|
|
A176997
|
|
Integers k such that 2^(k-1) == 1 (mod k).
|
|
13
|
|
|
1, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Old definition was: Odd integers n such that 2^(n-1) == 4^(n-1) == 8^(n-1) == ... == k^(n-1) (mod n), where k = A062383(n). Dividing 2^(n-1) == 4^(n-1) (mod n) by 2^(n-1), we get 1 == 2^(n-1) (mod n), implying the current definition. - Max Alekseyev, Sep 22 2016
These numbers were called "fermatians" by Shanks (1962). - Amiram Eldar, Apr 21 2024
|
|
REFERENCES
|
Daniel Shanks, Solved and Unsolved Problems in Number Theory, Spartan Books, Washington D.C., 1962.
|
|
LINKS
|
|
|
EXAMPLE
|
5 is in the sequence because 2^(5-1) == 4^(5-1) == 8^(5-1) == 1 (mod 5).
|
|
MATHEMATICA
|
m = 1; Join[Select[Range[m], Divisible[2^(# - 1) - m, #] &],
Select[Range[m + 1, 10^3], PowerMod[2, # - 1, #] == m &]] (* Robert Price, Oct 12 2018 *)
|
|
PROG
|
(PARI) isok(n) = Mod(2, n)^(n-1) == 1; \\ Michel Marcus, Sep 23 2016
(Python)
from itertools import count, islice
def A176997_gen(startvalue=1): # generator of terms >= startvalue
if startvalue <= 1:
yield 1
k = 1<<(s:=max(startvalue, 1))-1
for n in count(s):
if k % n == 1:
yield n
k <<= 1
|
|
CROSSREFS
|
Odd integers n such that 2^n == 2^k (mod n): this sequence (k=1), A173572 (k=2), A276967 (k=3), A033984 (k=4), A276968 (k=5), A215610 (k=6), A276969 (k=7), A215611 (k=8), A276970 (k=9), A215612 (k=10), A276971 (k=11), A215613 (k=12).
|
|
KEYWORD
|
nonn,changed
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|