Problems & Puzzles: Puzzles

Puzzle 558.- Goldbach decomposition & consecutive primes

JM Bergot sent the following nice puzzle:

One notices that in the Goldbach decomposition of 24, (5,19), (7,17), and (11,13) that there are six consecutive primes.

Q.  Is that the most possible?

Contributions came from Seiji Tomita, Tony D Noe, Fred Schalekamp, J. K. Andersen, Emmanuel Vantieghem, Jeff Heleen & Farideh Firoozbakht.

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Seiji wrote:

Case of 8 consecutive primes.
Smallest solution is
17+43=19+41
    =23+37
    =29+31


Case of 10 consecutive primes.

13+47=17+43
    =19+41
    =23+37
    =29+31

Case of 12 consecutive primes.

137+193=139+191
      =149+181
      =151+179
      =157+173
      =163+167

Case of 14 consecutive primes.

8021749+8021873=8021753+8021869
              =8021759+8021863
              =8021771+8021851
              =8021789+8021833
              =8021791+8021831
              =8021801+8021821


Case of 16 consecutive primes.

1071065111+1071065269=1071065123+1071065257
                    =1071065129+1071065251
                    =1071065137+1071065243
                    =1071065141+1071065239
                    =1071065153+1071065227
                    =1071065167+1071065213
                    =1071065179+1071065201

Case of 18 consecutive primes.

1613902553+1613902747=1613902561+1613902739
                    =1613902567+1613902733
                    =1613902573+1613902727
                    =1613902601+1613902699
                    =1613902621+1613902679
                    =1613902627+1613902673
                    =1613902643+1613902657
                    =1613902649+1613902651

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Tony wrote:

The number n=210 is famous for being the largest number such that every
prime p in the interval [n/2,n-2] produces a Goldbach partition n = p +
(n-p).  That is, 210-p is prime for every one of the 19 primes p in the
interval [105,208].  This is proved in the paper "An upper bound in
Goldbach's problem" by Deshouillers et al. (see
http://www.dms.umontreal.ca/~andrew/PDF/GoldbachII.pdf).  210 actually has
22 consecutive primes in the Goldbach partition because 97, 101, and 103
are primes in the (n-p) part of the partition.  Of course, 210 does not
produce the most consecutive primes.  I found 34 for 90090.

The Goldbach partitions of 210 are
(11, 199), (13, 197), (17, 193), (19, 191), (29, 181), (31, 179), (37,
173), (43, 167), (47, 163), (53, 157), (59, 151), (61, 149), (71, 139),
(73, 137), (79, 131), (83, 127), (97, 113), (101, 109), (103, 107)

Among the 38 primes that occur here, there are 22 consecutive primes 97 to
199. The puzzle did not say that the consecutive primes must form all
Goldbach partitions. If it is required that the consecutive primes occur in
all Goldbach partitions, then only n=8, 12, 18, 24, 30 have this property.

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Fred wrote:

I've found the following smallest solutions: 

# Consecutive primes   Sum    First and last Decomposition 
         6              24    (5,19)...(11,13) 
         8              60    (17,43)...(29,31) 
        10              60    (13,47)...(29,31) 
        12             330    (137,193)...(163,167) 
        14        16043622    (8021749,8021873)...(8021801,8021821) 

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Andersen wrote:

The Goldbach partitions of 30 also have six consecutive primes. It seems very
unlikely there is a number with more. A quick computation shows there are none
below 3*10^6. Work by Tomás Oliveira e Silva can extend the limit to 16*10^17.

The Goldbach conjecture verification project by Silva at
http://www.ieeta.pt/~tos/goldbach.html has shown that all even numbers n up to
16*10^17 have a Goldbach partition n = p+q with p <= 9341.
The primes are known to thin out regularly below 16*10^17 so there can be no
case where  all primes from p to n-p are symmetric around n/2 for p <= 9341
and 3*10^6 < n < 16*10^17. This means there is no number between 3*10^6 and
16*10^17 with consecutive primes in the Goldbach partitions.

It can also be shown by considering residue classes.
A computation shows that for every prime r < 10000, there is a prime p <
1500000 in every residue class modulo r (including 0 modulo r which has r
itself). This means there can be no n > 3*10^6 with a prime r < 10000 such
that n-p is prime for all primes p from r to n/2. At least one value of p will
make n-p divisible by r.
Combined with Silva's work (where 9341 < 10000), this shows there are no
solutions to the puzzle between 3*10^6 and 16*10^17.

If we only want a subset of the Goldbach partitions to have consecutive primes
then we want consecutive primes which are symmetric around n/2. The cases with
more consecutive symmetric primes than all smaller numbers are given by
A055381, or possibly A055380 in some cases. A055381 is "Smallest number such
that there are n symmetric (in the gap sense) primes on each side":
4, 9, 12, 30, 30, 165, 8021811, 1071065190, 1613902650, 1797595815015.

A055381(3) = 12 gives the puzzle example with 3 Goldbach partitions of 24.
A055381(5) = 30 corresponds to 5 Goldbach partitions of 60 with the
consecutive symmetric primes 30+/-k, for k = 1, 7, 11, 13, 17.
60 has a 6th Goldbach partition 7+53 with the non-consecutive prime 7 (11 is
not in the partitions).

