Problems & Puzzles: Puzzles

Puzzle 696 1801 Fred Schneider sent the following nice puzzle related to gaps over consecutive primes: 1801 is an interesting prime number.  If you check the next 8 prime numbers after it and take the differences, each (even) difference from 2 to 16 occurs exactly once.    P       Diff 1801 1811, 10 1823, 12 1831, 8 1847, 16 1861, 14 1867, 6 1871, 4 1873, 2 Q1. Can you find a number that starts a longer "covering set" of gaps like this? Q2.  Is there some maximum length that can be spoken to?

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Contributions came from Vicente Felipe Izquierdo, Abhiram R. Devesh, Jud McCranie, W. Edwin Clark, J. K. Andersen, Fred Schneider, Giovanni Resta, Jan van Delden, Emmanuel Vantieghem & Hakan Summakoglu.

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Vicente wrote:

La cadena más larga que he encontrado de momento es de 17 primos.
Te indico a continuación: el orden del primer primo, la longitud de la cadena de primos, el grupo de primos, la diferencia.
 
{7,4,{17,19,23,29},2}
{23,5,{83,89,97,101,103},2}
{68,6,{337,347,349,353,359,367},2}
{120,7,{659,661,673,677,683,691,701},2}
{279,8,{1801,1811,1823,1831,1847,1861,1867,1871},2}
{279,9,{1801,1811,1823,1831,1847,1861,1867,1871,1873},2}
{2685,10,{24113,24121,24133,24137,24151,24169,24179,24181,24197,24203},2}
{2685,11,{24113,24121,24133,24137,24151,24169,24179,24181,24197,24203,24223},2}
{6999,12,{70639,70657,70663,70667,70687,70709,70717,70729,70753,70769,70783,70793},2}
{70897,13,{894637,894643,894667,894689,894709,894713,894721,894731,894749,894763,894779,
894791,894793},2}
{658641,14,{9903149,9903169,9903191,9903193,9903211,9903227,9903241,9903247,9903251,
9903259,9903269,9903281,9903307,9903331},2}
{303953,15,{4317193,4317211,4317239,4317251,4317253,4317263,4317283,4317289,4317311,
4317319,4317323,4317347,4317361,4317377,4317403},2}
{727268,16,{11012317,11012341,11012369,11012371,11012389,11012401,11012411,11012437,
11012453,11012459,11012479,11012501,11012509,11012539,11012543,11012557},2}
{12922691,17,{235397039,235397053,235397081,235397101,235397131,235397153,235397161,235397171,
235397177,235397189,235397191,235397207,235397233,235397237,235397269,235397287,235397311},2}
 

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Abhiram wrote:

The smallest primes for each of the covering set are
 

Length 10 : 24113 [24113 , 24121 , 24133 , 24137 , 24151 , 24169 , 24179 , 24181 , 24197 , 24203 , 24223  ]
 

Length 11 : 503227 [503227 , 503231 , 503233 , 503249 , 503267 , 503287 , 503297 , 503303 , 503317 , 503339 , 503351 , 503359  ]
 

Length 12 : 894637 [894637 , 894643 , 894667 , 894689 , 894709 , 894713 , 894721 , 894731 , 894749 , 894763 , 894779 , 894791 , 894793 ] 
 

Length 13 : 9903149 [*149 , *169 ,*191 , *193 , *211 , *227 ,*241 , *247 , *251 , *259 , *269 ,*281 , *307 ,*331 ] 
 

Length 14 : 4317193 [*193 , *211 , *239 , *251 , *253 ,*263 , *283 , *289 , *311 , *319 , *323 , *347 , *361 ,*377 , *403 ]
 

Length 15 : 11012317 [*317, *341, *369, *371, *389, *401, *411, *437, *453, *459, *479, *501, *509, *539, *543, *557]

 

I could not find any prime numbers having a covering set length greater than 15 for n < 10^8
 

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Jud wrote:

I checked up to 1 trillion and the best one I found is 6955703129. The next 21 baps have each of 2, 4, 6, ... 42 once and only once. I submitted the solutions I found to OEIS: https://oeis.org/A227335. If more terms are found, they can be added.

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Clark wrote:

As to Q1: I found the prime 235397039 for which the successive differences of the next
16 primes give all of the even integers from 2 to 32. This is the longest such gap sequence
I found among the first 18 million primes.

 235397039 
 235397053 = 235397039 + 14 
 235397081 = 235397053 + 28 
 235397101 = 235397081 + 20 
 235397131 = 235397101 + 30 
 235397153 = 235397131 + 22 
 235397161 = 235397153 + 8 
 235397171 = 235397161 + 10 
 235397177 = 235397171 + 6 
 235397189 = 235397177 + 12 
 235397191 = 235397189 + 2 
 235397207 = 235397191 + 16 
 235397233 = 235397207 + 26 
 235397237 = 235397233 + 4 
 235397269 = 235397237 + 32 
 235397287 = 235397269 + 18 
 235397311 = 235397287 + 24 


I cannot say anything about Q2. 
 

