Problems & Puzzles: Puzzles

 Puzzle 920. An enigma related to A291582 Vicente Felipe Izquierdo sent the following nice puzzle-enigma related to the sequence A291582. He wrote: For each prime P calculate: - the sum of the non-prime numbers greater than P, until the next prime number. (SM) - the sum of the non-prime numbers less than P, until the prev prime number. (Sm) Let's call D the difference SM - Sm.   Example: P = 7, SM = 8 + 9 + 10 = 27 Sm = 6 D = 21   Let us form the ordered set of the frequency of appearance of the D differences of all the primes, in decreasing order of the observed frequency.Unexpectedly the resulting set is {30, 132, 306, 552, 870, 1260, 1722, ...} exactly the same set of values than the values for the sequence A291582!!!, discarding the very low frequencies (less than 4) and all the frequencies for negative values of D.   But the sequence A291582 was obtained last year by Craig Knecht while studying a very different issue: "Maximum number of 6 sphinx tile shapes in a sphinx tiled hexagon of order n"  Q. Can you send an explanation for this coincidence, or at least make some interesting comments?

Contributions came from Walter Trump, Jan van Delden and Emmanuel Vantieghem

***

Walter wrote:

The difference D can be calculated with triangular numbers T(n) = 1 + 2 + 3 + ... + n = (n²+n)/2 .
If (p – a) is the prime below p and (p + b) the prime above p then the sums are:
SM = (p + 1) + (p + 2) + (p + 3) + ... + (p + (b – 1))
Sm = (p – 1) + (p – 2) + (p – 3) + ... + (p – (a – 1))
D = SM – Sm = p(b – a) + 1 + 2 + 3 + ... + (b – 1)  + 1 + 2 + 3 + ... + (a – 1)
D = p(b – a) + T(b – 1) + T(a – 1)

In general D depends on the value of p. But if a = b then D is independent of the value of p.
These D’s occur very often, because there are no limits for the associated p.
Such primes with a = b are called balanced.
D = p(b – b) + T(b – 1) + T(b – 1) = 2T(b – 1)

b has to be a multiple of 6, if p > 5. Otherwise one of the three primes would be a multiple of 3 and we get a contradiction.

Smaller distances occur more frequently than larger ones.

Therefore we get the sequence of distances:  b(n) = 6, 12, 18, ..., 6n

And the sequence of differences:  D(n) = 30, 132, 306, ..., 2T(6n-1)
BUT 2T(6n-1) = (6n-1)²+(6n-1) = 36n² - 12n + 1 + 6n – 1 = 36n² - 6n = 6n(6n-1)    and this is the formula of A291582

***

Jan wrote:

Consider the three primes q=p-2k, p and r=p+2l. (p>3).

D = sum(x,x=p+1..r-1)- sum(x,x=q+1..p-1) can be written as:

D =  2(k^2+l^2)-(k+l)+2p(l-k) = f(p,k,l)

This expression is independent of p if k=l:

D = 2k(2k-1)

If k=1 mod 3 the numbers [q,p,r] mod 6 are [p-2,p,p+2] mod 6 = [p+4,p,p+2] mod 6.

If k=2 mod 3 the numbers [q,p,r] mod 6 are [p-4,p,p+4] mod 6 = [p+2,p,p+4] mod 6.

In both cases the numbers [q,p,r] form a complete odd residue set mod 6,

hence p,q or r is divisible by 3 and hence p,q,r are not consecutive primes.

So we are left with k=0 mod 3, say k=3m. But then D = 6m(6m-1) which are the numbers in A291582.

Since the probability on primegap 6m decreases with increasing m, it makes sense that larger D of this type occur less often.
It must be possible to show that given m the number of consecutive prime solutions of the form [p-6m,p,p+6m] are infinite, at least a heuristic could be given, something of the form C[m] p/ln(p)^3 would be my guess [Like in the derivation of the density of the twin primes].

The tricky bit in the question is that it might happen that D=f(p1,k1,l1)=f(p2,k2,l2) with k1<>l1 and k2<>l2.
This type of equation is much harder to fulfil. The number of solutions might be finite or infinite (with a very low density compared to the previously discussed case). But I see no direct way to decide which is which.

All I “know” is that if p2>p1  the part 2p1(l1-k1) versus 2p2(l2-k2) will be dominant for larger p1,p2, so |l2-k2| will  (probably) be smaller than |l1-k1|. Which means that p1 must have rather large primegaps l1,k1 in order to have a large number of solutions for such a D=f(p1,k1,l1).

***

Emmanuel wrote:

First of all : A291582(n)  is nothing else but  6n(6n-1).
This leads to a demystification of the puzzle : the sfinx's presence has nothing essential to do with our problem !
It gave only a geometrical interpretation of the number  6n(6n-1) ...

Vicente's observation is restricted to the fact that the positive values D with frequency >= 4 are all of the form  6n(6n-1).
Is that so ?
Well, let  ( p, q, r )  be a trio of consecutive primes.  Set  k = q-p  and  h = r-q.
Then : D = q+1 + q+2 +...+ q+h-1 - (q-1 + q-2 + ...+ q-k+1) = (h-k)q + (h(h-1)+k(k-1)/2.

If  h = k (which can only happen when  k  is of the form  6n), then D = 6n(6n-1) = A291582(n).
I think this situation is conjectured to happen infinitely many times.  This would imply Vicente's observation.

But, when k < h, the value  D  is reached only for a finite number of  q (the biggest being (D-7)/2 ).
But I'm not sure that there are at most four possible  q  that lead to the value  D (provided  D  is not of the form  6n(6n-1)).
Actually, I could not find a situation with  k < h  that gave more than three times a given value of  D.
The smallest such  D  with frequency 3 happens when  D = 13288305 with
(p1,q1,r1) = (738217, 738223, 738247)
(p2,q2,r2) = (1328797, 1328807, 1328827)
(p3,q3,r3) = (6644147, 6644149, 6644153).

Nevertheless, I'm sure such  D  with frequency > 3 exist !

***

 Records   |  Conjectures  |  Problems  |  Puzzles