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A329223 Poulet numbers (Fermat pseudoprimes to base 2) that are congruent to either 3 or 27 (mod 80) and each prime factor is congruent to 3 mod 80. 0

%I #17 Dec 01 2019 16:12:24

%S 51962615262396907,330468624532072027,2255490055253468347,

%T 18436227497407654507

%N Poulet numbers (Fermat pseudoprimes to base 2) that are congruent to either 3 or 27 (mod 80) and each prime factor is congruent to 3 mod 80.

%C If a term of this sequence is also a Carmichael number (A002997) and a Lucas-Carmichael number (A006972), then it would be a counterexample to Agrawal's conjecture, as Hendrick Lenstra and Carl Pomerance showed.

%C 330468624532072027 is the only Carmichael number below 2^64 that is a term of this sequence. However, it is not a Lucas-Carmichael number.

%C The sequence also includes: 68435117188079800987, 164853581396047908970027, 522925572082528736632187, 1820034970687975620484907, 4263739170243679206753787, 4360728281510798266333387, 28541906071781213329174507, 33833150661360980271172507, 84444874320158644422192427, 175352076630428496579381067, 270136290676063386556053067, 615437738523352001584590187, 3408560627000081376639770587, 11260257876970792445537580187.

%C No term with 5 prime factors (which would be congruent to 3 mod 80) is known to the author.

%C Are all terms also strong pseudoprimes to base 2 (A001262)?

%H H. W. Lenstra, and Carl Pomerance, <a href="http://www.aimath.org/WWN/primesinp/primesinp.pdf">Remarks on Agrawal's conjecture</a>, American Institute of Mathematics (2003), pp. 30-32.

%H Tomáš Váňa, <a href="http://web.ics.upjs.sk/svoc2009/prace/7/Vana.pdf">Agrawal's Conjecture and Carmichael Numbers</a>, student scientific conference, pp. 13-22.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Agrawal%27s_conjecture">Agrawal's conjecture</a>

%e 51962615262396907 is a term because it is a Fermat pseudoprime to base 2 and it is congruent to 27 (mod 80) and all of its prime factors (643, 154723, 522306163) are congruent to 3 mod 80.

%o (PARI) isok(n) = ((n%80==3) || (n%80==27)) && (Mod(2, n)^(n-1) == 1) || return(0); my(f=factor(n)[,1]); (#f > 1) && (#select(p->p%80==3, f) == #f);

%Y Cf. A001567.

%K nonn,more

%O 1,1

%A _Daniel Suteu_, Nov 08 2019

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Last modified May 1 05:44 EDT 2024. Contains 372148 sequences. (Running on oeis4.)