As you have already probably known, an Emirp is a prime whose reversal give a different prime i.e: 37 is an Emirp because 37 is prime and its reversal 73 is also a prime, 79 is also an Emirp. Now I have this 100 digit Emirp with a mind-boggling property (!!!): 31399 71973 78663 47113 91448 65157 72694 85891 75941 91229 38744 59187 76569 25789 74797 49143 19422 88961 13739 39731, write down that 100 digit Emirp into a 10-by-10 square, and you will find that all rows, columns, and the 2 main diagonals are all Emirps (!!!) and distinct (!!). I call such numbers a Magical Emirp. My question is: Are there only finitely many magical Emirps?
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6According to Sheldon Cooper, 73 is the best number because "73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3... and in binary 73 is a palindrome, 1001001, which backwards is 1001001." – Gregory Grant May 05 '15 at 19:45
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How did you find this example of a magical Emirp? – Gregory Grant May 05 '15 at 19:46
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1@GregoryGrant, yes I know that :) – Andrew G May 05 '15 at 19:47
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I found it by accident :D – Andrew G May 05 '15 at 19:47
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It is an open problem whether there are infinitely many Erimps. – wythagoras May 05 '15 at 19:53
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1I'm afraid this example was already know, see here: http://www.opm.mat.br/static/archive/opm_2013_1.pdf – Gregory Grant May 05 '15 at 19:53
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@GregoryGrant, I believe I found that about 3 years ago. – Andrew G May 05 '15 at 20:00
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1@AndrewG Here it is mentioned over five years ago: http://scienceblogs.com/principles/2011/01/20/dorky-poll-favorite-prime-numb/ – Gregory Grant May 05 '15 at 20:01
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@GregoryGrant I also have the larger one, but unfortunately the main number is just a prime and not an Emirp – Andrew G May 05 '15 at 20:12
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In any case if you came across that yourself that's quite impressive, even if somebody else had done it first. I don't really see how to search for such numbers easily. – Gregory Grant May 05 '15 at 20:13
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@wythagoras And the existence of a spmire is still debated amongst experts. – Thomas Poguntke May 05 '15 at 20:22
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@Engloutie you mean emirp ? – Andrew G May 05 '15 at 20:25
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@AndrewG No, it's just a joke related to the comment I responded to. Nevermind! – Thomas Poguntke May 05 '15 at 20:26
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A very heuristic argument says that there should be an infinite number. If we assume the primes are randomly distributed with $p$ having $\frac 1{\ln p}$ chance of being prime, we can ask what is the chance there is a Magical Emirp with $n^2$ decimal digits. You need two numbers of $n^2$ digits and $4n+2$ numbers of $n$ digits to be prime. We will take all the probabilities for $n$ digits to be $\frac 1{\ln 10^{n-1/2}}=\frac 1{(n-1/2)\ln 10}$ For $n$ large, we can ignore the $1/2$ and say the chance is $\frac 1{n \ln 10}$ This says the chance a random number of $n^2$ digits is a Magical Emirp is about $\frac 1{n^{4n+2}(n^2)^210^{2n+2}}$ As there are about $10^{n^2}$ numbers with $n^2$ digits we would expect $\frac {10^{n^2}}{n^{4n+2}(n^2)^210^{2n+2}}$ with $n^2$ digits. This grows without bound, so we would expect more and more of them each square number of digits. They still become rarer and rarer, so finding them is rather difficult. Plugging in $n=10$, there should be $10^{32}$ hundred digit ones. That is a lot, but a small fraction of the hundred digit numbers.
Ross Millikan
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