Problems & Puzzles: Puzzles

Puzzle 759. Palindrome primes in A.P.

Abhiram R. Devesh sent the following nice puzzle.
 

I came across an interesting section in Albert H. Beiler's Recreations in the Theory of Numbers, Second Edition, Dower Publications Inc, page 222. He talks about Palindromic primes in AP. I was able to generate palindromic primes in AP with lengths 3 and 4.

eg.
Length 3: p=727 and with d=30 : [p, p+d, p+2d] =  [727 ,757 ,787] are all palindromic primes (A244247 - Under review)

Length 4: p=10301 and with d=3030 : [p, p+d, p+2d, p+3d] =  [10301 ,13331 ,16361 ,19391 ] are all palindromic primes (A244273 - Under review)

 
Q. Does there exists the pair (p,d) that generates palindromic primes in AP of length 5 or above? or can we prove that it cannot exist?


Contributions came from J. K. Andersen, Jeff Heleen, Seiji Tomita, Jan van Delden, Shyam Sunder Gupta and Emmanuel Vantieghem

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Andersen wrote:

The maximal possible 10 palindromic primes in AP was found in 1999:
https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;672391b6.9904
(p, d) = (742950290870000078092059247, 1010100000000000)
The search team consisted of Harvey Dubner, Manfred Toplic, Tony Forbes, Jonathan Johnson, Brian Schroeder, and one Carlos Rivera!

Any 10 palindromes in AP must contain a digit which goes from 0 to 9.
An 11th palindrome is not possible in base 10.
 
The smallest start of 5 palprimes in AP is (p, d) = (1114111, 221220).
The smallest end is for (1150511, 112110).
 
In https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;ecc249e.0101
Harvey Dubner reported the first case of 6 consecutive palprimes in AP:
(1981856124216581891, 11100000000). It's a term of https://oeis.org/A229725.
 

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Jeff wrote:

For puzzle 759 there is already some great work done on this in Patrick DeGeest's fine
website at http://www.worldofnumbers.com/palprim2.htm#warut. Just scroll down to 
almost the bottom of the page and look for the heading 'Smallest Palprimes in 'Arithmetic Progression'
by Warut Roonguthai [ June 21-24, 1999 ]

Also on page http://www.worldofnumbers.com/palprim2.htm about 2/3 down the page is
work done by Harvey Dubner on palprimes in AP.
 

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Seiji wrote:

About Puzzle 759. Length 5: p=900010009, d=10202010.

{p+n*d, n=0..4} ={900010009, 910212019, 920414029, 930616039, 940818049}

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Jan wrote:

There should be solutions up to length n=10.

The difference d for APís with length>=q should be 0 mod q, q prime.

Furthermore to make sure that p+k.d is still a palindrome the digits in d should be small enough,
0<=d<=maxd, with (n-1)*maxd<=9.
 
I only investigated the situation where the digits of d are in {0,1}.
The prime p should have an odd number of digits, say 2k+1, otherwise p is
divisible by 11.

I searched for palindromic differences d, using digits {0,1} divisible by 2,3,5 and 7.
The 2 and 5 are necessary to make d end with a 0, so p+k.d is odd.

Some of my solutions using different patterns d:

n                                                 p                                      d
5                                     7190917                             101010
5                             10872427801                     1000100010
5                         1012717172101                 101010101010
5                     101270555072101             10101000101010
6                             92134043129                     1000100010
6                         1156021206511                       101010000
6                         7272836382727                 101010101010
6                     105433737334501             10101000101010
7                     118023131320811                 100010001000
7                     349263515362943                     1010100000
7                 15519112621191551             10101010101000
7                 13133112721133131           101010001010100
8                 13032102720123031           101010001010100
8             1701910118110191071       11011110111101100
9 1863101311009001131013681 11011110111101100000

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Syham wrote:

There are many pairs (p,d) that generates palindromic primes in AP of length 5 and 6. In fact there are many sets of consecutive Palindromic Primes in AP of length 5 and 6.
For example :
(1) Initial terms of smallest sets of consecutive palindromic primes in AP of length 5 are given by Sloanes OEIS A059122 as follows:

13201010231, 16831813861, 3026159516203, 303551090155303, 310917383719013, 336539111935633, 393398454893393, 705345191543507, 979863191368979, 10899190809199801 ..... and so on.
The common difference for above sets of 5 consecutive palindromic primes (palprimes) in arithmetic progression is given by Sloanes OEIS A229783 as under:
1110000, 810000, 8100000, 81000000, 81000000, 111000000, 111000000, 81000000, 81000000, 810000000, 810000000, 1110000000, 810000000, 810000000, 810000000, 810000000, 810000000, 1110000000, 1110000000, 810000000, 810000000, 810000000, 1110000000, 1110000000 ...
(2) Initial terms of smallest sets of consecutive palindromic primes in AP of length 6 are given by Sloanes OEIS A229864 as follows:
1981856124216581891 (common diff 11100000000) , 3450714222224170543, 7063314140414133607, 7833630234320363387, 100189099222990981001, 103223815141518322301, 105534664101466435501, 113509971404179905311, 121085819424918580121, 136175061101160571631, 136197647030746791631.... and so on.
 

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Emmanuel wrote:

This is what I obtained for puzzle 759 :

 
Five primes :
    (p,d) =  (7190917, 10101).
The prime p  is the smallest possible.

 
Six primes : 
    (p,d) = (19125452191, 10101000).  

 
Seven primes :
    (p,d) = (16313711111731361, 1000100010000).

 
Eight primes : 
    (p,d) = (169979300121003979961, 1010100000000).

 
For the lengths 6 to 8 there might exist solutions with smaller  p..

 
I spent a lot of time to find an AP with nine palprimes with  p < 10^23. Eilas, without succes.
However, I think there are solutions for an AP with ten primes.
More than ten is not possible : I think this is provable.

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