Problems & Puzzles: Puzzles

Puzzle 571.- P(a)=P(b) mod(P(c)

JM Bergot sent the following puzzle:

Rare must be solutions for P(a)=P(b) mod[P(c)] for all a,b,c being primes and P(c)>P(b).

As usual P(1)=2, P(2)=3, and so on...

Q. Can you send a few solutions, if there are any?

Contributions came from Jan van Delden, Torbjörn Alm, Jeff Heleen, Emmanuel Vantieghem, Jean-Christope Colin.

***

Jan wrote:

As I suspected the solutions are not “that rare” at all.

If b<c<a (if a<c then there are no solutions and a=b is trivial) then I found:

Upperbound a,  #primes of form P(prime), # Solutions
10,          5,        0
10^2,    26,      25
10^3,  169,    490
10^4,1230,11027

For instance 67=5 mod 31 and 83=3 mod 5, or P(19)=P(3) mod P(11) and P(23)=P(2) mod P(3).

If a<n then the number of combinations to check for our primes P(a), P(b), P(c) is about pi(n)^3/6=(n/ln(n))^3/6.
The number of solutions (based on these few data) is about n*2^(ln(n)/ln(10)-4).

***

Torbjörn Alm wrote:

I run up to a=200000, and there are 966768 solutions.
The higher a, the more solutions.

Lowest solution:

Solution: P(19) = P(3) mod [P(11)]
67 = 5 mod[31]

...

Solution: #966768
P(199999) = P(379) mod [P(105227)]
2750131 = 2609 mod[1373761]

***

Jeff Heleen wrote:

For puzzle 571 solutions are apparently not so rare.
With a, b, c, P(a), P(b) and P(c) all prime, these are
the first instances for each P(c) with all qualifying
P(b)'s:

a    b    c    P(a)    P(b)    P(c)
23    2    3    83    3    5
...
59    13    17    277    41    59

***

Emmanuel Vantieghem wrote:

I found many solutions.  In a few seconds,  my PC gave solutions for  j = 1 to 100 and  b = P(j), c = P(j+m)  and for  m = 1, 2, 3, and a few others.
Since there are too many solutions, I searched for  a, b, c  such that  P(a), P(b), P(c)  use only four digits.  Here are my best results :

288223823 = p15648307   = 211 (mod 2221) ; digits in use : 1, 2, 3, 8

611619191 = p31899451   = 211 (mod 2221) ; digits in use : 1, 2, 6, 9

777732223 = p40058819   = 211 (mod 2221) ; digits in use : 1, 2, 3, 7
99998999  = p5761403    = 191 (mod 211) ; digits in use : 1, 2, 8, 9
196161991 = p10878311   = 1171 (mod 17117) ; digits in use : 1, 6, 7, 9

I could not find an example with only three digits in use.

***

Jean-Christope Colin wrote:

I have counted up to P(a) = 10009 enclosed, 651 prime triplets (a,b,c) such as P(a)=P(b) mod[P(c)] and P(c)>P(b);
up to P(a)= 5000123, 4401 such prime triplets (a,b,c)

I notice too much P(a) is great, much they are a little of P(a) which possess good (b,c) but those P(a) have a lot of goob (b,c) : for exemple,for P(a) between 58000 and 59000, They are only 8 P(a) with good triplets but each of them have at minimun 9 (b,c) which make good (a,b,c) : namely

P(a)            a            a', a=P(a')        nbre(b,c)
58057        5879       774               9
58067        5881       775             11
58217        5897       776             14
58309        5903       777             16
58511        5923       778             13
58567        5927       779             10
58699        5939      780             14
58907        5953       781               9

and when I see the a' for the smallest P(a) I join you in my precedent e-mail (a'= 8, 9, 11, 12 ,13, 14, 16, 17, 18, 19, 20, 22...) , then It seems that almost all natural number a'>7 generate several prime triplets (a=P(a'), b, c) such as  P(a)=P(b) mod[P(c)] and P(c)>P(b). Thus, those triplets are quite far from rare

***

 Records   |  Conjectures  |  Problems  |  Puzzles