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Pillai's arithmetical function is simply $$P(n)=\sum_{i=1}^{n}\gcd{\left(n,i\right)}=\sum_{d|n}d\phi{\left(\frac{n}{d}\right)},$$ where $\phi\left(n\right)$ is Euler's totient function.

On the second page of this document, https://dmle.icmat.es/pdf/COLLECTANEAMATHEMATICA_1989_40_01_03.pdf, the author states that "$P(n)/n$ behaves like $6\log n/\pi^{2}$" (where $\log$ is the natural log).

If I understand correctly, this implies that $P(n)\approx 6n\ln n/\pi^{2}$. Of course, this is an approximation and not an upper bound. I did find, however, that the similar function $2n\ln n$ is a good upper bound and works for at least the first sixty positive integer values of $n$, except $1\le n\le3$. Unfortunately, I was unable to prove this for all positive integer values of $n$ (except $1$, $2$, and $3$). Is there any way of proving this bound? Or is there an even better upper bound that I am not aware of?

EDIT:

As Mindlack pointed out, values such as $n=\left(p_1\cdots p_k\right)^{2}$ go above this proposed upper bound. After some trial-and-error, I did find another possible upper bound ($\frac{5}{4}n\sqrt{n}$) that does work for $n=\left(3\times5\times7\times11\right)^{2}$, while $2n\ln{n}$ did not.

Glomeball
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    Well, $P(n)=\sum_{d|n}{\frac{n}{d}d\prod_{p|d}(1-p^{-1})}$ so $\frac{P(n)}{n}=\sum_{d|n}{\prod_{p|d}{(1-p^{-1})}}$ thus $\frac{P(n)}{n}=\prod_{p|n}{\left(1+v_p(n)(1-p^{-1})\right)}$.

    Now take $n=(p_1\cdots p_k)^2$, then $P(n) \geq n2^k$, and $2^k >> \ln{n}$, so something is wrong somewhere.

     – Aphelli Sep 11 '19 at 21:44
  • @Mindlack Hm, yes, you are right. $n=\left(3\times5\times7\times11\right)^{2}$ doesn't work. I suppose there is another upper bound then, as any constant factor before the $n\ln n$ can be shown to not work for a sufficiently large $n=\left(p_1\cdots p_{k}\right)^{2}$. – Glomeball Sep 11 '19 at 21:58
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    Maybe the approximation works actually on average? – Aphelli Sep 11 '19 at 22:01
  • @Mindlack Perhaps, but then it's not a very good upper-bound, is it? I do feel like the upper bound must be of a similar form, maybe something like $an\sqrt{n}$ or $an^{2}$ (where $a$ is a constant)? – Glomeball Sep 11 '19 at 22:03
  • Sometimes, you can’t have really good ones, because of irregularities (eg $\phi(n)/n$), so averaging isn’t that bad. – Aphelli Sep 11 '19 at 22:06
  • @Mindlack Oh, I think I found an upper bound $\frac{5}{4}n\sqrt{n}$ that does work, at least for $n=(3\times5\times7\times11)^{2}$, which is as far as Wolfram|Alpha is willing to calculate for me. However, I do think you are right that averaging isn't all that bad. – Glomeball Sep 11 '19 at 22:17

3 Answers3

2

I think I may have found an upper bound, namely $2n\sqrt{n}$. I may provide the solution if requested, but I'll leave two hints that should be enough to solve this problem: (1) $\tau\left(n\right)\le2\sqrt{n}$ (2) if the set of divisors of $n$ is $\{a_{1}, a_{2}, a_{3}, ..., a_{\tau\left(n\right)}\}$, what is an approximate value of $P\left(n\right)$?