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Jens wrote:

Q2. In puzzle 151 I found that 484511389338941 starts consecutive gaps of
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26.

k-1 increasing gaps like this corresponds to k consecutive primes
x^2+x+p for a starting prime p and x = 0 to k-1. If the k-tuple conjecture
is true then there are occurrences for arbitrarily large k.

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Fred wrote:

Q1. I implemented (in C++) an optimized sieve of Eratosthenes to compute these numbers.  

Implementation: I maintained an array of numbers based on the prime gap.  

If for instance, I found a gap of 6 and that was already set in my array.  I would remove all the numbers

from the array that were less than or equal to the number already in the six slot.  

For the primes themselves, I maintained a circular buffer.

One note: For performance reasons, I had the code clear the array if it found a gap of more than 80.

I think it's really unlikely that a solution of length 40 could be found in this range

given one could not be found more than 25.)
 

The n-th term in the sequence below has n gaps.  I sieved through about 30 trillion

but could find any new terms more than the 25th term around 10.6 trillion.
 

1 3

2 5

3 17

4 83

5 337

6 659

7 2,621

8 1,801

9 24,113

10 24,113

11 503,227

12 894,637

13 9,903,149

14 4,317,193

15 11,012,317

16 235,397,039

17 8,516,934,067

18 15,703,837,349

19 6,955,703,129

20 52,600,764,283

21 6,955,703,129

22 2,700,022,855,487

23 4,659,851,923,081

24 6,485,648,067,251

25 10,643,798,564,411
 

Q2. Luis Rodriguez alerted me to his Conjecture 69 (Dickson) and Puzzle 151.

If there's no limit on consecutive gap length, then there must be

no limit on this more general puzzle (of which the consecutive gap is one solution).

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Giovanni wrote:

My results are the following:

8 1801
9 24113
10 24113
11 503227
12 894637
13 9903149
14 4317193
15 11012317
16 235397039
17 8516934067
18 14990671117
19 6955703129
20 52600764283
21 6955703129
22 2700022855487
23 4659851923081
24 6485648067251
25 10643798564411

Where the 25 gaps following 10643798564411 are :
50, 28, 44, 46, 8, 4, 26, 12, 42, 30, 40, 38, 24, 18, 36, 6, 16, 20, 22, 2, 34, 32, 10, 14, 48.

If there are 26 such gaps, they occur after a prime larger than 10^14.

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Jan wrote:

Q1: I searched untill 6*10^9+6, to honour puzzle 696 and my electricity bill.

 

I found:

 

Startprime	Length
3	1
7	2
17	3
83	4
337	5
659	6
1801	8
2621	7
24113	10
120647	9
503227	11
894637	12
4317193	14
11012317	15
66140441	13
235397039	16

 

Q2:

 

Although the relative frequency of the gaps D(x;g) follow a simple pattern: P(gapsize=2k)=a.g^k, [with g depending on x, and a normalising constant to make the total probability 1],

finding a formula for the approximation of the relative frequencies of the lengths of a covering set is hard. Below is a graph where the relative frequency of the lengths i [Ri] are depicted versus the total number of different covering sets. [For x<=10^8].

 

We see that R2<R3 with R2 slowly increasing. We also have R4<{R5,R6} where R5 is about equal to R6.

 

If we look for a larger covering set, say length n, the probability that a gap is not in our target set [2..2n] increases as well as the probability that we hit upon a value in the target set twice (too soon).
There will also be more permutations of the values [2..2n] that are simply not allowed. Finding large solutions is probably not impossible but will take some (serious) computing time.

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Emmanuel wrote:

In the next small table, I give the first prime  p  of a set of  k+1  primes with differences equal to some permutation of the first  k even integers

       p                           k

   24113                       10

  503227                      11

  894637                      12

 9903149                     13

20091349                    14

21087751                    15

235397039                  16

8516934067                17

 

The next  p   (if exists) is greater than  10^10.

I think there is no reason to believe that the length  k  is bounded.  But I think the corresponding  p  will become gigantic.

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Hakan wrote:

I found 17 consecutive primes like this.
P               Diff
235397039, 
235397053, 14
235397081, 28
235397101, 20
235397131, 30
235397153, 22
235397161, 8
235397171, 10
235397177, 6
235397189, 12
235397191, 2
235397207, 16
235397233, 26
235397237, 4
235397269, 32
235397287, 18
235397311, 24

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