A125951 Exponents f(n), n = 1, 2, ..., in the infinite product 1 - z - z^2 - z^3 = Product_{n>=1} (1-z^n)^f(n).

1, 1, 2, 2, 4, 5, 10, 15, 26, 42, 74, 121, 212, 357, 620, 1064, 1856, 3209, 5618, 9794, 17192, 30153, 53114, 93554, 165308, 292250, 517802, 918207, 1630932, 2899434, 5161442, 9196168, 16402764, 29281168, 52319364, 93555601, 167427844

Offset: 1

Author

Barry Brent (barrybrent(AT)member.ams.org), Feb 04 2007

Keywords

• nonn

Comments

Let w = z + z^2 + z^3. Then 1 - z - z^2 - z^3 = 1 - 1w = (by the cyclotomic identity) Product_{n>=1} (1-w^n)^P(1,n), where P is the necklace polynomial. P is a counting function. Is f also a counting function?

Examples

			f(1) = f(2) = 1 because 1 - z - z^2 - z^3 = (1-z)^1 *(1-z^2)^1 * ....
		

References

• T. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, Theorem 14.8.
• Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 500.

Links

• David Broadhurst, Multiple Landen values and the tribonacci numbers, arXiv:1504.05303 [hep-th], 2015.

Crossrefs

Cf. A006206, A113788.

Programs

Sage

z = PowerSeriesRing(ZZ, 'z').gen().O(30)
r = (1 - (z + z**2 + z**3))
F = -z*r.derivative()/r
[sum(moebius(n//d)*F[d] for d in divisors(n))//n for n in range(1,24)]

Formula

Let r(n) be the coefficient of z^n in 1 - z - z^2 - z^3, so that r(0) = 1 and r(n) = 0 for n>3. Let F(k) satisfy the recurrence n r(n) + sum_{k=1}^n r(n-k)F(k) = 0. Let mu be the usual Möbius function. Then f(n) = (1/n) sum_{d|n} mu(n/d) F(d) (so that n*f(n) is the Möbius inverse of F(n).)