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
***
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