Problems & Puzzles: Puzzles

Puzzle 935. pi+SOD(pi)=p(i+n)

Paolo Lava sent the following nice puzzle:

For each natural integer n, find the first p(i) such that
p(i)+SOD(p(i)) = p(i+n)

He sent the first seven terms:

n p(i) p(i+n)
1 11 13
2 19 29
3 37 47
4 59 73
5 97 113
6 55787 55819
7 26699 26731

Q1. Is this sequence finite or infinite?
Q2. Extend this Table so far as possible.

 

Contributions came from Giovanni Resta, Paolo Lava, Paul Cleary, Pierandrea Formusa.

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Giovanni wrote on Dec 16, 2018:

I found these entries

8 663959
9 6974477
10 86966771
11 1095975857
12 4649574977
13 349685387573
14 988839709939
15 86396869388567

In principle it seems that the series can be infinite, but finding 
further terms becomes difficult. On the average the gap between a prime 
number p = p(i) and q=p(i+n) is about n*log(p). On the other hand, the 
sum of digits of p is at most about log10(p)*9 which is less than 
4*log(p).So, as n increases (and p increases as well) becomes more and 
more difficult to find a sequence of n+1 primes whose gap is less than 
4*log(p).

For example, for the 15-term p = 86396869388567, we have sod(p)=92 while 
around that magnitude of primes usually primes 15 primes apart are 
separated by an averages gap greater than log(p)*15 > 480.

The gaps following 86396869388567 are 2, 4, 6, 2, 6, 4, 6, 6, 14, 16, 
14, 4, 2, 4, 2 (their sum is indeed 92) so these primes form a very 
tight cluster. To find further terms on the table requires to find 
longer tight cluster of large primes.

***

Paolo wrote on 17 Dec 2018:

Up to now I reached n=11
 

1 11 13

2 19 29

3 37 47

4 59 73

5 97 113

6 55787 55819

7 26699 26731

8 663959 663997

9 6974477 6974521

10 86966771 86966821

11 1095975857 1095975913

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Paul wrote on Dec 17, 2018:

Q2..  using the same format.

 

8 , 663959 , 663997

9 , 6974477 , 6974521

10 , 86966771 , 86966821

11 , 1095975857 , 1095975913

12 , 4649574977 , 4649575039

 

Calculated as far as the prime 35419661603 before stopping, with no n = 13 found.

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Pierandrea wrote on Dec 20, 2018:

Q2:

 
n=8 p(i)=663959 p(i+n)=663997

 
PARI code to get this result:
forprime (p=1, 1000000, if(p+sumdigits(p)==nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(p+1)+1)+1)+1)+1)+1)+1)+1),print(p);break));

 
-----------------------------------------------------
 

 
n=9 p(i)=6974477 p(i+n)=6974521

 
PARI code to get this result:
forprime (p=1, 8000000, if(p+sumdigits(p)==nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(p+1)+1)+1)+1)+1)+1)+1)+1)+1),print(p);break));

 
-----------------------------------------------------

 
n=10 p(i)=86966771 p(i+n)=86966821

 
PARI code to get this result:
forprime (p=1, 100000000, if(p+sumdigits(p)==nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(p+1)+1)+1)+1)+1)+1)+1)+1)+1)+1),print(p);break));
print("fine")
 

 
-----------------------------------------------------

 
n=11 p(i)=1095975857 p(i+n)=1095975913

 
PARI code to get this result:
forprime (p=1, 1100000000, if(p+sumdigits(p)==nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(nextprime(p+1)+1)+1)+1)+1)+1)+1)+1)+1)+1)+1),print(p);break));

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