Problems & Puzzles: Conjectures

Conjecture 41. Pe's Conjecture

Joseph Le Pe sent (Feb 7, 2006) the following conjecture, in order to be published in these pages:

Undoubtedly, you know of Gilbreath's conjecture, which states that the
leading entries of the rows (except the first row) of the table of
successive absolute differences of the primes is always equal to 1.

Consider instead the table of successive absolute differences of the squares
of the prime numbers. I conjecture that all but a finite number of the
leading entries of the rows of this table are prime numbers.
For a fuller development, see: http://tinyurl.com/bcrsp
 

The same day I informed to Joseph this:

Is it original from you? As a matter of fact up to the row 9000th the last composite is the number 27; it appears in the row 1165 and is the 131th of them. No other composite appears in my Ubasic code running up to the row 9000. At the end the emerging leading primes are just one of the following four: 3, 5, 11 & 13. Have you devised an explanation of this behavior?
 

He relplied this:

...as far as I know, it is originally from me. I checked R. Guy's 2005 "Unsolved Problems in Number Theory" book, specifically the section on Gilbreath's conjecture. It is not there, and I would think that if my conjecture had been made as of last year, it would have appeared as a note in the book.... No, I have not. I can think of no obvious reason why those primes should occur at the end. My suspicion is that my conjecture is at least as difficult as Gilbreath's.

Surprisingly Wilfrid Keller sent to me (Feb. 9, 2006) the following note:

Carlos... despues de tanto tiempo! Me acorde de ti cuando vi este mensaje de Steven Harvey:

http://groups.yahoo.com/group/primenumbers/message/17645 .

Como aparentemente te habías dedicado a este problema, pense que podría
interesarte el siguiente comentario que acabo de mandarle a Steven:

...

Only a few hours ago I saw in the PrimeNumbers forum your hint pointing to a
modified Gilbreath conjecture. Very nice indeed.

I played around with it for a while, and noticed thereby that the conjecture
might probably be stated in a more specific way:

CONJECTURE. For all N > 1165, the leading terms in the table T can only
assume one of the prime values 3, 5, 11, 13.


Some numerical experimentation led me to observe -- apart from confirming
the more specific conjecture for all N up to N = 12000 -- that
(surprisingly) these values occur in a quite regular proportion, which is
approximately 1/3, 1/3, 1/6, 1/6, respectively
.

So, for N = 1201 to 12000 we have the frequencies and respective
proportions:

3 5 11 13
3641 3584 1762 1813
0.337 0.332 0.163 0.168

Unfortunately I cannot discern any pattern for these occurences that might
allow a proof (in case they should repeat periodically). In any case, an interesting entertainment.

Finally I wrote to Joseph this last idea:

Have you devised any kind of connection between this your conjecture and the ideas posed in my Puzzle 274?

What I want to mean is that Sierpinski triangles devised by Thomas, at the end were not only linked to the prime numbers but to a more general sequence of numbers, prime numbers included; more over, the Sierpinski triangles are just a part of a square matrix while the Gilbreath's conjecture is in the remaining space.

Perhaps something similar may occur with your new object: perhaps it contains even more info still hide.
 

Questions

1. Can you prove the Pe's conjecture?

2. Can you explain why on the large run apparently only emerge the primes 3, 5, 11 & 13?

3. Can you explain the statistic calculated by Keller for these primes?


 


Contributions came from Wilfrid Keller, T. D. Noe, J. K. Andersen & Patrick Capelle:

***

Keller wrote:

Terms of the sequence defined by Joseph Pe as "the leading elements of the rows of the table T" will be denoted by a(n), n = 1, 2, 3, ... .

So we have a(1) = 4, a(2) = 5, a(3) = 11, a(4) = 3, a(5) = 37, and so on. The following observations have been reported (or are being added here):

(1) For 2 <= n <= 1165, a(n) is a prime for all but 131 indices
n, including a(1165) = 27. Among the prime values in this
interval there are only 84 which are different from 3, 5, 11
or 13.

(2) For 1165 < n < 35408, the only values (prime or composite)
assumed by a(n) are 3, 5, 11, 13.

Other compact intervals with only these values had occurred
earlier (whithout being noticed):

n = 7-22 (length 16), n = 84-245 (length 162),
n = 742-1081 (length 340)

(3) Quite surprisingly (to me), it turned out that

a(35408) = 582349 = 29.43.467, a(35409) = 582341 = 661.881;
a(35412) = 75 = 3.5^2, a(35413) = 59, a(35414) = 27 = 3^3,
a(35415) = 19, a(35416) = 582045 = 3.5.38803,
a(35417) = 582037 (prime), followed by many occurrences
like that (which will be specified below) .

