Problems & Puzzles: Collection 20th

Coll.20th-001. A Puzzle-sequence

On March 10, 2018, G. L. Honaker posed the following puzzle sequence in the middle of a mailing list:

Puzzle Sequence: 7, 19, 79, 491, 2423, ... . Find the next term.

BTW, this puzzle sequence was already published in his site here.

The puzzle was solved by Fausto Morales eleven days later, on March 21, 2018: The next term is 8147. Here is his explanation:

Having observed that, for n=1 to 5, the nth term has always been the lowest prime such that the 1st n consecutive gaps between consecutive primes following it are precisely each of the first n odd primes plus 1, my proposed solution is 8147, because the next 6 gaps above it are as follows:

8161 - 8147 = 14 = 13 + 1
8167 - 8161 = 6 = 5 + 1
8171 - 8167 = 4 = 3 + 1
8179 - 8171 = 8 = 7 + 1
8191 - 8179 = 12 = 11 + 1
8209 - 8191 = 18 = 17 + 1”

G. L. Honaker has accepted the Mr. Morales solution.

Q. So, the only remaining question is this: find the next term if it exists or prove that it does not exist.

Contributions came from Emmanuel Vantieghem, Jan van Delden and Giovanni Resta.

***

Emmanuel wrote:

Suppose there is a next term  p.

Then  p  is the first member of a set  P = {p, p1, p2, p3, p4, p5, p6, p7}
of eigth consecutive primes with successive gaps {g1, g2, g3, g4, g5, g6, g7}
which are a permutation of  {4, 6, 8, 12, 14, 18, 20}.

I examined all possible permutations and these was my observation :

whatever  p is mod 3 (1 or 2), there is allways some  i <= 7 such that  p+g1+ ... +gi (= pi)  is divisible by 3.

So, there is no next term !

...

Maybe I have been too quick (which is never OK).
It could be possible that the 'next' term is a prime  p  such that the ten successive primes
have gaps a permutation of  {4, 6, 8, 12, 14, 18, 20, 24, 30, 32}.
I'll give it a try, but I'm afraid i will run out of time ...

***

Jan wrote:

There is no next term.

The consecutive gaps g[i] must be in the set {4,6,8,12,14,18,20}.

Consider a permutation of this set: [g[i]] and a starting prime p[1]. The primes are now defined by p[i+1]-p[i]=g[i].
So one could write p[i+1]=p[1]+sum(g[j],j=1..i). Or in words, each prime is now the cumulative sum of the gaps in the permutation+ a starting prime. This list of cumulative sums must be admissible mod 3, mod 5 and mod 7.

If p[1]<=347 then the set of consecutive gaps contains the number 2.

If one wants to find a solution, we must have a residue set mod p (p=(2),3,5,7) such that:

• The number of residues is smaller than p [We must be able to choose a p[1] such that p[i+1]<>0 mod p for all i] (I)
• If there are p-1 residues each unequal to 0 mod p, than p[1] must be 0 mod p. But p=3,5,7 is not possible.
So if 0 is not present, the number of free residues must be smaller than p-1 in order to be able to choose a starting prime p[1].(II)

The permutation [4,6,8,12,14,18,20] defines gaps relative to p[1]: [4,10,18,30,44,62,82].

Modulo 3 it reads: [1,1,0,0,2,2,1], so the residue set mod 3={0,1,2}, which is a full residue set. (I)

The permutation  [8,6,12,18,14,4,20] defines gaps relative to p[1]: [8,14,26,44,58,62,82].
Modulo 3 it reads: [2,2,2,2,1,2,1], so the residue set mod 3={1,2}, p must be 0 mod 3, but p=3 is not a solution.

A check reveals that there are no permutations that comply to rule (I) and rule (II) modulo 3.
It is possible to give a proof, without computing all permuations, which uses the fact that 82 is always present, which is 1 mod 3.

***

Giovanni wrote:

The next term does not exist.

To have a next term, the gaps between the primes must be
equal to a permutation of 1 + the first 7 odd primes

1 + {3, 5, 7, 11, 13, 17, 19} = {4, 6, 8, 12, 14, 18, 20}.

If the first prime (the solution) is p, the following
primes are determined by the gaps, like p+4, p+4+6, p+4+6+8,
etc. if the gaps are sorted as above.

These gaps, modulo 3, are equal to {1, 0, 2, 0, 2, 0, 2}.

It is easy to see that the partial sums of these gaps,
whatever is their permutation,
always contain at least a 1 and at least a 2 modulo 3.

For the case above, the sums are equal to
{1, 1+0, 1+0+2, 1+0+2+0, ...} =
{1,1,3,3,5,5,7} == {1,1,0,0,2,2}.

If the gaps are in another order, say {0,2,2,1,0,0,2},
the resulting sums are {0,2,4,5,5,5,7} = {0,2,1,2,2,2,1}.

Now, the first prime of the sequence (the solution we
are looking for) is greater than 3 and thus it is
either equal to 1 or equal to 2 mod 3.

But this means that building the other 7 primes starting from
p and using the gaps always result in a number which is 1+2==0
or 2+1 ==0 mod 3, thus a multiple of 3 that cannot be a prime.

Hence there cannot be a next term.

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