|
%I #8 Feb 19 2021 20:10:00
%S 1,1,1,2,4,6,10,15,26,42,74,121,212,357,620,1064,1856,3209,5618,9794,
%T 17192,30153,53114,93554,165308,292250,517802,918207,1630932,2899434,
%U 5161442,9196168,16402764,29281168,52319364,93555601,167427844,299841117,537357892,963641588,1729192432
%N Number of orbits of length n under the map whose periodic points are counted by A001643.
%C 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.
%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)*A001643(n/d).
%e 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.
%o (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))
%o a(n) = (1/n)*sumdiv(n, d, moebius(d)*a001643(n/d)); \\ _Michel Marcus_, Sep 11 2017
%Y Cf. A001642, A060164, A060165, A060166, A060167, A060169, A060170, A060171, A060171.
%K easy,nonn
%O 1,4
%A _Thomas Ward_, Mar 13 2001
%E More terms from _Michel Marcus_, Sep 11 2017
|
