Problems & Puzzles: Puzzles

Puzzle 907. 10|p(n)^2 + p(n+1)^2

Vic Bold sent the following nice puzzle.

The sum of the squares of two consecutive primes is divisible by 10 for the ten successive pairs from 61,67 to 101,103.

 
Pn (pn)^2+(pn+1)^2
61 8210
67 9530
71 10370
73 11570
79 13130
83 14810
89 17330
97 19610
101 20810
103  
 

By mi side I (CR) could get a longer list composed of 34 primes P such that P and his nextprime(P)=Q are such that (P^2+Q^2)@10=0:

35951863 35951869 35951887 35951891 35951897 35951899 35951917 35951939 35951953 35951959 35951963 35951969 35951977 35952011 35952017 35952019 35952023 35952029 35952067 35952089 35952097 35952109 35952113 35952131  35952143 35952149 35952167 35952209 35952223 35952239 35952247 35952269 35952307 35952311


Q. Send your longer list?


Contributions came from Jan van Delden and Jeff Heleen

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Jan wrote:

The longest list I found consists of 41 expressions of the type p^2+q^ = 0 mod 10, using 42 consecutive primes, starting at: 3233679837239.


I searched until 1000*2^32. I used two sieves. The first one on [1,2^32] and created a second one [A*2^32,(A+1)*2^32] using the first. In theory I could apply this technique until 2^64. In practice it is still too slow.  I think I must learn how to incorporate PrimeSieve into my programs in c. .

 

Actually I think that the lengths of the lists found are rather large. For each of the final 4 digits of a prime p, {1,3,7,9}, only two make sure that for a second prime q we have p^2+q^2=0 mod 10. So a first thought could be to suppose that the probability that two consecutive primes p,q fulfill p^2+q^2=0 mod 10 is best estimated by 1/2. But that means that the probability for a sequence of say 35 primes (under the assumption that the events are independent) is about (1/2)^34. Which is very small compared to the size of the first prime found by Carlos especially because there are fewer primes than natural numbers “of that size”. Or the other way around: apparently these probabilities are not independent or incorrect.

 

A rather small test suggests that the frequencies of each of the 4 last digits are about equal.

The conditional probability of q given p depends on p and q, so the probabilities are not independent.


Moreover: sum(P(q|1))>sum(P(q|7)) > sum(P(q|3))>sum(P(q|9)). The first 3 being larger than 1/2.

 

Maybe I should study random walks in graphs.

***

Jeff wrote:

For puzzle 907 I found a 36-expressions length solution, 37 total primes:
 
45149845529      45149845547      45149845579      45149845633      45149845651      45149845667      45149845679      45149845717      45149845721      45149845723      45149845739      45149845813      45149845831      45149845843      45149845849      45149845883      45149845889      45149845907      45149845921      45149845927     
45149845991      45149845997      45149846029      45149846087      45149846089      45149846113      45149846141      45149846173      45149846249      45149846267      45149846299      45149846303      45149846311      45149846317      45149846341      45149846393      45149846411

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