Problems & Puzzles: Puzzles

Puzzle 481. E.E-p=q JM Bergot sent the following nice puzzle: One notices that subtracting 17 from nine consecutive even numbers squared--6 through 22--leaves a prime difference.  Q. Can you find a sequence of more even numbers squared with some one prime subtracted to give all differences primes?

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Contributions came from Jacques Tramu, Maximilian Hasler, J. K. Andersen, Torbjörn Alm, Gennady Gusev, Jan van Delden & Farideh Firoozbakht.

I'm posting only their respective highest score:

Length	prime	Starting even	Who
13	227	16	Gennady
18	837077	916	Farideh
21	2004917	2536	Hasler, Torbjörn Alm
23	208305767	14552	Andersen, Tramu

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Gennady added:

I considered the sum (instead of differences) of squares of even numbers and prime, I got the next interesting result: 19 primes for prime 163 + squares of i: even i from 2 to 38.

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Jan van Delden wrote:

For each n, I computed the set of primes p such that (2n)^2-p is prime. I computed the intersection of these sets for consecutive n until this intersection became empty and took one step back.  The result for each n is the maximal obtainable length L for this n and the p-value(s) that define it.  If I found a solution (2n,L,p) I discarded solutions of the form (2(n+k),L-k,p). I would like to stress that my procedure therefore doesn’t find the first opportunity to find a certain suboptimal length L, but I didn’t find these other solutions interesting. However this procedure ensures no bigger length then the maximum already found is possible (in the searched range).

2n	L	P	p/(2n)^2
4	3	5	0.31
6	9	17	0.47
10	7	83	0.83
14	10	167	0.85
16	13	227	0.89
32	5	983	0.96
34	4	587,1097	0.51
44	8	677	0.35
48	11	2273	0.99
58	6	3323	0.99
74	12	5297	0.97
916	18	837077	1.00
1016	14	1001447	0.97
1088	16	1027067	0.87
1842	19	3063023	0.90
2672	15	7116023	1.00
4478	17	19884707	0.99

 

 

 

 

 

 

 

 

 

 

At first (until  2n=1500) I used  every 2<p<(2n)^2-2 in my search. Then I realized that if an lower bound for the length L is known  the first few p<=L can be skipped (or p<L if searching for an non­decreasing length), because: (2(n+p))^2 = (2n)^2 mod p. I didn’t use this fact however, because I wanted to find the first occurrences of other L as well.

Then I saw that the primes actually get quite close to (2n)^2.  In fact linear regression shows a remarkable good fit (R^2=0.999) for the line: p=0.909*(2n)^2. I therefore decided to  use p>3n^2, with the risk of missing out on a few. I searched until 2n=7000 and found no other lengths L. The last 3 entries in the table could therefore be not the first occurrences.

For the values of L which increase with increasing n (grayed),  it seems like L=4+ln(4n^2) is a good fit. The fit for the other values in the table above:  L=ln(4n^2). I didn’t find a good heuristic for this yet.

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