|
| |
|
|
A300949
|
|
Carmichael numbers whose prime factors form an arithmetic progression.
|
|
1
|
|
|
|
1729, 2465, 29341, 294409, 1152271, 1857241, 6189121, 19384289, 56052361, 64377991, 118901521, 172947529, 216821881, 228842209, 625482001, 775368901, 1213619761, 1299963601, 2301745249, 4562359201, 8346731851, 9293756581, 9624742921, 9701285761, 11346205609, 13079177569, 13946829751, 14386156093
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
All terms < 10^18 have three prime factors. There are terms with more, e.g., 97888020200929464481 = 34471 * 91921 * 149371 * 206821, 147681255946700149193521 = 214831 * 572881 * 930931 * 1288981, and 2393527068197020059464161 = 431047 * 1149457 * 1867867 * 2586277.
A term with 3 prime factors is of the form (p-d)p(p+d), where p-d, p and p+d are prime, and p-d-1 | d(2d+3), p-1 | d^2, and p+d-1 | d(2d-3). Thus for each d there are only finitely many possible p that make this work. Note that 6|d, see A262723.
Conjecture: if n is a Carmichael number and lpf(n)gpf(n)(lpf(n)+gpf(n))/2 = n, then (lpf(n)+gpf(n))/2 is prime; and thus n has exactly three prime factors. Such numbers n form a proper subsequence of this sequence, also subsequence of A262723. - Charles R Greathouse IV and Thomas Ordowski, Mar 17 2018. Edited by Max Alekseyev, Mar 17 2018
Proof of the above conjecture: Say n = paq with 2 < p < q being primes and a = (p+q)/2, with (a,p!)=1. If n is a Carmichael number, then pa == 1 (mod q-1), so p^2 + pq == 2 (mod q-1), so p^2 + p == 2 (mod q-1). In particular, p^2 + p - 2 >= q-1, which implies that (p+1)^2 > q. Say a has k prime factors, so that a >= (p+2)^k. But a < q, so q > (p+2)^k. Thus, (p+1)^2 > q > (p+2)^k. This implies k=1. - Carl Pomerance (in a letter to the second author), Mar 18 2018
Note: this does not exclude the existence of the Carmichael numbers m = pq(p+q)/2 with more than three prime factors, where p and q are prime. - Thomas Ordowski, Mar 19 2018
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
29341 = 13*37*61 is a Carmichael number, and [13, 37, 61] is an arithmetic progression of length 3 and with common difference of 37 - 13 = 61 - 37 = 24. We have 37 = (13 + 61)/2.
|
|
|
MAPLE
|
# using Richard Pinch data file carmichael-16
cfile:= "carmichael-16":
Res:= NULL:
do
S:= readline(cfile);
if S = 0 then break fi;
L:= map(parse, StringTools:-Split(S));
R:= sort(L[2..-1]);
d:= (R[-1]-R[1])/(nops(R)-1);
if R = [seq(i, i=R[1]..R[-1], d)] then
Res:= Res, L[1];
fi
od:
fclose(cfile):
Res;
|
|
|
MATHEMATICA
|
Select[Cases[Range[1, 10^7, 2], _?(And[Mod[#, CarmichaelLambda@ #] == 1, ! PrimeQ[#]] &)], Length@ Union@ Differences@ FactorInteger[#][[All, 1]] == 1 &] (* Michael De Vlieger, Mar 17 2018, after Artur Jasinski at A002997 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,changed
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