(4) Computing a(n) up to n = 48000 (the highest limit accepted
by my UBASIC program) it appeared that the values 3, 5, 11, 13
do not stop being assumed more frequently than others. The com-
plete counts (starting from n = 2) and proportions are:

3 5 11 13 others

15293 15159 7600 7707 2240

0.3186 0.3158 0.1583 0.1606 0.0467

(5) All of the 2240 recorded (not necessarily different) "other" val-
ues a(n) also satisfy a(n) = 3 (mod 8) or a(n) = -3 (mod 8).
Thus, (odd) values that do not share this property, like 7, 9,
15, 17, 23, 25, 31, 33, ... did never occur (so far).

Instead, the values 3, 5, 11, 13, 19, 21, 27, 29, 35, 37, 43,
45, 51, 53, 59, 61, 67, 69, 75, 77, 83, 85, 91, 93, 99 (which
include twin primes like 3, 5; 11, 13; 59, 61) have all been
observed without exception (to that limit).

(6) Strikingly, in the higher region of n >= 35408, relatively
large pairs of values with difference 6 and consecutive indices
n occur quite frequently, like the above

a(35408) = 582349 = 29.43.467, a(35409) = 582341 = 661.881;
a(35416) = 582045 = 3.5.38803, a(35417) = 582037 (prime).

Note that these two pairs of values are quite nearby each other.
We observe remarkable "clusters" of such "nearby" pairs. On the
other hand, there are again regions (of smaller extent), where
only values 3, 5, 11, 13 occur, like n = 40122-45349 (length
5228). According to the above statistics, in the long run these
four values seem still to dominate (so far?!).

In any case, the conjecture that I had prematurely established, is cer- tainly wrong in the original form, and I now tend to believe that Pe's conjecture might not withstand either.

Here is another numerical remark. Considering the next pair of "pos- sible" odd values v = 19, 21, we find:

a(n) = 19 for n = 49, 64, 66, 68, 70, 405, 419, 423, 425, 427, 429,
433, 449, 453, 457, 548, 741, 1092, 1114, 1120, 35415,
35421, 35429, 35483, 35491, 35495, 35554, 35572, 35602, ...

a(n) = 21 for n = 6, 387, 395, 467, 471, 475, 479, 483, 485, 491,
493, 526, 528, 536, 547, 549, 557, 706, 1108, 1117, 1129,
35436, 35478, 35486, 35492, 35522, 35546, 35571, 35579, ...

One might ask: Does each of v = 19 and v =21 occur infinitely often?
In the affirmative, Pe's conjecture would also be revoked.

***

T. D. Noe wrote:

Call 4,9,25,49,... the first row. I computed 10^6 rows. The conjecture "For all N > 1165, the leading terms in the table T can only assume one of the prime values 3, 5, 11, 13" fails first in the 35408th row, which starts with 582349.

The great majority of the first 10^6 rows begin with 3, 5, 11, or 13. Here are the statistics: 3 (299265), 5 (299131), 11 (151976) and 13 (151995).
The are 929233 rows beginning with a prime number.

***

J. K. Andersen wrote:

Row 1 is the squares of primes.
Let r(n) be the leading entry in row n.
There are many other values than 3, 5, 11, 13 after row 1165.
The first is r(35408) = 582349 = 29*43*467.
The large values are clustered in groups with a decreasing tendency.

The values up to r(1000000) which are larger than all preceding values:
r(1) = 4
r(2) = 5
r(3) = 11
r(5) = 37
r(23) = 107
r(31) = 669
r(368) = 12397
r(35408) = 582349
r(114223) = 13179573
r(405706) = 86490227
r(919762) = 406453141

3, 5, 11 and 13 remain common and are the only values in some long intervals.
The longest such intervals in the first million rows are:
34242 from row 1166 to 35407
66628 from row 47595 to 114222
215128 from row 190578 to 405705
13014 from row 447892 to 460905
449186 from row 470576 to 919761
All these intervals except the shortest are followed by a record r(n).

The first million rows have 902367 numbers in {3,5,11,13}:
299265 3's, 299131 5's, 151976 11's, 151995 13's There is no apparent sign that other values are disappearing.
There are 97633 other values: 26866 primes and 70767 composites.
The last is r(999998) = 21.

