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URL: https://oeis.org/A125208

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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.
3
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, 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, 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
OFFSET
1,6
FORMULA
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!).
Sum_k T(n,k)*p^k = Sum_k A125205(n,k)*p^(n*(n-1)/2-k)*(1-p)^k
EXAMPLE
Triangle begins:
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;
...
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.
PROG
(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)))) }
CROSSREFS
Cf. A125205, A125206, A125209 (row-reversed version), A125210 (dual version).
Sequence in context: A238414 A195151 A271024 * A333351 A193140 A319993
KEYWORD
sign,tabf
AUTHOR
Max Alekseyev, Jan 09 2007
STATUS
approved