A055381(6) corresponds to 165+/-k, for k = 2, 8, 14, 16, 26, 28
A055381(7): 8021811+/-k, for k = 10, 20, 22, 40, 52, 58, 62
A055381(8): 1071065190+/-k, for k = 11, 23, 37, 49, 53, 61, 67, 79
A055381(9): 1613902650+/-k, for k = 1, 7, 23, 29, 49, 77, 83, 89, 97
A055381(10):1797595815015+/-k, for k = 2, 4, 38, 64, 74, 76, 94, 118, 142, 152

If n = p+p is allowed as one of the Goldbach partitions then we may have to
consider A055380, "Smallest prime such that there are n symmetric (in the
gap sense) primes on each side":
5, 18731, 683783, 98303927, 60335249959, 1169769749219

These numbers are larger than A055381 and there are theoretical reasons to
think it will continue to be so. I guess the only contribution by A055381 to
the record setters is A055381(1) = 5 which gives 10 = 5+5 = 3+7 as the
smallest with three consecutive primes (3, 5, 7).

***

Emmanuel wrote:

There are many sequences of length  2k  of consecutive primes  { p(m+1), p(m+2), …, p(m+2k) }  such that all numbers of the form p(m+j)+p(m+2k+1-j)  are the same for  j = 1, 2, …,k.  Here are some examples :

 k = 8 : {17,19,23,29,31,37,41,43}; sum = s = 60
: {149,151,157,163,167,173,179,181} ; s = 330
: {677,683,691,701,709,719,727,733} ; s = 1410

                . . . . . . . . .

 k = 10 : {60,13,17,19,23,29,31,37,41,43,47} ; s= 60 (! remarkable !)
: {139, 149, 151, 157, 163, 167, 173, 179, 181, 191} ; s = 330

                . . . . . . . . .

k = 12 : {137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193}; s = 330 (also remarkable !)
: {55787, 55793, 55799, 55807, 55813, 55817, 55819, 55823, 55829, 55837, 55843, 55849}; s = 111636

                . . . . . . . . .

 k = 14 : {8021749, 8021753, 8021759, 8021771, 8021789, 8021791, 8021801, 8021821, 8021831, 8021833, 8021851, 8021869, 8021869, 8021873} ; s = 16043622

 k = 16 : {1071065111, 1071065123, 1071065129, 1071065137, 1071065141, 1071065153, 1071065167, 1071065179, 1071065201, 1071065213, 1071065227, 1071065239, 1071065243, 1071065251, 1071065257, 1071065269} ; s = 2142130380

 k = 18 : {1613902553, 1613902561, 1613902567, 1613902573, 1613902601, 1613902621, 1613902627, 1613902643, 1613902649, 1613902651, 1613902657, 1613902673, 1613902679, 1613902699, 1613902727, 1613902733, 1613902739, 1613902747} ; s = 3227805300, 

For  k = 20, the first prime must be greater than 28x10^9.

There is a heuristic argument for the conjecture that, for every (even)  k, there is an  m  with the desired property.  Indeed, the above results suggest that the probability for the set

                   {  p(m+j)+p(m+2k+1-j)  |  j = 1, 2, ..., k  }

to consist of only one element is not zero (but undoubtedly very small when  k  is big).  Hence, by Murphy’s Law ...

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Jeff wrote:

For puzzle 558: More than 6 consecutive primes
are possible.

# of
consec.      prime                                    Goldbach
primes       range                                    number
8              17...43                                    60
10             13...47                                   60
12             137...193                                330
14             8021749...8021873                  16043622
16             1071065111...1071065269        2142130380
18             1613902553...1613902747        3227805300

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Farideh wrote:

30 = 7 + 23 = 11 + 19 = 13 + 17 ,  6 consecutive primes.

 

60 = 13  + 47 = 17 + 43 = 19 + 41 = 23 + 37 = 29 + 31;  13, 17, 19, 23, 29, 31, 37, 41, 43 & 47

are ten  consecutive primes.

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Carlos Rivera wrote:

Seiji, Emmanuel & Jeff reached to the largest set (18) of consecutive primes solutions for this puzzle. The non-discovered by them set of 20 primes, is hidden in the sequence  A055381 reported by J. K. Andersen above. In A055381 we can read that Donovan Johnson, in Mar 09 2008, reported the key  number behind the 20 primes set for this puzzle: 1797595815015. Twice this number is the Goldbach number we need to generate the 20 primes set.

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J. K. Andersen wrote:

Tony's interpretation allows Goldbach partitions where only one of the primes
is part of the consecutive primes. In that case, large solutions can be found
with large primorials.
If q and p#-q are primes then (q, p#-q) is a Goldbach partition of p#.
For any p, all primes between p#-p and p#-p^2 are of form p#-q for a prime q
(there may be additional such primes below p#-p^2).
This means p# is likely to have many Goldbach partitions with consecutive
primes a little below p#.
997#-q gives 1095 consecutive primes for primes q = 1447, 1453, 1867, 1931,
2657, ..., 1099409, 1100807, 1101253, 1102427, 1102747.
The next prime is 997#-1102837 where 1102837 = 1009*1093.
Marcel Martins' Primo has proved all the primes.
It is possible there are other Goldbach partitions of 997# with more than 1095
consecutive primes.

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