

A267500


Number of fixed points or cycles of autobiographical numbers (A267491 ... A267498) in base n.


10




OFFSET

2,1


COMMENTS

For n>=5, it appears that a(n)=2^(n3)+2*n^217*n+43. This formula is correct for 5<=n<=11, but may not be true for larger n.


REFERENCES

Antonia Münchenbach and Nicole Anton George, "Eine Abwandlung der ConwayFolge", contribution to "Jugend forscht" 2016, 2016


LINKS

Table of n, a(n) for n=2..11.
Andre Kowacs, Studies on the Pea Pattern Sequence, arXiv:1708.06452 [math.HO], 2017.


FORMULA

a(n) = 2^(n3) + 2*n^2  17*n + 43, for 5<=n<=11.


EXAMPLE

In base two there are only two fixedpoints, 111 and 1101001.
In base 3, there are 7 fixedpoints: 22, 10111, 11112, 100101, 1011122, 2021102, 10010122 and 1 cycle of length 3 with 2012112, 1010102, 10011112.
In base 10, there are 109 fixedpoints, 31 cycles of length 2 (62 numbers) and 10 cycles of length 3 (30 numbers).


CROSSREFS

Cf. A047841, A267491, A267492, A267493, A267494, A267495, A267496, A267497, A267498, A267499, A267500, A267502.
Sequence in context: A213135 A037187 A248061 * A300016 A166542 A316967
Adjacent sequences: A267497 A267498 A267499 * A267501 A267502 A267503


KEYWORD

nonn,base,more


AUTHOR

Antonia Münchenbach, Jan 27 2016


STATUS

approved



