/(E*Pi)] x + Erf[1/Sqrt[2]]], {x, 0, 20}], x] * Range[0, 20]! (* _Vaclav Kotesovec_, Jan 07 2014 *)

%t m = 22; A[_] = 0;

%t Do[A[x_] = Integrate[Exp[A[x] + A[x]^2/2], x] + O[x]^m, {m}];

%t CoefficientList[1 + A[x], x]*Range[0, m-1]! (* _Jean-François Alcover_, Sep 29 2019 *)

%o (PARI) seq(n)={my(p=x+O(x*x^(n%2))); for(i=1, n\2, p=intformal(1 + intformal(1/(1-p)))); Vec(serlaplace(p))} \\ _Andrew Howroyd_, Sep 19 2018

%Y Cf. A001147, A091481.

%K nonn,eigen

%O 1,4

%A _Christian G. Bower_, Apr 01 1998

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