A157164 1/product(1 - a(n)*(x^n)/n!, n=1..infinity) = 1 + sum((x^k)/k, k=1..infinity) = 1 - log(1-x).

1, -1, -1, -8, -16, -74, -526, -6768, -30024, -291072, -2805408, -29134896, -374664720, -5276228736, -67692115440, -1404936248064, -16610113920768, -301439595923712, -5110200130727808, -103584959322338304

Offset

1,4

Links

Table of n, a(n) for n=1..20.

W. Lang Recurrences for the general problem.

Formula

Recurrence I: With P(n,m) the set of partitions of n with m parts:

a(n)= (n-1)! - sum(sum(M1(n;vec(e))*product(a(j)^e(j),j=1..m), p=(1^e(1),...,n^e(n)) from P(n,m)), m=2..n) for n>=2, with sum(j*e(j),j=1..n)=n, sum(e(j),j=1..n)=m for the partition p of n with m parts. Input a(1)=0!=1. See the array A008284(n,m) for the cardinalities of the sets P(n,m). The M1 numbers are given, for any partition p=(1^e(1),...,n^e(n)) from P(n,m) by M1(n;vec(e)):=n/product(j!^e(j), j=1..n). See the array A036038.

Recurrence II: a(n)= -(n-1)!*((1+(-1)^n) + sum(d*(a(d)/d!)^(n/d),d|n with 1<d<n)) + sum(((-1)^(m-1))*(m-1)!*sum(M2(n;vec(e)), p from P(n,m)), m=1..n-1), n>=2; a(1)=0!=1. The set P(n,m) has been defined in recurrence I above. M2(n;vec(e):=n!/product(e(j)!, j=1..n).

Recurrence II (rewritten, due to email from V. Jovovic, Mar 10 2009. See the link for the general derivation):

a(n) = -(n-1)!*sum(d*(a(d)/d!)^(n/d),d|n with 1<=d<n) + A(n), with the e.g.f. log(1-log(1-x)) for the sequence A(n); hence A(n)=A089064(n)=[1,0,1,1,8,26,194,1142,...].

Example

Recurrence I: a(4) = 6 - 4*a(1)*a(3) - 6*a(2)^2 -12*a(1)^2*a(2) - 24*a(1)^4 = 6-4*(-1)-6-12*(-1) = -8.
Recurrence II: a(4) = -6*(1+1+ 2*(-1/2)^2) + 0!*6 - 1*(8+3) + 2*6 = -8.
Recurrence II (rewritten): a(4) = -6*(1+2*(-1/2)^2) +1 = -8.

Crossrefs

A137852 (with different signs) for exp(x) = 1/product(1 - a(n)*(x^n)/n!, n=1..infinity).

Sequence in context: A282785 A278747 A218066 * A131539 A331419 A271080

Adjacent sequences: A157161 A157162 A157163 * A157165 A157166 A157167

Keyword

sign,easy

Author

Wolfdieter Lang Aug 10 2009

Status

approved