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A125210
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.
2
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, 6, -15, 0, 20, 45, -18, -330, 60, 2445, -6485, 8712, -7260, 3925, -1350, 270, -24, 7, -21, 0, 35, 105, 42, -980, -1950, 11760, 12355, -182721, 589281, -1128820, 1502550, -1471305
OFFSET
1,2
FORMULA
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!).
Sum_k T(n,k)*q^k = Sum_k A125205(n,k)*(1-q)^(n*(n-1)/2-k)*q^k
Sum_k T(n,k)*q^k = Sum_k A125206(n,k)*q^(n*(n-1)/2-k)*(1-q)^k
EXAMPLE
Triangle begins:
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;
...
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.
PROG
(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)))) }
CROSSREFS
Cf. A125205, A125206, A127258 (row-reversed version), A125208 (dual version).
Sequence in context: A141486 A184344 A144243 * A098434 A375479 A212634
KEYWORD
sign,tabf
AUTHOR
Max Alekseyev, Jan 09 2007
STATUS
approved