A300949 Carmichael numbers whose prime factors form an arithmetic progression.

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

Offset: 1

Author

Robert Israel and Thomas Ordowski, Mar 16 2018

Keywords

• nonn

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

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (calculated using data from Claude Goutier; terms 1..3841 from Max Alekseyev)
• Claude Goutier, Compressed text file carm10e22.gz containing all the Carmichael numbers up to 10^22.
• R. G. E. Pinch, Tables relating to Carmichael numbers.

Crossrefs

Cf. A002997, A087788, A262723.

Programs

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 *)