A060171 Number of orbits of length n under a map whose periodic points seem to be counted by A006953.

12, 54, 80, 30, 24, 5400, 0, 990, 1568, 636, 24, 2720, 0, 240, 5704, 510, 0, 3835776, 0, 26724, 3600, 108, 24, 89760, 0, 240, 1064, 120, 24, 113569300, 0, 510, 11752, 0, 264, 278281640

Offset: 1

Author

Thomas Ward, Mar 13 2001

Keywords

• easy
• nonn

Comments

The sequence A006953 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(3) = 80 since a map whose periodic points are counted by A006953 has 12 fixed points and 252 points of period 3, hence 80 orbits of length 3.
		

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. A006953, A060164, A060165, A060166, A060167, A060168, A060169, A060170, A060172, A060173.

Programs

PARI

a(n) = sumdiv(n, d, moebius(d)*denominator(bernfrac(2*n/d)/(2*n/d)))/n; \\ Michel Marcus, Sep 10 2017

Formula

a(n) = (1/n)* Sum_{d|n} mu(d)*A006953(n/d).