cp's OEIS Frontend

A125208 Irregular triangle read by rows: T(n,k) (n>=1, 0<=k<=n(n-1)/2) is such that Sum_k T(n,k)*p^k gives the expectation of the number of connected components after deleting every edge of the complete graph on n labeled vertices with probability p.

%I A125208 #15 Mar 13 2022 19:01:59
%S A125208 1,1,1,1,0,3,-1,1,0,0,4,3,-6,2,1,0,0,0,5,0,10,-10,-15,20,-6,1,0,0,0,0,
%T A125208 6,0,0,15,-5,0,-60,25,90,-90,24,1,0,0,0,0,0,7,0,0,0,21,-21,35,0,-105,
%U A125208 0,-105,420,0,-630,504,-120,1,0,0,0,0,0,0,8,0,0,0,0,28,-28,0,56,35,-168,112,-280
%N A125208 Irregular triangle read by rows: T(n,k) (n>=1, 0<=k<=n(n-1)/2) is such that Sum_k T(n,k)*p^k gives the expectation of the number of connected components after deleting every edge of the complete graph on n labeled vertices with probability p.
%F A125208 G.f.: Sum_{n,k} T(n,k)*x^n/(p^(n*(n-1)/2)*n!) = H(x,p)*exp(H(x,p)) where H(x,p)=Sum_{n=1..oo} x^n/(p^(n*(n-1)/2)*n!).
%F A125208 Sum_k T(n,k)*p^k = Sum_k A125205(n,k)*p^(n*(n-1)/2-k)*(1-p)^k
%e A125208 Triangle begins:
%e A125208   1;
%e A125208   1, 1;
%e A125208   1, 0, 3, -1;
%e A125208   1, 0, 0,  4, 3, -6,  2;
%e A125208   1, 0, 0,  0, 5,  0, 10, -10, -15, 20, -6;
%e A125208    ...
%e A125208 Sum_k T(3,k)*p^k = 1+3*p^2-p^3 is the expectation of the number of connected components in a complete graph on 3 labeled vertices where every edge is removed with probability p.
%o A125208 (PARI) { H=sum(n=0,6,x^n/p^(n*(n-1)/2)/n!); A=H*log(H); for(n=1,6,print(Vecrev(p^(n*(n-1)/2)*n!*polcoeff(A,n,x)))) }
%Y A125208 Cf. A125205, A125206, A125209 (row-reversed version), A125210 (dual version).
%K A125208 sign,tabf
%O A125208 1,6
%A A125208 _Max Alekseyev_, Jan 09 2007