proposed

approved

editing

proposed

It seems that 2^(n-2) <= a(n) < 2^(n-1) for n > 1.

It seems that 2^(n-2) <= a(n) < 2^(n-1) for n > 1. All terms are part of a cycle under x -> f(x) mod 2^L. For example, 1 = f(0), 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 bit-by-bit complement to each other.

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.

10 26 is a term because iterating the map on 10 26 gives, in binary format, 1010 11010 -> 1100101 1010100101 -> 10011111011010, 1101111111001011010, which ends with 101011010.

proposed

editing

editing

proposed

It seems that 2^(n-2) <= a(n) < 2^(n-1) for n > 1. All terms are part of a cycle under x -> f(x) mod 2^L. For example, 1 = f(0), 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 bit-by-bit complement to each other.

10 is a term because iterating the map on 10 ('1010' gives, in binary format) gives: 10 , 1010 -> 101 1100101 -> 10202, which is '10011111011010' in binary format ending , which ends with '1010'.

proposed

editing

editing

proposed

from sympy import isprime; R = []

R = []

for i in range(0, 10034):

proposed

editing

editing

proposed

allocated for Ya-Ping Lu

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

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

1,3

It seems that 2^(n-2) <= a(n) < 2^(n-1) for n > 1. 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.

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

10 is a term because iterating the map on 10 ('1010' in binary format) gives: 10 -> 101 -> 10202, which is '10011111011010' in binary format ending with '1010'.

(Python)

from sympy import isprime; R = []

for i in range(0, 100):

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 = ', ')

Cf. A002522, A350130, A350590, A352973.

allocated

nonn,base

Ya-Ping Lu, Jun 07 2022

approved

editing

allocated for Ya-Ping Lu

allocated

approved