%N Carmichael numbers whose prime factors form an arithmetic progression.
%C 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.
%C 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.
%C 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
%C 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
%C 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
%H Max Alekseyev, <a href="/A300949/b300949.txt">Table of n, a(n) for n = 1..3841</a>
%H R. G. E. Pinch, <a href="http://www.chalcedon.demon.co.uk/rgep/cartable.html">Tables relating to Carmichael numbers</a>
%e 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.
%p # using Richard Pinch data file carmichael-16
%p cfile:= "carmichael-16":
%p Res:= NULL:
%p S:= readline(cfile);
%p if S = 0 then break fi;
%p L:= map(parse, StringTools:-Split(S));
%p R:= sort(L[2..-1]);
%p d:= (R[-1]-R)/(nops(R)-1);
%p if R = [seq(i,i=R..R[-1],d)] then
%p Res:= Res, L;
%t 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 *)
%Y Cf. A002997, A087788, A262723.
%A _Robert Israel_ and _Thomas Ordowski_, Mar 16 2018