There are 42747 different values: 7240 primes and 35507 composites.
After 3, 5, 13, 11, the most common values are:
4678 21's, 4673 19's, 1409 27's, 1362 29's, 1245 35's, 1228 37's.

Primes above 3 are on the form 6k-1 or 6k+1.
From this it follows (details omitted) that:
In row 2, all numbers after the first two are divisible by 24.
Then the same holds for all following rows.
The second number in all rows except row 1 is 8 or 16 (mod 24).
The first number in all rows except row 1 is 3 or 5 (mod 8).
That means r(n) is either on the form 8k+3 or 8k+5 for n>1. r(1)=4.

499922 of the first 1000000 values are 3 (mod 8). 500077 are 5 (mod 8).
The smallest missing values on these forms in the first 1000000 rows are:
2067, 2693, 5275, 5467, 5523, 5931, 5995, 6051, 6125, 6149.
I wonder whether all 8k+3 and 8k+5 values will eventually occur.
It may look bad for Pe's conjecture.

***

Patrick Capelle wrote (please notice an attempt of explanation of the Q3) :

 

 

4

9

25

49

121

…

Row 1

5

16

24

72

48

…

Row 2

11

8

48

24

72

…

Row 3

3

40

24

48

24

…

Row 4

37

16

24

24

24

…

Row 5

21

8

0

0

24

…

Row 6

13

8

0

24

24

…

Row 7

…

…

…

…

…

…

 
The table T can be divided into two principal parts :
1. The right part, concerning the sequence of squares of odd prime numbers :
The absolute difference between the squares of two successive odd prime numbers is always even (see row 1).
This difference gives always a composite number because |q2p2| = |(q+p).(q-p)|  and  q - p > 1 when p and q are odd primes.
Hence, if we except the first number of the first row, all the numbers of this row are even and composite.
For the following rows, it means that all the numbers will be even when we go from the second column to the right of the table T.
It doesn't mean necessarily that below the first row (first column excluded), all the even numbers will be composite.
For that, we need another constatation, which is a consequence of the fact that the square of an odd number is always equal to 1(mod 8) :
the absolute difference between the squares of two successive odd prime numbers is always a multiple of 8.
It implies that it is also the case when the odd prime squares are not consecutive.
It implies also that below the first row all the entries of the right part ot this table T will be of the form 8n, with n >= 0.
From the right to the left, the absolute successive differences will finally reduce (not linearly) the number of different even numbers and the size of them, such that at the end we will have more chance to find the first multiples of 8 (zero included) in the second column : 0, 8, 16, 24, ...
 
2. The left part, concerning the square of the prime number 2 :
By the fact that the initial sequence of prime squares is beginning by the even number 4, the difference between 9 and 4 will give 5 (which is one of the emerging primes of the first column !), and after that all the absolute differences between the even numbers (of the second column) and the first odd numbers obtained will give only odd numbers for the rest of the first column.
Hence, the odd (and accidentally prime) numbers of this table T must be searched only in the first column.
The table T of successive absolute differences is like an infinite ocean of even (and composite) numbers of the form 8n, bordered on the left by a column of odd numbers.
 
 
Why the emerging odd numbers of the first column are the numbers 3, 5, 11 and 13 ?
Probably because there are emerging even numbers in the second column (i don't have checked), such that the absolute difference with any of those four prime numbers will give one of them.
We must try to search these even repetitive numbers among the first values of the form 8n, n >= 0, and it's only after the 1165th row that we can obtain them exclusively.
There are only three even numbers having the property to give only numbers of this set of prime numbers (i.e., 3, 5, 11 and 13) after an absolute difference : 0, 8 and 16.
 
|0 - 3| = 3                                                     
|0 - 5| = 5                                    
|0 - 11| = 11                               
|0 - 13| = 13  
                              
|8 - 3| = 5  
|8 - 5| = 3 
|8 - 11| = 3
|8 - 13| = 5
 
|16 - 3| = 13
|16 - 5| = 11
|16 - 11| = 5
|16 - 13| = 3
 
In total, we have 12 possibilities.
As you can see, the numbers 0 and 16 are neutral for the repartion of the results 3, 5, 11 and 13 : we don't have more chance to have a prime number than another.
It's only the number 8 which is responsible for the fact that the numbers 3 and 5 are finally more represented than the numbers 11 and 13.
If we have statistically an equal repartition for 0, 8 and 16 in the second column (starting from the 1166th row), we will obtain the proportions 4/12 - 4/12 - 2/12 - 2/12  for the primes 3, 5, 11 and 13 respectively (results of W. Keller). If it is not the case, we must observe a reduction in the number of emerging even numbers of the second column and/or an assymetric repartition for them (giving the same proportions or nearly the same). 
 
