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%I #9 Mar 08 2020 16:43:51
%S 1,1,1,1,1,1,1,1,1,1,1,2,3,3,2,1,1,1,1,1,1,2,4,5,7,8,9,8,7,5,4,2,1,1,
%T 1,1,1,1,2,4,6,9,14,22,29,33,37,43,43,37,33,29,22,14,9,6,4,2,1,1,1,1,
%U 1,1,2,4,6,10,16,29,46,69,99,141,183,230,277,319,342,352,342,319,277,230,183,141,99,69,46,29,16,10,6,4,2,1,1,1
%N Triangle read by rows: T(n,k) = number of unlabeled bicolored graphs having 2n nodes and k edges, which are invariant when the two color classes are interchanged. Here n >= 0, 0 <= k <= n^2.
%D R. W. Robinson, Numerical implementation of graph counting algorithms, AGRC Grant, Math. Dept., Univ. Newcastle, Australia, 1978.
%H Andrew Howroyd, <a href="/A123548/b123548.txt">Rows n=0..20, flattened</a> (rows 0..7 from R. W. Robinson)
%e Triangle begins:
%e n = 0
%e k = 0 : 1
%e ************************ total ( 2n = 0) = 1
%e n = 1
%e k = 0 : 1
%e k = 1 : 1
%e ************************ total ( 2n = 2) = 2
%e n = 2
%e k = 0 : 1
%e k = 1 : 1
%e k = 2 : 1
%e k = 3 : 1
%e k = 4 : 1
%e ************************ total ( 2n = 4) = 5
%e n = 3
%e k = 0 : 1
%e k = 1 : 1
%e k = 2 : 1
%e k = 3 : 2
%e k = 4 : 3
%e k = 5 : 3
%e k = 6 : 2
%e k = 7 : 1
%e k = 8 : 1
%e k = 9 : 1
%e ************************ total ( 2n = 6) = 16
%o (PARI)
%o permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m}
%o edges(v, t) = {prod(i=2, #v, prod(j=1, i-1, my(g=gcd(v[i], v[j])); t(2*v[i]*v[j]/g)^g )) * prod(i=1, #v, my(c=v[i]); t(2*c)^(c\2)*if(c%2, t(c), 1))}
%o Row(n) = {my(s=0); forpart(p=n, s+=permcount(p)*edges(p, i->1+x^i)); Vecrev(s/n!)}
%o { for(n=0, 6, print(Row(n))) } \\ _Andrew Howroyd_, Mar 08 2020
%Y Row sums give A122082.
%Y Cf. A008406.
%K nonn,tabf
%O 0,12
%A _N. J. A. Sloane_, Nov 14 2006
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