A060168 Number of orbits of length n under the map whose periodic points are counted by A001643.

1, 1, 1, 2, 4, 6, 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, 299841117, 537357892, 963641588, 1729192432

Offset: 1

Author

Thomas Ward, Mar 13 2001

Keywords

• easy
• nonn

Comments

The sequence A001643 seems to record the number of points of period n under a map. The number of orbits of length n for this map gives the sequence above.

Examples

			u(7) = 10 since a map whose periodic points are counted by A001643 would have 1 fixed point and 71 points of period 7, hence 10 orbits of length 7.
		

Links

• Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
• Yash Puri and Thomas Ward, A dynamical property unique to the Lucas sequence, Fibonacci Quarterly, Volume 39, Number 5 (November 2001), pp. 398-402.

Crossrefs

Cf. A001642, A060164, A060165, A060166, A060167, A060169, A060170, A060171, A060171.

Programs

PARI

a001643(n) = if(n<0, 0, polcoeff(x*(1+2*x+4*x^3+5*x^4+6*x^5)/(1-x-x^2-x^4-x^5-x^6)+x*O(x^n), n))
a(n) = (1/n)*sumdiv(n, d, moebius(d)*a001643(n/d)); \\ Michel Marcus, Sep 11 2017

Formula

a(n) = (1/n)* Sum_{ d divides n } mu(d)*A001643(n/d).

Extensions

More terms from Michel Marcus, Sep 11 2017