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#1 |
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Mar 2018
10228 Posts |
Is 479 the only example of prime p(k) such that reversed is 2*p(k+1)?
974=2*487 infact I did not found another example...is this due to the rarefaction of primes? |
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#2 |
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Aug 2006
3×1,993 Posts |
It's hard to guess, but I would not be surprised if there were infinitely many.
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#3 |
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Mar 2018
2×5×53 Posts |
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#4 | |
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Feb 2017
Nowhere
4,643 Posts |
Quote:
The most significant digit (MSD) d of pk would have to be even, so d = 2, 4, 6, or 8. Since d < 9, we can say that the MSD of pk+1 is either d or d+1. If d > 4, 2*pk+1 would have more digits than pk, so the only possibilities are d = 2 or d = 4. If 2*pk+1 == 2 (mod 10), then pk+1 == 1 (mod 5). If 2*pk+1 == 4 (mod 10), then pk+1 == 2 (mod 5). So, either pk has MSD d = 2, and pk+1 == 1 (mod 10), or pk has MSD d = 4, and pk+1 == 7 (mod 10). These conditions (especially the MSD being either 2 or 4) narrow things down a bit... |
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#5 | |
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Jun 2003
116748 Posts |
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Only 4...7 will work. Here too, it will have to be 4[579]...7 If 2nd digit < 5, doubling will leave the MSD as 8, which means the reverse can't be a prime, therefore 2nd digit >= 5. Also 4...7 x 2 = 9...[odd]4. Since reverse (4[odd]...9) and 4...7) are consecutive primes, they'll share the second digit (this is not hard and fast rule, but once the numbers get big enough this will hold). |
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#6 |
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Mar 2018
53010 Posts |
do you guess that below 10^14 there could be another prime of this type?
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#7 |
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Feb 2017
Nowhere
110438 Posts |
As an exercise in writing Pari scripts, I told Pari to look at numbers of the form 45...9, 47...9 and 49...9 up to the limit 50 million, which was as far as I could go and stay under the primelimit I had set. If the number formed by reversing the decimal digits was between 2 and 2.05 times the original number, I had it check whether the original number and half the reversed number were both (pseudo)prime. If they were, I told it to find the difference in prime counts between the original and half the reversed number. If this difference was less than 100, I told Pari to print the original and reversed numbers, and the difference in prime counts.
I'm not sure what (if anything) the results indicate. 45319 91354 32 458819 918854 51 4547909 9097454 48 4597919 9197954 69 479 974 1 4759 9574 2 47459 95474 28 47659 95674 17 47869 96874 48 47969 96974 44 4797959 9597974 65 47025049 94052074 67 47387749 94778374 86 47525059 95052574 68 49199 99194 37 49499 99494 23 495199 991594 55 4946989 9896494 70 4969399 9939694 28 49199389 98399194 14 |
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#8 | |
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Mar 2018
2·5·53 Posts |
Quote:
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#9 |
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Aug 2006
3·1,993 Posts |
Highly unlikely. If I had to guess I'd say a few % chance you'd find one below 10^50. To check more carefully you'd want to look at the contributions of the major primes and then calculate by digit. The most important primes here are those dividing b(b-1) with b = 10 since you're using decimal, so 2, 3, and 5.
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#10 | |
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Mar 2018
2×5×53 Posts |
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#11 |
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Aug 2006
3·1,993 Posts |
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