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.

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

Links

Table of n, a(n) for n=1..83.

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

Adjacent sequences: A125205 A125206 A125207 * A125209 A125210 A125211

Keyword

sign,tabf

Author

Max Alekseyev, Jan 09 2007

Status

approved