In conclusion, i would say that this phenomenon can appear when you have at least a part of the prime numbers concerned in the begin of the process and a pattern of emerging even numbers (in the second column) whose repartition will perpetuate the repetition of the same prime numbers in the first column.
After enough rows, we will have always the same small even numbers in the second column by a process of reduction of the number of different even numbers after many successive absolute differences. Why did this process stop at a step giving three numbers (0, 8 and 16) ? Why 3 ? It's like a machine stopped too soon. By replacing 4 by 1 in the initial sequence, we could have a situation quite different, with less emerging even numbers or assymetric repartition of them. But it means, in the context of this problem, that we must consider the number 1 as a prime number ...
 
I am conscious that my attempt of explanation is not giving all the answers and that i am only replacing a conjecture about prime numbers by another one concerning emerging numbers in the second column. But perhaps the problems could be easier to treat only by proposing conjectures about the right part of the table T (where the prime squares have the same nature because they are all odd). Moreover, it is possible that some problems or conjectures about prime numbers are still not solved because one does not dare to to put away the number 2 and to approach separately the study of the odd prime numbers.
 
A lot of questions are still remaining in my mind. I give here only some of them :
  • Beginning with the sequence of k-powers of prime numbers, is each term of the absolute difference between k-powers (see the first row) one of the eventual emerging numbers of its own column ? It's the case when k = 1 or 2 for at least the first column, but is it the case for the other columns and for k > 2 ? Is it also the case when we start with any sequence of increasing terms (i mean not necessary powers of prime numbers) ?
  • Beginning with the sequence of k-powers of odd prime numbers, are all the eventual emerging even numbers of the form f(k).n, with n = 0, .., k ? Is there a simple relation between k and the number of emerging elements ? How to predict f(k) ?
  • Concerning the number of prime leading elements for the sequence of k-powers of prime numbers, I give here a table which is an extension of the list proposed by Joseph L. Pe (i have used the same Mathematica program). Can you explain why the number of prime leading elements is decreasing when k > 1  is increasing (on a large run, for a number N of rows) ? Can you propose a formula giving the decreasing values in function of k ? 

 

 

Number of prime leading elements in function of power k

 

1

2

3

4

5

6

7

8

9

10

Value of N

1000

0

897

319

136

94

71

41

43

23

29

2000

0

1869

602

265

182

117

89

64

51

57

3000

0

2869

883

405

252

182

139

102

78

87

4000

0

3869

1111

547

321

249

170

137

97

113

5000

0

4869

1335

689

425

313

216

164

129

124

6000

0

5869

1560

814

489

364

264

190

151

146

                       
 

 

***

J. L. Pe wrote, (Feb, 27, 06):

 

I just arrived from the Philippines a few days ago. It was with interest that I read the reader contributions to my conjecture #41, especially those of J. K. Andersen, who computed 10^6 rows of the prime squares absolute difference Table and provided very detailed statistics. It seems that there is less evidence for the conjecture than what we previously thought.

However, the new computations provide evidence for a new (although weaker) conjecture that I make below:

For any N, in the first N rows of the prime squares absolute difference table, the number of prime leading terms is greater than the number of composite leading terms.

In particular, Andersen has found that among the first 10^6 rows, there are 70,767 composites, with the rest being primes.

***

Joseph L. Pe wrote on March 31, 2006:

Here's a new idea on Conjecture #41.

Preliminary investigations have led me to make the following generalization of the Gilbreath's and Pe's conjectures: For a fixed positive integer n, let T(n) be the table of successive absolute differences of the n-th powers of primes. Then the number of k-almost prime leading elements, 0 <= k < n, is greater than the number of leading elements that are not of this form.

Recall that a number is called k-almost prime if the sum of the exponents in its prime factorization equals k. Thus, a 0-almost prime equals 1, a 1-almost prime is a prime number, and a 2-almost prime is a semi-prime. If n = 1, we have a weak form of Gilbreath's conjecture, and if n = 2, we have Pe's conjecture.

***

 

 

 

 

 


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