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Hello I have a basic number theory question.

  • I want to find a list of primes of the form a, a + d, a + 2d, ... , a + 5d
    • So a sequence of at least 6 or greater if I want to select a = 101 then what would I choose as my d to get a sequence of primes (they don't have to be consecutive).
    • I am quite lost. Maybe there isn't an answer and it's not possible? I don't know how to find d in this case.
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    If you allow general a and d then it is definitely possible. It is famous theorem which says that you can find arbitrary long sequence of primes in arithmetic progression. I don't know what happens if a=101. Maybe you can write a computer code to just check for different values of d? – 2017-01-28
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    Here is the example: https://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem#Numerical_work $a$ is a bit greater than 101 though – 2017-01-28
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    When you asked this question a little while ago you were given the sequence $\{7,37,67,97,127,157\}$ . What was wrong with that? – 2017-01-28
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    it doesn't contain $101$. – 2017-01-28
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    Yes, but why not to start with $a=199$? This is easy. You said in your deleted question [here](http://math.stackexchange.com/questions/2118315/finding-a-sequence-of-primes) that any $3$-digit $a$ would be fine. – 2017-01-28
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    I'm interested in finding or seeing if its possible to find a sequence starting with a = 101 (again if its possible).. If that question can't be answered can I atleast know how to find sequences of this form? – 2017-01-28
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    In your deleted question you wanted $101+107n$. By Dirichlet, this *has* infinitely many primes (you said no, because $101+107=208$ is not prime, but think a bit. Infinitely many primes does *not* mean, that consecutive sequence members are prime). Why did you delete your question instead of editing it? – 2017-01-28
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    There isn't a simple way of generating such sequences, at least, not one I know of. the long, known examples are produced by computer searches. The Green-Tao Theorem, which guarantees existence, does not provide a way to construct them. – 2017-01-28
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    Maybe you are interested in [this question](http://math.stackexchange.com/q/2135409/11206) – 2017-02-08

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A well-known sequence $a_n:=a+nd$ producing at least $6$ primes with $3$-digit $a$ (this was your requirement here) is $a_n=199+210n$, i.e., $$ 199, 409, 619, 829,1039,1249,1459,1669,1879,2089, $$ so that the first consecutive $10$ sequence members are prime numbers. With $a=101$ you can search for such a $d$ by computer. I do not know of a conceptual way to find such a $d$.

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The list of first 10 such $d$ that $101+nd$ is prime for $n=0,..,5$: $[3990,5610,98220,101640,108780,122790,176550,312210,331590,333840]$

Generated by a simple Haskell program:

isPrime :: Integer -> Bool
isPrime k = null [ x | x <- [2..floor(sqrt(fromIntegral k))], k `mod` x  == 0]

good :: Integer -> Integer -> Bool
good a d = all id [isPrime (a + n * d) | n <- [0..5]]

main = print $ take 10 $ filter (good 101) [1..]
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    The OP wanted "at least $6$ or greater" primes in a row, so this is fine, but $a=199$ gives $10$ instead of $6$. Can you find a $d$ for $a=101$ with $10$ primes in a row? – 2017-01-28
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    I don't see any sense in that :) Of course if in the above program you replace [0..5] by [0..9] and "take 10" by "head" it'll eventually output such $d$ if it exists (which I can't prove), but on my machine this already took ten minutes with no result. It is possible to optimize considerably, but this is already a programming question, not math. – 2017-01-28
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    I was just interested in how large $d$ would be. Apparently it is not a straightforward calculation (by a simple Haskell program). – 2017-01-28
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    For length 7 the smallest $d$ is 1683990. For 10 it is probably beyond of brute-force possibilities. – 2017-01-28
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    You are right, of course, to ask why one would want to know this. – 2017-01-28
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    @DietrichBurde Maybe you are interested in [this question](http://math.stackexchange.com/q/2135409/11206) – 2017-02-08