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The sequence:

$3, 11, 13, 17, 19, 29, 37, 41, 53, 59, 61, 67, 83, 97, 101, 107, 113, 131, 137, 139, 149, 163, 173, 179, 181, 193, 197, 211, 227, 257, 269, 281, 293, 313, 317, 347, 349, 353, 373, 379, 389, 401, 409, 419, 421, 443, 449, 461, 467, 491, 509, 521, 523, 541, 547, 557, 563, 569, 577, 587, 593, 613, 617, 619, 653, 659, 661, 677, 701, 709, 757, 761, 769, 773, 787, 797, 809, 821, 827, 829, 853, 857, 859, 877, 883, 907, 929$...

$n = 2^x + 2^y + 1$

$x,y$ in this sequence.

Is there case that $n$ is prime with this sequence?

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    $\begingroup$ What is that sequence? $\endgroup$ – Gerry Myerson May 25 '13 at 5:54
  • $\begingroup$ @Ethan, I think the question is, if $x$ and $y$ are in the given sequence, can $2^x+2^y+1$ be prime? $\endgroup$ – Gerry Myerson May 25 '13 at 5:54
  • $\begingroup$ @Gerry Myerson,thanks. $\endgroup$ – miket May 25 '13 at 5:55
  • $\begingroup$ @Ethan,is 0 in this sequence? $\endgroup$ – miket May 25 '13 at 5:59
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    $\begingroup$ So, miket, why didn't you tell us what you already knew about the sequence? Don't you think it would help you get an answer, if you had let us in on the secret of A226014? $\endgroup$ – Gerry Myerson May 25 '13 at 6:09
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Such pairs do exist, and they are precisely the columns of the following matrix:

$$\left( \begin{array}{ccccccccccccccc} 3 & 3 & 11 & 37 & 53 & 59 & 179 & 179 & 197 & 227 & 353 & 421 & 449 & 467 & 853 \\ 37 & 317 & 17 & 67 & 83 & 797 & 293 & 509 & 227 & 509 & 587 & 787 & 659 & 653 & 907 \\ \end{array} \right)$$

The minimal prime obtained in this way is, by inspection, $2^{11}+2^{17}+1=133121$.

Edit: I forgot to check the diagonal! That adds the single pair $(3,3)$, corresponding to $$2^3+2^3+1=17.$$ This brings us to a total of $16$ solutions.

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  • $\begingroup$ Very interesting.Thanks! $\endgroup$ – miket May 25 '13 at 6:19
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The integer $17$ is prime,

The integer $3$ appears in your sequence,

Also

$2^3+2^3+1=17$

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    $\begingroup$ I suspect OP wants $x\ne y$, but it's hard to get any information from that source. $\endgroup$ – Gerry Myerson May 25 '13 at 6:10
  • $\begingroup$ Thanks! $x$ can be equal to $y$. $\endgroup$ – miket May 25 '13 at 6:19

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