A352973 -id:A352973 - cp's OEIS frontend

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, 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

Offset: 1

Ya-Ping Lu, Jun 07 2022

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, 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, 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

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

Cf. A002522, A350130, A350590, A352973.

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 = ', ')

cp's OEIS Frontend

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

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