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Dirchlet's theorem states that there are infinitely many primes of the form an+b , where n is a natural number, when gcd(a,b)=1.

Let a number which is the product of two distinct primes with the same number of
digits (for example 11 . 13 , or 103 . 997) called brilliant. (I introduced this definition on the site Alpertron - look at number theory problems. I hope the name brilliant has no other meaning in mathematics)

Now, my conjecture is that there are infinitely many brilliant numbers of the form an+b , n a natural number , if gcd(a,b)=1.

Motivation : Consider the following sequence

a(1) = 6 (the least brilliant number)

a(k+1) = the least brilliant number of the form na(k) + 1 for all k>=1.

A proof of my conjectur would ensure, that this sequence goes on for ever. If wished, I can show the sequence as for as I computed it.

The first few numbers in my sequence (after 6) and their factorizations are :

187 [11, 1; 17, 1]

2993 [41, 1; 73, 1]

395077 [461, 1; 857, 1]

6321233 [1721, 1; 3673, 1]

2022794561 [39367, 1; 51383, 1]

12136767367 [103289, 1; 117503, 1]

5510092384619 [2172089, 1; 2536771, 1]

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    $\begingroup$ What is your question? "Is my conjecture true"? $\endgroup$ – user98602 Dec 15 '13 at 16:00
  • $\begingroup$ Yes, sorry for the long introduction. $\endgroup$ – Peter Dec 15 '13 at 16:02
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    $\begingroup$ "I hope the name brillant has no other meaning in mathematics." I wasn't even aware that it has any other meaning in English. Most people who see it (in English) will think you mistyped the word brilliant. $\endgroup$ – KCd Dec 15 '13 at 16:05
  • $\begingroup$ OK fixed it, but this should not be the main problem. $\endgroup$ – Peter Dec 15 '13 at 16:10
  • $\begingroup$ By the way, Dario Alpern misunderstood me. He also defines squares of a prime to be brilliant, which I do not. $\endgroup$ – Peter Dec 15 '13 at 16:12

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