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%I #11 Mar 13 2022 19:02:33
%S 1,2,-1,3,-3,0,1,4,-6,0,4,3,-6,2,5,-10,0,10,15,-18,-60,130,-105,40,-6,
%T 6,-15,0,20,45,-18,-330,60,2445,-6485,8712,-7260,3925,-1350,270,-24,7,
%U -21,0,35,105,42,-980,-1950,11760,12355,-182721,589281,-1128820,1502550,-1471305
%N Irregular triangle read by rows: T(n,k) (n>=1, 0<=k<=n(n-1)/2) is such that Sum_k T(n,k)*q^k gives the expectation of the number of connected components in a random graph on n labeled vertices where every edge is present with probability q.
%F G.f.: Sum_{n,k} T(n,k)*x^n/((1-q)^(n*(n-1)/2)*n!) = H(x,1-q)*exp(H(x,1-q)) where H(x,p)=Sum_{n=1..oo} x^n/(p^(n*(n-1)/2)*n!).
%F Sum_k T(n,k)*q^k = Sum_k A125205(n,k)*(1-q)^(n*(n-1)/2-k)*q^k
%F Sum_k T(n,k)*q^k = Sum_k A125206(n,k)*q^(n*(n-1)/2-k)*(1-q)^k
%e Triangle begins:
%e 1;
%e 2, -1;
%e 3, -3, 0, 1;
%e 4, -6, 0, 4, 3, -6, 2;
%e 5, -10, 0, 10, 15, -18, -60, 130, -105, 40, -6;
%e ...
%e Sum_k T(3,k)*q^k = 3-3*q+q^3 is the expectation of the number of connected components in a random graph on 3 labeled vertices where every edge is present with probability q.
%o (PARI) { H=sum(n=0,6,x^n/(1-q)^(n*(n-1)/2)/n!); B=H*log(H); for(n=1,6,print(Vecrev((1-q)^(n*(n-1)/2)*n!*polcoeff(B,n,x)))) }
%Y Cf. A125205, A125206, A127258 (row-reversed version), A125208 (dual version).
%K sign,tabf
%O 1,2
%A _Max Alekseyev_, Jan 09 2007