Problems & Puzzles: Puzzles

Puzzle 573.- Primes & its digital roots

Here I'm asking by runs of K consecutive primes having the same digital root. Example. For K=7, all these consecutive primes (18410243 18410261 18410279 18410333 18410351 18410369, 18410387) have the same digital root (5).

Q1. Send your a largest run.

Regarding the following recursive formula: p+dr(p) = q; q+dr(q) = r,... where p, q, r, ... are primes,

Q2. How far and why that formula can go?

Contributions came from J. K. Andersen, Farideh Firoozbakht, Torbjörn Alm, Hakan Summakoğlu, Emmanuel Vanieghem, Jan van Delden.

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J. K. Andersen wrote:

Q1. The digital root is the value modulo 9. The longest run at (2006)
http://users.cybercity.dk/~dsl522332/math/congruent-primes.htm#mod9
is 12 primes with digital root 7, starting at 4016465016163.

Q2. The sequence of digital roots is determined by the first number.
A prime sequence cannot continue after the first odd digital root, because the next number will be even. The longest possible sequence is four primes with digital roots 2, 4, 8, 7. The first case is 227, 229, 233, 241. The first with consecutive primes is 2657, 2659, 2663, 2671.

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Farideh Firoozbakht wrote:

All these 8 consecutive primes  716230591,716230609,716230699,716230717,

716230771,716230807,716230843 & 716230861  have the same digital root (7).

Torbjörn Alm wrote:

Searching up to 2000000000 I found 5 chains of size 8 besides numerous chains of size 7 etc.

This is the largest one:

Run length: 8  primitive root= 4
1324001281  1324001299  1324001317  1324001353  1324001389  1324001407  1324001443  1324001461

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Hakan Summakoğlu wrote

For K=8, all these consecutive primes

(716230591 , 716230609 , 716230699 , 716230717, 716230771 , 716230807 , 716230843 , 716230861) have the same digital root (7).

Emmanuel Vanieghem wrote:

My longest run is  10930792111, ... ,10930792399 (9 consecutive primes with digital root 7).
 

About Q2, the longest sequence involves four primes at most.

Proof : if  p  is prime and  p+dr(p)  too, then  dr(p) = 2, 4  or  8.

   If  dr(p) = 8, then  dr(p+dr(p)) = dr(16) = 7  and  q+dr(q)  cannot be prime.

   If  dr(p) = 4, then  dr(p+dr(p)) = 8, whence  r = q+dr(q)  can be prime, but  s = r+dr(r)  not.

   If  dr(p) = 2, then  dr(p+dr(p)) = 4, whence  q, r, and  s  can be prime, but  s+dr(s)  not.
 

The situation of four primes occurs first  for  p = 227 (the others being 229, 233 and 241).

Jan van Delden wrote:

Q1: The primes 716230591..716230861 give a run of 8 consecutive primes with digital root 7.

 

Q2: The maximal length is 4, with p equal to 47 mod 90. [In theory, I didn’t find one yet, length 3: [11,13,17]].

 

For a prime p, the digital root dr(p) is equal to 1,2,4,5,7 or 8 and dr(p)=p mod 9. [Except for 3 where dr(3)=3]
Since we want q=p+dr(p) prime, we should have dr(p) equal to either 2,4 or 8, so p is either 2,4 or 8 mod 9.

 

If p=2 mod 9, q=p+dr(p)=4 mod 9, r=q+dr(q)=8 mod 9 and s=r+dr(r)=7 mod 9 continuing we get 5,1,2 and we get a cyclic group of moduli.
However 2,4,8, etc. are small gaps so we could investigate the last digit of the primes involved. Modulo 10 our p should be either 1,3,7 or  9.

 

If p=2 mod 9 with the last digit equal to 7 we get a length of 4 with ending digits: [7,9,3,1] and the next term would end in 8. Starting p with a different last digit would end sooner with last digit 5. Starting with p=4 (or p=8 mod 9) would make the sequence shorter because the last digit would become even 1 (or 2) step(s) sooner.

 

Our “champion” p should therefore be 47 mod 90. Giving primes of the form [47,49,53,61] mod 90. I searched until 10^10 and didn’t find a solution.

For “large” p it will be hard to find such a solution because we need a sequence with small gaps [2,4,8].

Later Jan accepted that he had a bug in his code that skipped the solution 227, 229, 233, 241 & others.

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