|
|
A213382
|
|
Numbers n such that n^n mod (n + 2) = n.
|
|
9
|
|
|
1, 4, 7, 13, 16, 19, 31, 37, 49, 55, 61, 67, 85, 91, 109, 121, 127, 139, 157, 175, 181, 193, 196, 199, 211, 217, 235, 247, 265, 289, 301, 307, 313, 319, 325, 337, 379, 391, 397, 409, 415, 445, 451, 469, 487, 499, 517, 535, 541, 571, 577, 589, 595, 631, 667, 679
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Equivalently, numbers n such that (n^n+2)/(n+2) is an integer. Derek Orr, May 23 2014
It was conjectured that A176003 is a subsequence.
Terms that do not appear in A176003: 16, 61, 193, 196, 313, 397, 691, 729, 769 ...
The conjecture is correct: verify the cases 1 and 3, then it suffices to show that (3p-2)^(3p-2) = 3p-2 mod 3 and mod p. Mod 3 the congruence is 1^(3p-2) = 1, and mod p the congruence is (-2)^(3p-2) = -2 which is true by Fermat's little theorem. - Charles R Greathouse IV, Sep 12 2012
a(62) = 729 is the first number not congruent to 1 mod 3. - Derek Orr, May 23 2014
|
|
LINKS
|
|
|
EXAMPLE
|
A213381(n) = 7^7 mod 9 = 7, so 7 is in the sequence.
|
|
MATHEMATICA
|
Select[Range[700], PowerMod[#, #, #+2]==#&] (* Harvey P. Dale, Oct 03 2015 *)
|
|
PROG
|
(Python)
for n in range(999):
x = n**n % (n+2)
if x==n:
print(n, end=", ")
|
|
CROSSREFS
|
Cf. A213381 : a(n) = n^n mod (n+2).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|