4
$\begingroup$

I am hoping someone can estimate the number of primes that come up this way: take a number $L,$ then let $$ C = \operatorname{lcm} (1,2,3,\ldots,L). $$ We know that $C$ has quite a lot of divisors; indeed, this is just a variant of Ramanujan's Superior Highly Composite Numbers, in that the exponent of some prime $p$ is roughly proportional to $1/\log p.$ Indeed, for this case, the exponent is $$ \left\lfloor \frac{\log L}{\log p} \right\rfloor \; \; .$$

For every divisor $d |C,$ let $q = 1 + d.$ I would like an estimate of the number of $q$ that come up prime. For example, with $L=4,$ we get $C=12$ and divisors $1,2,3,4,6,12,$ thence primes $2,3,5,7,13.$ Below is a table for $L \leq 61;$ if $L$ is as above, and $P$ is the count of primes, the number ``log ratio'' is $(\log P) / (\log L).$ This and other calculations suggest that $P$ is just short of exponential in $L,$ just a little slower growth than $e^{L / \log L}.$

 lcm of up to:  4 divisors : 6  prime count:  5 log ratio: 1.160964047443681
 lcm of up to:  5 divisors : 12  prime count:  8 log ratio: 1.292029674220179
 lcm of up to:  7 divisors : 24  prime count:  13 log ratio: 1.318123223061841
 lcm of up to:  8 divisors : 32  prime count:  15 log ratio: 1.30229686520284
 lcm of up to:  9 divisors : 48  prime count:  22 log ratio: 1.406794046107798
 lcm of up to:  11 divisors : 96  prime count:  36 log ratio: 1.494443472618428
 lcm of up to:  13 divisors : 192  prime count:  66 log ratio: 1.633425911446773
 lcm of up to:  16 divisors : 240  prime count:  81 log ratio: 1.584962500721156
 lcm of up to:  17 divisors : 480  prime count:  148 log ratio: 1.763796674277192
 lcm of up to:  19 divisors : 960  prime count:  252 log ratio: 1.877922798412615
 lcm of up to:  23 divisors : 1920  prime count:  446 log ratio: 1.945568555358454
 lcm of up to:  25 divisors : 2880  prime count:  660 log ratio: 2.016927706518875
 lcm of up to:  27 divisors : 3840  prime count:  905 log ratio: 2.065616479359747
 lcm of up to:  29 divisors : 7680  prime count:  1638 log ratio: 2.197974766132367
 lcm of up to:  31 divisors : 15360  prime count:  2912 log ratio: 2.322837836698491
 lcm of up to:  32 divisors : 18432  prime count:  3578 log ratio: 2.360987534410382
 lcm of up to:  37 divisors : 36864  prime count:  6661 log ratio: 2.438168109993427
 lcm of up to:  41 divisors : 73728  prime count:  12344 log ratio: 2.536890418395409
 lcm of up to:  43 divisors : 147456  prime count:  23060 log ratio: 2.670917389347842
 lcm of up to:  47 divisors : 294912  prime count:  42735 log ratio: 2.769445392395919
 lcm of up to:  49 divisors : 442368  prime count:  63329 log ratio: 2.840855381850106
 lcm of up to:  53 divisors : 884736  prime count:  118262 log ratio: 2.942014853177503
 lcm of up to:  59 divisors : 1769472  prime count:  221800 log ratio: 3.018864087268213
 lcm of up to:  61 divisors : 3538944  prime count:  417192 log ratio: 3.148065898896646

So that's the question, can anyone give a reasonable estimate on $P,$ possibly conditional on open conjectures? Oh, not by the way, what I really want to know is closer to the product of all those primes; in case of interest, see https://math.stackexchange.com/questions/1334767/the-maximal-size-of-between-varphin-divided-by-lambdan/1336917#1336917

$\endgroup$
  • $\begingroup$ A typo perhaps: do you not mean "the number log ratio is log(L)/log(P)" ? $\endgroup$ – Thomas Sauvaget Jun 25 '15 at 19:30
  • $\begingroup$ @ThomasSauvaget, I put in some parentheses. I like to use log and sin without parentheses around the argument, but i can see how that may be ambiguous in this expression. $\endgroup$ – Will Jagy Jun 25 '15 at 19:34
  • $\begingroup$ I don't have an account on Math Stackexchange at the moment, but the conjecture of Wythagoras that motivated your question follows quickly from Theorem 1 in this paper of Erdos--Pomerance--Schmutz: math.drexel.edu/~eschmutz/PAPERS/lambda.pdf $\endgroup$ – so-called friend Don Jun 25 '15 at 23:35
  • $\begingroup$ @so-calledfriendDon, good, that seems to do it. $\endgroup$ – Will Jagy Jun 26 '15 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.