

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
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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 (pd)p(p+d), where pd, p and p+d are prime, and pd1  d(2d+3), p1  d^2, and p+d1  d(2d3). Thus for each d there are only finitely many possible p that make this work. Note that 6d, 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 q1), so p^2 + pq == 2 (mod q1), so p^2 + p == 2 (mod q1). In particular, p^2 + p  2 >= q1, 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

Max Alekseyev, Table of n, a(n) for n = 1..3841
R. G. E. Pinch, Tables relating to Carmichael numbers


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 carmichael16
cfile:= "carmichael16":
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

Cf. A002997, A087788, A262723.
Sequence in context: A306478 A051388 A033181 * A198775 A154729 A083737
Adjacent sequences: A300946 A300947 A300948 * A300950 A300951 A300952


KEYWORD

nonn


AUTHOR

Robert Israel and Thomas Ordowski, Mar 16 2018


STATUS

approved



