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%I #32 Nov 25 2020 14:15:50
%S 1,1,3,11,43,180,804,3763,18331,92330,478795,2547885,13880832,
%T 77284220,439146427,2543931619,15010717722,90154755356,550817917537,
%U 3421683388385,21601986281226,138548772267326,902439162209914,5967669851051612,40053432076016812
%N Number of connected multigraphs with n edges and rooted at two indistinguishable vertices whose removal leaves a connected graph.
%C This sequence counts the CDE-descendants of a single edge A-Z.
%C [C]onnect: different nodes {P,Q} != {A,Z} may form a new edge P-Q.
%C [D]issect: any edge P-Q may be dissected into P-M-Q with a new node M.
%C [E]xtend: any node P not in {A,Z} may form a new edge P-Q with a new node Q.
%C These basic operations were motivated by A338487, which seemed to count the CDE-descendants of K_4 with edge A-Z removed.
%D Technology Review's Puzzle Corner, How many different resistances can be obtained by combining 10 one ohm resistors? Oct 3, 2003.
%H Joel Karnofsky, <a href="https://web.archive.org/web/20111123142636/http://www.cs.nyu.edu/~gottlieb/tr/2003-oct-3.pdf">Solution of problem from Technology Review's Puzzle Corner Oct 3, 2003</a>, Feb 23, 2004.
%e The a(3) = 3 CDE-descendants of A-Z with 3 edges are
%e .
%e A A A
%e ( ) / /
%e o o - o o - o
%e | / \
%e Z Z Z
%e .
%e DCC DD DE
%e .
%o (PARI) \\ See A339065 for G.
%o InvEulerT(v)={my(p=log(1+x*Ser(v))); dirdiv(vector(#v,n,polcoef(p,n)), vector(#v,n,1/n))}
%o seq(n)={my(A=O(x*x^n), g=G(2*n, x+A,[]), gr=G(2*n, x+A,[1])/g, u=InvEulerT(Vec(-1+G(2*n, x+A,[1,1])/(g*gr^2))), t=InvEulerT(Vec(-1+G(2*n, x+A,[2])/(g*subst(gr,x,x^2)))), v=vector(n)); for(n=1, #v, v[n]=(u[n]+t[n]-if(n%2==0,u[n/2]-v[n/2]))/2); v} \\ _Andrew Howroyd_, Nov 20 2020
%Y Cf. A180414, A337517, A338487, A339038, A339045, A339065.
%K nonn
%O 1,3
%A _Rainer Rosenthal_, Nov 18 2020
%E a(7)-a(25) from _Andrew Howroyd_, Nov 20 2020
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