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A214305
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Fermat pseudoprimes to base 2 with two prime factors.
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11
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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