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Let $k$ be any positive integer then how to prove that the sequence $$Q_k=10^{\lceil k \log_{10}(n) \rceil }+n^{k-1}$$ Contains infinitely many primes?

It seems like because if you look at some small values of $k$ we can think of infinude of primes $$Q_2=\{11,12,13,104,105,106,107,108,109,1010,1011,1012,1013 \cdots \}$$ $$Q_3=\{11,14,109,116,1025,10251036,1049,1064,1081,10100 ,111331 \cdots\}$$

Is this true? If false then how many terms are prime in the general $Q_n$ sequence?

Where this came from: Consider the sequence of of numbers formed by concatenating (keeping side by side) $n$ and $n^k$ . For $k=2$. The numbers would be $11 ,24,39,416,525,636 ,749,864 \cdots$ Let they form a sequence $s(n,k)$ . Thus $$s(n,2)= \{ 11 ,24,39,416,525,636 ,749,864 \cdots \}$$
Similarly $$s(n,3)=\{11,28,327,464,5125 \cdots \}$$ We can easily see that $$s(n,k)=n \cdot 10^{\lceil k \log(n) \rceil }+n^k$$ The sequence in question $Q_k$ is just $$Q_k= \frac{s(n,k)}{n}$$

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  • $\begingroup$ this seems interesting, how this question came up? Is it a subproblem of something else? or just experimenting with sequences of numbers? $\endgroup$ – clark May 21 '14 at 4:53
  • $\begingroup$ @clark just experimenting with sequences of numbers $\endgroup$ – Shivam Patel May 21 '14 at 4:54
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    $\begingroup$ Could be, but there are, in essence, no sequences that depend on a single variable ( your $n$ ) where it can be proved that infinitely many primes occur. Plenty of functions of two variables, any integer quadratic form that cannot be factored for example, as $x^2 + y^2$ for example. $\endgroup$ – Will Jagy May 21 '14 at 5:08
  • $\begingroup$ Is there any significant to writing Qn in this form? It seems for straightforward to simplify to Qn = $n^k$ + n $\endgroup$ – beanshadow May 21 '14 at 5:08
  • $\begingroup$ @beanshadow there is a meaning ,so let me edit the question $\endgroup$ – Shivam Patel May 21 '14 at 5:18
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The exponent $\lceil k\log_{10}n\rceil$ is constant for large stretches of $n$, so the question can be rephrased:

For each integer $k>0$ are there infinitely many integers $e$ with at least one prime $10^e+n^{k-1}$ with $10^{e/k}\le n<10^{(e+1)/k}$?

For $k=1$ this is

Are there infinitely many integers $e$ with $10^e+1$ prime?

which is probably false (these are generalized Fermat primes with base 10, so $e$ must be a power of 2 and the expected number of primes is just 2, 11 and 101). For $k=2$ this is

Are there infinitely many integers $e$ with at least one prime $10^e+n$ with $10^{e/2}\le n<10^{(e+1)/2}$?

which is true under stronger-than-currently-known results on prime gaps.

For larger $k$ this is surely unknown, since it requires finding primes in quadratic or higher polynomials.

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  • $\begingroup$ thankyou for throwing light on the question. Can you tell me which is the closest result we have got to this? $\endgroup$ – Shivam Patel May 22 '14 at 3:22
  • $\begingroup$ @ShivamPatel: The conjecture on generalized Fermat numbers is folklore. the best current result on prime gaps is Baker-Harman-Pintz 2001, "The difference between consecutive primes, II" (which is 'nearly' good enough, though even RH does not suffice). Relevant to the $k>2$ case is the Bateman–Horn–Stemmler conjecture. $\endgroup$ – Charles May 22 '14 at 11:07
  • $\begingroup$ I am not able to understand the -The Bateman–Horn–Stemmler conjecture " analogue here? $\endgroup$ – Shivam Patel May 23 '14 at 6:10
  • $\begingroup$ @ShivamPatel: Read the Bateman, Horn, and Stemmler paper (or the shorter Bateman and Horn paper), or at least the Wikipedia article, and compare to what I wrote in my answer. $\endgroup$ – Charles May 23 '14 at 13:35
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    $\begingroup$ You need only $f_1$, which is your $Q_k$. $\endgroup$ – Charles May 23 '14 at 14:27

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