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A354832 Integers m such that iterating the map f(x) = x^2 + 1 on m generates a number ending with m in binary format.

%I A354832 #10 Jul 31 2022 13:56:48
%S A354832 0,1,2,5,10,26,37,90,165,421,933,1957,4005,8101,8282,24666,40869,
%T A354832 106405,237477,286810,811098,1286053,3383205,5005402,11771813,
%U A354832 28549029,38559834,105668698,239886426,296984485,833855397,1313628250,3461111898,7756079194,9423789989
%N A354832 Integers m such that iterating the map f(x) = x^2 + 1 on m generates a number ending with m in binary format.
%C A354832 It seems that 2^(n-2) <= a(n) < 2^(n-1) for n > 1.
%C A354832 All terms are part of a cycle under x -> f(x) mod 2^L. For example, 5 = f(2), 10 = f(5) mod (2^4), 26 = f(5), 37 = f(10) mod (2^6), and 90 = f(5) mod (2^6).
%C A354832 It takes 2 iterations for a term in the sequence to generate a number ending with the term itself in binary format. Endings of the numbers in the 2 iterations, m1 -> m2 -> m1, for the number of binary digits (d) up to 10 are given below. Note that m1 and m2 are bit-by-bit complement to each other, due to the fact that f(f(x)) = x mod 2^L as pointed out by Kevin Ryde in Discussion.
%C A354832 d     m1 or m2 (bin)      m2 or m1 (bin)       m1 (decimal)
%C A354832 --  ------------------  ------------------  ------------------
%C A354832 1            0 (m1/m2)           1 (m2/m1)  a(1) = 0; a(2) = 1
%C A354832 2           10 (m1)             01 (m2)     a(3) = 2
%C A354832 3          010 (m2)            101 (m1)     a(4) = 5
%C A354832 4         1010 (m1)           0101 (m2)     a(5) = 10
%C A354832 5        11010 (m1)          00101 (m2)     a(6) = 26
%C A354832 6       011010 (m2)         100101 (m1)     a(7) = 37
%C A354832 7      1011010 (m1)        0100101 (m2)     a(8) = 90
%C A354832 8     01011010 (m2)       10100101 (m1)     a(9) = 165
%C A354832 9    001011010 (m2)      110100101 (m1)     a(10)= 421
%C A354832 10  0001011010 (m2)     1110100101 (m1)     a(11)= 933
%e A354832 26 is a term because iterating the map on 26 gives, in binary format, 11010 -> 1010100101 -> 1101111111001011010, which ends with 11010.
%o A354832 (Python)
%o A354832 R = []
%o A354832 for i in range(0, 34):
%o A354832     t = 2**i; L = []
%o A354832     while t not in L: L.append(t); t = (t*t + 1) % 2**(i+1)
%o A354832     {R.append(j) for j in {L[-1], L[-2]} if j not in R}
%o A354832 R.sort(); print(*R, sep = ', ')
%Y A354832 Cf. A002522, A350130, A350590, A352973.
%K A354832 nonn,base
%O A354832 1,3
%A A354832 _Ya-Ping Lu_, Jun 07 2022