

A354832


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


0



0, 1, 2, 5, 10, 26, 37, 90, 165, 421, 933, 1957, 4005, 8101, 8282, 24666, 40869, 106405, 237477, 286810, 811098, 1286053, 3383205, 5005402, 11771813, 28549029, 38559834, 105668698, 239886426, 296984485, 833855397, 1313628250, 3461111898, 7756079194, 9423789989
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OFFSET

1,3


COMMENTS

It seems that 2^(n2) <= a(n) < 2^(n1) for n > 1.
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).
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 bitbybit complement to each other, due to the fact that f(f(x)) = x mod 2^L as pointed out by Kevin Ryde in Discussion.
d m1 or m2 (bin) m2 or m1 (bin) m1 (decimal)
   
1 0 (m1/m2) 1 (m2/m1) a(1) = 0; a(2) = 1
2 10 (m1) 01 (m2) a(3) = 2
3 010 (m2) 101 (m1) a(4) = 5
4 1010 (m1) 0101 (m2) a(5) = 10
5 11010 (m1) 00101 (m2) a(6) = 26
6 011010 (m2) 100101 (m1) a(7) = 37
7 1011010 (m1) 0100101 (m2) a(8) = 90
8 01011010 (m2) 10100101 (m1) a(9) = 165
9 001011010 (m2) 110100101 (m1) a(10)= 421
10 0001011010 (m2) 1110100101 (m1) a(11)= 933


LINKS



EXAMPLE

26 is a term because iterating the map on 26 gives, in binary format, 11010 > 1010100101 > 1101111111001011010, which ends with 11010.


PROG

(Python)
R = []
for i in range(0, 34):
t = 2**i; L = []
while t not in L: L.append(t); t = (t*t + 1) % 2**(i+1)
{R.append(j) for j in {L[1], L[2]} if j not in R}
R.sort(); print(*R, sep = ', ')


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



