11

Let $\pi(n)$ be the prime-counting function and $\lambda(n)$ the Carmichael-function.

Does $$\pi(n)=\lambda(n)$$ hold for infinite many positive integers $n$ ?

I have no idea for an approach other than just brute force. The solutions I got so far :

2 3 4 10 14 39 124 322 365 1086 3283 5205 16978 41899 53774 64730 64850 157165 481476 881787 1207317 3523898 9559815
Calvin Khor
  • 36,192
  • 6
  • 47
  • 102
Peter
  • 86,576
  • 4
    It should be interesting as we know the prime counting function is upper bound by Euler totient plus the number of distinct prime divisors. – Roddy MacPhee Aug 02 '21 at 11:31
  • One notes that an even number $n=2m$ can't be listed if $m$ is by Bertrand's postulate. Okay at least if $m$ is odd. – Roddy MacPhee Aug 02 '21 at 16:30
  • 3
    I'm interested in this question! No proofs availiable but I can continue your list for $n \le 5 \cdot 10^9$: 70116568, 107111654, 115373055, 190462249, 195636807, 211581819, 360263259, 514275679, 514279251, 1394197300, 1394208500, 2295707465, 4923612845 – Martin Hopf Aug 03 '21 at 17:03
  • The hard part in brute force is making the sieve as we know that any multiplier $n$ with $\lambda(m)$ divisible by $\lambda(n)$ implies $m$ and $nm$ can't both work. – Roddy MacPhee Aug 04 '21 at 23:51
  • The largest solution I know so far is : $$555\ 033\ 996\ 739$$ with value $$21\ 347\ 124\ 656$$ – Peter Aug 06 '21 at 18:07
  • Here is a reformulation of the problem. Given for $\forall m \in [p_n,p_{n+1})$, where $p_n$ - is the $n$-th prime, we have $\pi(m)=n$, how often can we find a $p_n<q<p_{n+1}$ such that $\lambda(q)=n$? This book, page 308, has a remark

    <<$n$ is a Carmichael number iff if it is of the form $p_{n_1}p_{n_2}...p_{n_k}$, with $p_{n_i}$ different primes such that $p_{n_i}-1\mid n-1$, for $\forall i=1..k$.>>

     – rtybase Aug 12 '21 at 14:06
  • New record : $$4086622775746\ 145950268980$$ – Peter Aug 12 '21 at 14:50
  • 2
    @Peter: You should add that sequence in OEIS. – Jose Arnaldo Bebita Dris Aug 13 '21 at 10:48
  • @ArnieBebita-Dris I do not like OEIS for several reasons. I allow everyone to add it , even without mentioning my nickname , mentioning "by Peter" is also OK. – Peter Aug 14 '21 at 07:39
  • New record solution : $$[30109570421550, 1003652347380]$$ – Peter Aug 14 '21 at 09:02
  • 2
    New record solution : $$[12069948025343772,335276334037326]$$ – Peter Aug 16 '21 at 15:15

1 Answers1

5

I can't prove if there exists infinite matches.

However, one can find large solutions $\lambda(n) = \pi(n)$ by considering an integer ratio $r \approx \Large \frac{n}{\pi(n)}$. Chances are good to find $rq$ with $q$ is either a prime or a semi-prime, such that $\lambda(rq) = \pi(rq)$.

Let us consider $r = 42$. To find $n$, the Riemann $R(x)$ function is a good way to approximate:

$\Large \frac {n}{\pi(n)}$$\approx 42 \ $ with $\ n \approx 4.84777065654 \cdot 10^{18}$

Searching in that range the exact value for $\pi(n)$ has only to be determined once, then the primes within that range can be counted. $\lambda(n)$ has only to be calculated if $n$ is a multiple of $42$.

After all, six solutions can be spotted in this range:

$$\lambda(n) = \pi(n) = m$$

        n                       m
==========================================

4847770656544884894     115423110870116304


4847770656544922694     115423110870117204


4847770656544965702     115423110870118230


4847770656544978806     115423110870118542


4847770656544979562     115423110870118560


4847770656544980906     115423110870118590
Martin Hopf
  • 638
  • The big question is whether this works also in larger ranges. But nevertheless, I think, this deserves the bounty. – Peter Aug 17 '21 at 12:40