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Note: Posted in MO since it is unanswered in MSE.

Analogous to Chebyshev's race between primes, I examined the race between primes in the greatest prime factors, GPF, of natural numbers. Similar to the regular prime race, in the GPF race, the proportion of prime of both forms is roughly $50\%$ each. However, unlike the regular prime race where the race is equal (A038691) or changes lead between primes of the form $4k+1$ and $4k+3$ (A096628) in the GPF race, $4k+3$ dominates. For $n \le 3 \times 10^{10}$, there are $14970975209$ numbers where the GPF is of the form $4k+1$ and $15029024755$ numbers where $GPF$ is of the form $4k+3$ and up till this point, there is not a single instance where primes of the form $4k+1$ equals or takes the lead in the race.

Let $f_1(n)$ and $f_3(n)$ be the number of natural numbers $\le n$ in which the GPF is of the form $4k+1$ and $4k+3$ respectively. The difference between the number of GPFs on these two forms appeared to increase in a very orderly manner as shown in the graph below.

enter image description here

Question 1: Does the GPF race ever become equal or change the lead i.e. is there an $n$ such that $f_1(n) = f_3(n)$?

Clearly, the graph is sub-linear. I applied different curve fitting models to this data to have an idea of the growth rate. The best fit was given by

$$ f_3(n) - f_1(n) \approx an^b $$

where $a \approx 0.1785$ and $b \approx 0.8122$. This experimental model has a very high $R^2 = 0.9999958$. If this is close to the true growth rate then, prime of the form $4k+1$ will always lag behind in the GPF race.

Question 2: What is the true growth rate of $f_3(n) - f_1(n)$?

Nilotpal Sinha
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    Wow, a huge search range (+1) ! – Peter Apr 09 '22 at 08:17
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    The sum of the above numbers is $36$ less then $8\cdot 10^9$ , but there are only $33$ powers of $2$ (including $1$) upto $8\cdot 10^9$. What do I miss ? – Peter Apr 09 '22 at 08:30
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    @Peter Very acute observation. Actually I was checking for primes of the form $6k \pm 1$ so I started the search from $p_4 = 7$. I modified the same code to search for primes of the form $4k /pm + 1$ but forgot to initialize to $p_2 = 3$. – Nilotpal Sinha Apr 09 '22 at 09:09
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    OK, although $5$ is also a prime of the form $6k\pm 1$. Do you know the first crossover of the regular prime race ? A fair coin could not have given this number of heads/tails, but I think, the Chebycheff bias is similarily significant, so it might be too early to draw conslusions. – Peter Apr 09 '22 at 09:34
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    @Peter For the regular prime race, crossover happen quite frequently with the smallest at $n = 2946$. More: http://oeis.org/A096628 – Nilotpal Sinha Apr 09 '22 at 11:39
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    More precisely, at the $2946$th prime, which is $n=26861$. There are several other places below $8\cdot10^9$ as well: see the middle of page 2 of my paper with Granville. – Greg Martin Apr 09 '22 at 17:34
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    $f_3(x)-f_1(x) = - \sum_{p\le x} \chi_{-4}(p) \psi(\frac xp,p)$, where $\chi_{-4}$ is the nontrivial Dirichlet character modulo $4$ and $\psi(t,y)$ is the number of integers up to $t$ whose prime factors are all at most $y$ (the "$y$-friable" numbers). This is turn is asymptotically $- \sum_{p\le x} \chi_{-4}(p) \rho(\frac{\log x}{\log p}-1)$, where $\rho$ is the Dickman–de Bruijn function related to friable numbers. One could try finding an explicit formula for this using Perron's formula, treating the $\rho$ factor as a smoothing term.... – Greg Martin Apr 09 '22 at 17:50
  • @GregMartin A very interesting paper. Thanks for sharing :) – Nilotpal Sinha Apr 10 '22 at 09:40
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    @GregMartin How do you know this is the asymptotic? – reuns Apr 11 '22 at 06:24
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    @NilotpalSinha since you have the data, can you share a histogram of the GPFs? Curious how many numbers have GPF less than a certain number. – sku May 15 '22 at 03:06
  • @sku Yes, I have the data for $n \le 77258000000$ will share them shortly. Any idea how I share an excel or csv file in MSE? – Nilotpal Sinha May 15 '22 at 04:28
  • Put on gdrive and share link – sku May 15 '22 at 04:35
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    @GregMartin One problem with using the asymptotic for $\psi(\frac{x}{p},p)$ to obtain an asymptotic for $\sum_{p\leq x} \chi(p)\psi(\frac{x}{p},p)$ is that the error terms might not "cancel out" nicely.

    For example, consider those primes $p>\sqrt{x}$. For each such prime, there are exactly $[\frac{x}{p}]$ integers $\leq x$ divisible by $p$. Thus we end up with $$\sum_{\sqrt{x}<p\leq x} \chi(p)\left[\frac{x}{p}\right]=x\sum \frac{\chi(p)}{p}+\sum \chi(p)\left{\frac{x}{p} \right}$$

    It seems very hard to handle the term $$\sum_{\sqrt{x}<p\leq x} \chi(p)\left{\frac{x}{p} \right}.$$

     – Eric Naslund May 25 '22 at 03:47
  • @NilotpalSinha: I wanted to ask if you could try fitting the specific model $f_3(n)-f_1(n) \approx Cn^{0.75}$ and see how this does? – Eric Naslund May 25 '22 at 15:29
  • @EricNaslund The data seems to suggest that the true exponent is somewhere close to $0.75$. For $n = 104253000000$, $\frac{\ln(f_3(n) - f_1(n))}{\ln n} \approx 0.744893792393631$ and this ratio of the logarithms have an increasing trend so it is possible that it approaches $3/4$. Later, I will fit a proper function of the form $Cn^{3/4}$ see how the constant $C$ is behaving. Do you have any heuristics to suggest that the exponent must be $3/4$? – Nilotpal Sinha May 25 '22 at 16:28
  • @NilotpalSinha: Unfortunately there was a mistake in the approach I used, I'm not sure how to handle the small primes, so I am not sure if $n^{0.75}$ is correct. – Eric Naslund May 25 '22 at 19:00
  • @EricNaslund May be you can share the intuition/heuristics that initially led you to believe that the true exponent might be $0.75$ and high the difficulty in this approach – Nilotpal Sinha May 27 '22 at 12:03
  • @EricNaslund Thanks for putting a bounty on this question. No answers so far indicate this is a really hard problem. – Nilotpal Sinha Jun 03 '25 at 09:33

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