mersenneforum.org An extension of Legendre's conjecture
 Register FAQ Search Today's Posts Mark Forums Read

 2022-09-07, 00:57 #1 Charles Kusniec     Aug 2020 Guarujá - Brasil 1528 Posts An extension of Legendre's conjecture Let dsq(n) be the function that returns the first digit of the fractional part of the square root of n. dsq(n) = floor((sqrt(n)-floor(sqrt(n)))*10) = https://oeis.org/A023961. dsq(n) = {0, 1, 2, 3, 4, 5, 6, 7, 8, or 9}. On the other hand, admit that there is a proof of Legendre's conjecture, and that this proof states that 1. exist at least 1 prime number between a square number and its following oblong number https://oeis.org/A307508, and 2. exist at least 1 prime number between an oblong number and its following square number https://oeis.org/A334163. It is simple to demonstrate that 1. all numbers between a square number and its following oblong number (including them) have 0<=dsq(n)<=4, and 2. all numbers between a oblong number and its following square number (excluding them) have 5<=dsq(n)<=9. Considering that a. the distribution of digits dsq(n) is uniform, linear and infinite always growing along the number line; b. the distribution of prime numbers along the number line follows a curve where the gap between prime numbers production is finite. Then, from this infinite gap (or before?) the production of primes is higher than the size of each of the 10 dsq´s value between squares. So, the conjecture is: from one point onward between two square numbers there are at least 10 prime numbers where each one has a possible 0<=dsq(n)<=9. (each dsn value has at least one prime). That is, from this largest possible prime gap number, all 10 subintervals of dsq(n) between two squares have at least one prime number. If this conjecture is right, then it is possible to find a sequence of subsequent prime numbers of any size where all these prime numbers have the same desired dsq(n). Last fiddled with by Dr Sardonicus on 2022-09-07 at 12:48 Reason: fix title
 2022-09-07, 02:19 #2 Charles Kusniec     Aug 2020 Guarujá - Brasil 2·53 Posts There is no impediment to extend this reasoning to the other digits infinitely, so we can have segments with minimum multiples of 10 primes for each digit we increase. That is why every non-square square root is irrational and never ends..... The primes too!
2022-09-07, 15:29   #3
Charles Kusniec

Aug 2020
Guarujá - Brasil

10610 Posts

Quote:
 Originally Posted by Charles Kusniec Then, from this infinite gap (or before?) the production of primes is higher than the size of each of the 10 dsq´s value between squares.
Better wording:

From the point where the production of primes reaches the maximum gap, they will continue to be distributed only over the 10 possible dsq values.

2022-09-07, 17:10   #4
Charles Kusniec

Aug 2020
Guarujá - Brasil

1528 Posts

Quote:
 Originally Posted by Charles Kusniec Better wording: From the point where the production of primes reaches the maximum gap, they will continue to be distributed only over the 10 possible dsq values.
From this point onwards, the difference $$(dsq(prime) - dsq(subsequent \ prime)) \ mod \ 10$$ will be 0 or 1.

Then we will have the same algorithm occurring for the 2nd digit and so on.

For each combination of digits we will have a beginning and ending integer value. They form what we can call "prime filling sectors between squares".

As the difference of the distance between squares are the odd numbers that is a linear function, and the distribution of the numbers by the possible dsq is also linear, then I imagine that the values limiting the "filling sectors" obey to a quadratic function.

Last fiddled with by Charles Kusniec on 2022-09-07 at 17:12

 2022-09-08, 00:44 #5 CRGreathouse     Aug 2006 5,987 Posts A slightly stronger conjecture is that for any sufficiently large x, there is a prime between x and x + sqrt(x)/5. A table of prime gaps suggests that this probably holds for x > 360653 and a bit of calculation should give sharper bounds for both conjectures.
 2022-09-08, 00:46 #6 CRGreathouse     Aug 2006 10111011000112 Posts This should also be provable with any number of standard prime gap bounds, e.g., Dusart's.
2022-09-08, 01:43   #7
Dr Sardonicus

Feb 2017
Nowhere

10111110011012 Posts

Quote:
 Originally Posted by CRGreathouse This should also be provable with any number of standard prime gap bounds, e.g., Dusart's.
AFAIK the conjecture that there is a prime in (n2, (n+1)2) has not been proven. A result of Chen proves that for sufficiently large n there is a number in the interval that is either a prime or a product of two primes.

There are also results that "short intervals" (x, x + xθ) or (x - xθ, x) contain primes for "sufficiently large" x for values of θ slightly larger than 1/2.

What is suspected (this is related to Cramér's conjecture) is that there is a prime between n and something like n + log2(n) for sufficiently large n.

 2022-09-09, 00:16 #8 CRGreathouse     Aug 2006 176316 Posts OK, I said silly things -- of course we can't prove a conjecture stronger than Legendre at our current technology. Our best exponent is 0.525 and that's a long way from 0.5. I think what I meant is that various bounds should let us show that this holds for large regions. The best ones for this type of verification would not be the Dusart-style bounds but the linear bounds; there are quite good ones available now. But still that's only enough for finite regions, not the whole result.

 Similar Threads Thread Thread Starter Forum Replies Last Post RomanM Miscellaneous Math 2 2021-12-22 14:11 Hugo1177 Miscellaneous Math 1 2021-01-13 22:15 Nick Number Theory Discussion Group 0 2017-01-07 13:15 davieddy Puzzles 7 2007-09-04 12:50 Numbers Math 2 2005-09-07 15:22

All times are UTC. The time now is 18:08.

Tue Nov 29 18:08:24 UTC 2022 up 103 days, 15:36, 0 users, load averages: 0.96, 1.08, 1.06