Problems & Puzzles: Puzzles

 Puzzle 653 R(p) is n^n. In one of the the Prime Curios! pages from Caldwell & Honaker, Jr. I found this question, by Kulsha: The reversal of 61277761 is 88. There should be infinitely many primes of this form? Q. Find the next cases to R(p)=n^n. _______ R(n) means here "reverse of n"

Contributions came from Giovanni Resta, Farideh Firoozbakht, Jeff Ellen, W. Edwin Clark, J. K. Andersen, Thomas Ritschel, Dan Dima & Emmanuel Vantieghem

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Giovanni wrote:

Tthe next prime of the form R(n^n) is R(5569^5569), a prime of 20861 digits. I obtained this result many years ago, so I do not remember the boundary of my search. [has it been published?] I did not.[Prime or PSP?] Just a probable prime (according to Mathematica implementation notes):
"Mathematica first tests for divisibility using small primes, then uses the Miller-Rabin strong pseudoprime test base 2 and base 3, and then uses a Lucas test." The number has not a "form" that allows a fast certification, and I'm not up-to-date in the game of proving primality of large numbers.
b.t.w. No other solutions up to 16800.

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Farideh wrote:

5569 is the smallest prime p such that reversal(p^p) is prime (probable prime).

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Jeff wrote:

For puzzle 653 I have checked up to n=412 with no prime found. When n=413 I get an overflow in UBasic.

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W. E. Clark wrote:

It  turns out that the next value of n such that R(n^n) is (a probable) prime is 5569 and was found by Farideh Firoozbakht in June 2010 see http://oeis.org/A178329

I checked all n up to 11100 and found only the two values n = 8 and n = 5569
such that R(n^n) is prime.

I used Maple 16's probabilistic primality test isprime to check the primality of R(5569^5569).  This is a 20861 digit number and it took 31 minutes to return true.

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J. K. Andersen wrote:

The only other solution for n up to 25000 is a 20861-digit probable prime:
R(9290666754...4901031171) = 5569^5569

I found it with PrimeForm/GW but http://oeis.org/A178329 shows
Farideh Firoozbakht already reported it in 2010.

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Thomas wrote:

I've tested all n<=6000 using a combination of a simple Pari script (for doing the reversal) and PFGW.
The next case is found for p=9290666...1031171 = R(5569^5569), having 20861 decimal digits...

Meanwhile I extended my search up to n=43200, e.g. numbers of about 200,000 decimal digits.  Since doing the reversal in Pari was too demanding I modified the procedure to generating n^n in Pari, doing a simple string reversal in Perl and trial factoring and PRP testing with PFGW.
n's which are multiples of 3 or 10 were already discarded in the "Pari" stage.
And before the reversal the Perl script checked for numbers starting with an even digit (or "5"), for which the reverse would have trivial factors 2 (or 5).
All remaining candidates were fed into PFGW.

During this extended search I found another PRP which qualifies for the next member of the sequence:
R(34351^34351) = 9290666...4681311 (having 155815 decimal digits). It's found to be Fermat and Lucas PRP, but is obviously too large for a genuine primality test. Yes, appart from the default PRP test I ran an additional "-tc" test in PFGW:

9290666...4681311 is 3-PRP! (1819.1215s+57.2249s)
9290666...4681311 is Fermat and Lucas PRP! (14745.1088s+10.0309s)

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Dan wrote:
There is no other such solution for n <= 5569
5569^5569 has already 20,861 digits and it is a lot of time consuming
to test primality for larger values with Pari code.

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Emmanuel wrote:
R(n^n)  is composite for all   such that  8 < n <  5663. I aborted the computation because, for bigger   the eventual primality of  R(n^n)  will be too hard to prove.

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Afterwards, seeing the results of the other puzzlers, I went to reread my results, I realized that I did not noticed that the solution 5569 was an output of my search (done with Mathematica).

Anyhow, there is no further solution up to 5663.

About Thomas' result : Mathematica found  R(34351^34351) prime after more than 28 hours

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