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A125951 Exponents f(n), n = 1, 2, ..., for the infinite product 1 -z - z^2 - z^3 = Product_{n>=1} (1-z^n)^f(n). 0

%I

%S 1,1,2,2,4,5,10,15,26,42,74,121,212,357,620,1064,1856,3209,5618,9794,

%T 17192,30153,53114,93554,165308,292250,517802,918207,1630932,2899434,

%U 5161442,9196168,16402764,29281168,52319364,93555601,167427844

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

%C 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?

%D T. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, Theorem 14.8.

%H David Broadhurst, <a href="http://arxiv.org/abs/1504.05303">Multiple Landen values and the tribonacci numbers</a>, arXiv:1504.05303 [hep-th], 2015.

%F 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).)

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

%K nonn

%O 1,3

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

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Last modified August 28 00:41 EDT 2015. Contains 261101 sequences.