%I #11 Feb 19 2021 20:10:00
%S 1,1,1,2,3,4,7,10,17,26,44,68,115,184,306,500,835,1374,2301,3822,6409,
%T 10718,18028,30280,51077,86130,145641,246370,417600,708246,1203069,
%U 2045010,3480408,5927660,10105819,17241140,29439580,50302162,86012630,147166248,251963055,431633348
%N Number of orbits of length n under the map whose periodic points are counted by A001641.
%C The sequence A001641 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.
%H Y. Puri and T. Ward, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL4/WARD/short.html">Arithmetic and growth of periodic orbits</a>, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
%H Yash Puri and Thomas Ward, <a href="http://www.fq.math.ca/Scanned/39-5/puri.pdf">A dynamical property unique to the Lucas sequence</a>, Fibonacci Quarterly, Volume 39, Number 5 (November 2001), pp. 398-402.
%F a(n) = (1/n)* Sum_{ d divides n } mu(d)*A001641(n/d).
%e a(7) = 7 since a map whose periodic points are counted by A001641 would have 1 fixed point and 50 points of period 7, hence 7 orbits of length 7.
%o (PARI) a001641(n)=if(n<0, 0, polcoeff(x*(1+2*x+4*x^3)/(1-x-x^2-x^4)+x*O(x^n), n))
%o a(n) = sumdiv(n, d, moebius(d)*a001641(n/d))/n; \\ _Michel Marcus_, Sep 10 2017
%Y Cf. A001641, A060164, A060165, A060167, A060168, A060169, A060170, A060171, A060171.
%K easy,nonn
%O 1,4
%A _Thomas Ward_, Mar 13 2001
%E More terms from _Michel Marcus_, Sep 10 2017