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Numbers k such that repeated application of the Pisano period eventually gives 24.
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%I #29 Aug 04 2024 20:50:38

%S 2,3,4,6,7,8,9,12,13,14,16,17,18,19,21,23,24,26,27,28,29,32,34,36,37,

%T 38,39,42,46,47,48,49,51,52,53,54,56,57,58,59,63,64,67,68,69,72,73,74,

%U 76,78,79,81,83,84,87,91,92,94,96,97,98

%N Numbers k such that repeated application of the Pisano period eventually gives 24.

%C This sequence is infinite. A number n is a fixed point if the Pisano period of n is equal to n. The trajectory of k is the sequence of values the Pisano period takes on under repeated iteration, starting at k and leading to a fixed point; this sequence is the sequence of integers such that the trajectory leads to 24.

%H B. Benfield and O. Lippard, <a href="https://arxiv.org/abs/2404.08194">Fixed points of K-Fibonacci Pisano periods</a>, arXiv:2404.08194 [math.NT], 2024.

%H J. Fulton and W. Morris, <a href="https://eudml.org/doc/204918">On arithmetical functions related to the Fibonacci numbers</a>, Acta Arith., 2(16):105-110, 1969.

%H E. Trojovska, <a href="https://doi.org/10.3390/math8050773">On periodic points of the order of appearance in the Fibonacci sequence</a>, Mathematics, 2020.

%e a(1)=2 because 2 is the smallest number with Pisano period trajectory terminating at 24: pi(2)=3, pi(3)=8, pi(8)=12, pi(12)=24.

%o (Sage)

%o L=[]

%o for i in range(2,101):

%o a=i

%o y=BinaryRecurrenceSequence(1,1,0,1).period(Integer(i))

%o while a!=y:

%o a=y

%o y=BinaryRecurrenceSequence(1,1,0,1).period(Integer(a))

%o if a==24:

%o L.append(i)

%o print(L)

%Y Cf. A001175.

%K nonn

%O 1,1

%A _Oliver Lippard_ and _Brennan G. Benfield_, Aug 04 2024