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Number of orbits of length n under the map whose periodic points are counted by A061685.
0

%I #7 Feb 19 2021 20:10:00

%S 1,4,99,6272,876725,232419936,105471170140,76095730062464,

%T 82555139387847312,128928209221144677400,279860608037771819829980,

%U 820360089598849358326307904,3169977309466844379463315722484

%N Number of orbits of length n under the map whose periodic points are counted by A061685.

%C Old name was: A061685 appears to count the periodic points for a certain map. If so, then this is the sequence of the numbers of orbits of length n for that map.

%H Y. Puri and T. Ward, <a href="http://www.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 J.-M. Sixdeniers, K. A. Penson and A. I. Solomon, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL4/SIXDENIERS/bell.html">Extended Bell and Stirling Numbers From Hypergeometric Exponentiation</a>, J. Integer Seqs. Vol. 4 (2001), #01.1.4.

%H Thomas Ward, <a href="http://web.archive.org/web/20081002082625/http://www.mth.uea.ac.uk/~h720/research/files/integersequences.html">Exactly realizable sequences</a>. <a href="/A091112/a091112.pdf">[local copy]</a>.

%F If b(n) is the (n+1)th term of A061685, then a(n) = (1/n)*Sum_{d|n}mu(d)b(n/d).

%e b(1)=1, b(2)=9, b(3)=298. Hence a(3)=(1/3)(b(3)-b(1))=99.

%Y Cf. A061685.

%K nonn

%O 1,2

%A _Thomas Ward_, Feb 24 2004

%E Name clarified by _Michel Marcus_, May 14 2015