Glomeball
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    actually, for any real $ \delta > 0$ there is a maximum, achieved for some $n,$ for $\frac{p(n)}{n^{1 + \delta}}.$ In this sense, everything is very similar to the Colossally Abundant Numbers of Ramanujan, later Alaoglu and Erdos (who provided the name, in 1944). https://en.wikipedia.org/wiki/Colossally_abundant_number – Will Jagy Sep 12 '19 at 02:23
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    Please provide the proof for the upper bound $2n\sqrt{n}$. See also https://math.stackexchange.com/questions/3393406/another-gcd-problem – lhf Oct 14 '19 at 14:36
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A reliable way, with simple programming, to get unusually large values for a function that is number theoretic multiplicative is just to take $n$ from a small $m$ and $$ n = \operatorname{LCM} (1,2,3,4,5,...,m-1,m) $$

Note that this increases only when $m$ is a prime or prime power. One version of the Prime Number Theorem is the fairly good approximation $\log n \approx m \; . \;$

I've done that, my conclusion is that $P(n) > \; n \; ( \log n )^k$ for any fixed positive $k$ and large enough $n$ in this sequence. As I've mentioned, this sequence of $n$ act roughly as the colossally abundant numbers.

=========================

Full printout for $m \leq 25$

                 1.5 m 2   pillai 3  n  2 =  2     -1.106277801788494
                 2.5 m 3   pillai 15  n  6 =  2 3     1.57114840063342
   3.333333333333333 m 4   pillai 40  n  12 =  2^2 3     1.322705324263649
   6.000000000000003 m 5   pillai 360  n  60 =  2^2 3 5     1.27110600234983
   11.14285714285714 m 7   pillai 4680  n  420 =  2^2 3 5 7     1.340489752783667
   13.92857142857144 m 8   pillai 11700  n  840 =  2^3 3 5 7     1.381138463847476
   19.49999999999998 m 9   pillai 49140  n  2520 =  2^3 3^2 5 7     1.443196002229409
   37.22727272727266 m 11   pillai 1031940  n  27720 =  2^3 3^2 5 7 11     1.555505596788277
   71.59090909090902 m 13   pillai 25798500  n  360360 =  2^3 3^2 5 7 11 13     1.675517976485081
   85.90909090909101 m 16   pillai 61916400  n  720720 =  2^4 3^2 5 7 11 13     1.711618190328735
   166.7647058823523 m 17   pillai 2043241200  n  12252240 =  2^4 3^2 5 7 11 13 17     1.832281489005753
   324.7523219814228 m 19   pillai 75599924400  n  232792560 =  2^4 3^2 5 7 11 13 17 19     1.954844200385714
   635.3849777897412 m 23   pillai 3401996598000  n  5354228880 =  2^4 3^2 5 7 11 13 17 19 23     2.075907921212419
   917.7783012518491 m 25   pillai 24569975430000  n  26771144400 =  2^4 3^2 5^2 7 11 13 17 19 23     2.146284910434013

===============================

Abbreviated printout for $m \leq 64$

                 1.5  m  2     -1.106277801788494
                 2.5  m  3     1.57114840063342
   3.333333333333333  m  4     1.322705324263649
   6.000000000000003  m  5     1.27110600234983
   11.14285714285714  m  7     1.340489752783667
   13.92857142857144  m  8     1.381138463847476
   19.49999999999998  m  9     1.443196002229409
   37.22727272727266  m  11     1.555505596788277
   71.59090909090902  m  13     1.675517976485081
   85.90909090909101  m  16     1.711618190328735
   166.7647058823523  m  17     1.832281489005753
   324.7523219814228  m  19     1.954844200385714
   635.3849777897412  m  23     2.075907921212419
   917.7783012518491  m  25     2.146284910434013
   1180.000673038088  m  27     2.194463499259528
    2319.31166769555  m  29     2.313778030559765
   4563.806829981581  m  31     2.433171054297125
   5324.441301645173  m  32     2.462405192118611
   10504.97878432697  m  37     2.579664357765774
   20753.73857391433  m  41     2.69603025320276
   41024.83206471433  m  43     2.812189849790792
   81176.79536209513  m  47     2.927413728375297
   118643.0086061401  m  49     2.994577324543895
   235047.4698800856  m  53     3.108318781606997
    466111.084338476  m  59     3.220881584455215
   924581.0033599157  m  61     3.333132364266017
     1056664.0038399  m  64     3.356425839582893

===============================

Will Jagy
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0

I worked out Ramanujan's construction for this, similar to the Colossally Abundant numbers named by Alaoglu and Erdos in 1944. Ramanujan had worked those out earlier, bu they were not printed in his article (1915).

