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I was messing around in Excel at the end of work today and made a table where the $(i,j)$ entry $a_{i,j}$, for $j \geq i$, is 1 exactly when $i$ and $j$ are coprime (see snapshot of a portion of the table below):

enter image description here

My questions are:

  • Is there an explanation for the highlighted rectangular pattern that keeps popping up (albeit in different orientations)?
  • What about the numbers where $a_{*j}$ is $0$ for "most" values of $j$. They seem to be sandwiched between twin primes or at least right next to a prime. Is there any significance there?
  • Another question that popped out of this, are there infinitely many solutions for $\phi(n) = \phi(n+1)$? I also calculated $\phi(n)$ directly at the top of my table and noticed that the pattern for solutions seemed somewhat erratic. Is there a theorem related to this at all?

These questions are rather open ended and I apologize. Just trying to see if there is any merit to my curiosity today.

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    $\begingroup$ Note that numbers $n$ such that $\phi(n)=\phi(n+1)$ are tabulated, with references to the literature, at oeis.org/A001274 $\endgroup$ – Gerry Myerson Jul 12 '13 at 11:15
  • $\begingroup$ Thanks for the reference. Exactly what I was looking for. $\endgroup$ – Patrick Jul 13 '13 at 1:12
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Twin primes are always either side of a multiple of 6 (except for the first twin primes, 3 and 5). Why is that?
I think the numbers with most blank space below them are multiples of 6. Why would most numbers have a factor in common with any multiple of 6?
I don't know about your last question. None the less, playing around with maths is (a) fun and (b) how a lot of maths gets discovered.

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  • $\begingroup$ You are right, given three consecutive numbers $p_n, m, p_{n+1}$ (with $p_{n} > 5$) say, then one of them must be a multiple of $3$. Clearly that means $m$ is. Similarly one of $p_{n}$ and $m$ are divisible by $2$, thus $m = 6k$ for some integer $k$. I want to accept your answer for giving that hint but I still want to somehow keep the answer open for discussion about the other two questions. Is there a way to do that? $\endgroup$ – Patrick Jul 12 '13 at 11:13
  • $\begingroup$ I don't know exactly how these work. Thanks for the upvote anyway :) $\endgroup$ – Empy2 Jul 12 '13 at 15:28

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