A214305 Fermat pseudoprimes to base 2 with two prime factors.

341, 1387, 2047, 2701, 3277, 4033, 4369, 4681, 5461, 7957, 8321, 10261, 13747, 14491, 15709, 18721, 19951, 23377, 31417, 31609, 31621, 35333, 42799, 49141, 49981, 60701, 60787, 65077, 65281, 80581, 83333, 85489, 88357, 90751, 104653, 123251, 129889, 130561

Offset: 1

Author

Marius Coman, Jul 12 2012

Keywords

• nonn

Comments

This sequence is the same as A050217 for the first 60 terms and starts to differ at the 61st.

Conjecture: For any biggest prime factor of a Poulet number p1 with two prime factors, there exists a series with infinitely many Poulet numbers p2 formed this way: p2 mod (p1 - d) = d, where d is the biggest prime factor of p1. Note: it can be seen that the Poulet numbers divisible by 73 bigger than 2701 (7957, 10585, 15841, 31609, etc.) can be written as 1314*n + 73 as well as 2628*m + 73.

Conjecture: Any Poulet number p2 divisible by d can be written as (p1 - d)*n + d, where n is a positive integer, if there exists a smaller Poulet number p1 with two prime factors divisible by d.

Note: This conjecture can't be extrapolated for Poulet numbers p1 with more than two prime factors; for instance, if p1 = 561 = 3*11*17, there indeed are bigger Poulet numbers divisible by 17 (such as 1105 and 4369) that can be written as 544*n + 17, but there also exist such numbers that can't be written this way, e.g., 2465. But the first conjecture can be extrapolated.

Conjecture: For any biggest prime factor of a Poulet number p1 exists a series with infinitely many Poulet numbers p2 formed this way: p2 mod (p1 - d) = d, where d is the biggest prime factor of p1.

For each prime p, there are only a finite number of q such that p*q is here. See A085014. Sequence A180471 lists the factors of terms of this sequence. - T. D. Noe, Sep 20 2012

Numbers n = p*q such that n divides 2^(p-1)-1 and 2^(q-1)-1, where p,q are primes; thus 2^gcd(p-1,q-1) == 1 (mod n). - Thomas Ordowski, Aug 27 2016

These are semiprimes p*q such that 2^(p+q-2) == 1 (mod p*q). Proof: 2^(p-1) == 1 (mod p) and 2^(q-1) == 1 (mod q), so 2^((p-1)*(q-1)) == 1 (mod p*q), and (p-1)*(q-1) = (p*q-1)-(p+q-2). - Amiram Eldar and Thomas Ordowski, Apr 02 2021

Examples

			Few examples for the first 4 Poulet numbers with two prime factors:
For p1 = 341 = 11*31, the following Poulet numbers p2 for which p2 mod 310 = 31 were obtained: 2821, 4371, 4681, 10261 etc.
For p1 = 1387 = 19*73, the following Poulet numbers p2 for which p2 mod 1314 = 73 were obtained: 2701, 7957, 10585, 15841 etc.
For p1 = 2047 = 23*89, the following Poulet numbers p2 for which p2 mod 1958 = 89 were obtained: 31417, 35333, 60787, 62745 etc.
For p1 = 2701 = 37*73, the following Poulet numbers p2 for which p2 mod 2628 = 73 were obtained: 7957, 10585, 15841 etc.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 3299 terms from T. D. Noe)
• Marius Coman, Conjecture that states that a Mersenne number with odd exponent is either prime either divisible by a 2-Poulet number, 2015.
• Eric Weisstein's World of Mathematics, Poulet Number.
• Index entries for sequences related to pseudoprimes

Crossrefs

Subsequence of A050217.

Cf. A001567.

Programs

Mathematica

SemiPrimeQ[n_Integer] := If[Abs[n] < 2, False, (2 == Plus @@ Transpose[FactorInteger[Abs[n]]][[2]])]; Select[Range[200000], SemiPrimeQ[#] && PowerMod[2, #-1, #] == 1 &] (* T. D. Noe, Jul 12 2012 *)

PARI

list(lim)=my(v=List());forprime(p=31,lim\11, forprime(q=11,min(p-1,lim\p), if(Mod(2,p)^(q-1)==1 && Mod(2,q)^(p-1)==1, listput(v,p*q)))); if(lim>=1093^2,listput(v,1093^2)); if(lim>=3511^2,listput(v,3511^2)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Sep 20 2012