We pick a real number $\delta > 0.$ We are going to find the positive integer $n$ that gives the maximum of $$ \frac{P(n)}{n^{1 + \delta}}. $$ The number is defined by its prime factorization. For any prime $p,$ the exponent is defined to be $$ k = \left\lfloor \frac{p - p^\delta}{(p-1)(p^\delta - 1)} \right\rfloor $$ Once $p$ becomes too large, the exponent $k$ comes out to be zero.

Now, the reason we can make a nice, ordered list of these numbers is that we can solve for the first (the largest) $\delta > 0,$ given a prime $p$ and a required exponent $k.$

$$ \delta = \frac{\log (kp-k+p) - \log (kp-k+1)}{\log p} $$

I think I will just show the deltas in order, then the resulting numbers:

0.5849625007211564           2           1
0.4649735207179270           3           1
0.4150374992788436           2           2
0.3652123889719708           5           1
0.3219280948873623           2           3
0.3181232230618409           7           1
0.3062702284434951           3           2
0.2696644729485751          11           1
0.2630344058337939           2           4
0.2549471261506066          13           1
0.2341137559590370          17           1
0.2287562508385780           3           3
0.2284802521951116           5           2
0.2263517560368875          19           1
0.2223924213364479           2           5
0.2140550189604739          23           1
0.2006819335059107          29           1
0.1971139143454234          31           1
0.1950190875406737           7           2
0.1926450779423957           2           6
0.1881907993877789          37           1
0.1833482899316373          41           1
0.1826583386441380           3           4
0.1811791768147107          43           1
0.1772534359825800          47           1
0.1721960126588626          53           1

==================================================================

jagy@phobeusjunior:~$ ./Pillai_Colossal_read 
   1:  1 =   1
   2:  2 =   2
   3:  6 =   2 3
   4:  12 =   2^2 3
   5:  60 =   2^2 3 5
   6:  120 =   2^3 3 5
   7:  840 =   2^3 3 5 7
   8:  2520 =   2^3 3^2 5 7
   9:  27720 =   2^3 3^2 5 7 11
  10:  55440 =   2^4 3^2 5 7 11
  11:  720720 =   2^4 3^2 5 7 11 13
  12:  12252240 =   2^4 3^2 5 7 11 13 17
  13:  36756720 =   2^4 3^3 5 7 11 13 17
  14:  183783600 =   2^4 3^3 5^2 7 11 13 17
  15:  3491888400 =   2^4 3^3 5^2 7 11 13 17 19
  16:  6983776800 =   2^5 3^3 5^2 7 11 13 17 19
  17:  160626866400 =   2^5 3^3 5^2 7 11 13 17 19 23
  18:  4658179125600 =   2^5 3^3 5^2 7 11 13 17 19 23 29
  19:  144403552893600 =   2^5 3^3 5^2 7 11 13 17 19 23 29 31
  20:  1010824870255200 =   2^5 3^3 5^2 7^2 11 13 17 19 23 29 31
  21:  2021649740510400 =   2^6 3^3 5^2 7^2 11 13 17 19 23 29 31
  22:  74801040398884800 =   2^6 3^3 5^2 7^2 11 13 17 19 23 29 31 37
  23:  3066842656354276800 =   2^6 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41
  24:  9200527969062830400 =   2^6 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41
  25:  395622702669701707200 =   2^6 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 43
  26:  18594267025475980238400 =   2^6 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47
  27:  985496152350226952635200 =   2^6 3^4 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53

==================================

Will Jagy
  • 146,052