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