login
This site is supported by donations to The OEIS Foundation.

 

Logo

Invitation: celebrating 50 years of OEIS, 250000 sequences, and Sloane's 75th, there will be a conference at DIMACS, Rutgers, Oct 9-10 2014.

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A125951 Exponents f(n), n = 1, 2, ..., for the infinite product 1 -z - z^2 - z^3 =Product_{n=1}^{\infty} (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}^{\infty} (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}^{\infty} (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.

%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 M\"obius function. Then f(n) = (1/n) sum_{d|n} \mu(n/d) F(d) (so that n*f(n) is the M\"obius 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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .

Last modified September 3 03:19 EDT 2014. Contains 246369 sequences.