The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
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
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
Sequence in context: A002094 A212709 A264867 * A115725 A305577 A196630
KEYWORD
nonn,base
AUTHOR
Ya-Ping Lu, Jun 07 2022
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified May 28 19:24 EDT 2024. Contains 372919 sequences. (Running on oeis4.)