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I did the specialization for the $m's$ in the multiplication formula for the Gamma function, see the identity (4) in page 250 of Apostol, Introduction to Analytic Number Theory Springer (1976) as the divisors $d\mid n$ of an integer $n\geq 1$ to get, after one uses $\prod_{d\mid n}d=n^{\sigma_0(n)/2}$, where $\sigma_0(n)=\sum_{d\mid n}1$ counts the number of divisors, and with $\sigma(n)=\sum_{d\mid n}d$ denoting the sum of divisors function

$$\frac{\prod_{d\mid n} d^{ds} }{\prod_{d\mid n}\Gamma(ds)} \left( \prod_{d\mid n}\Gamma \left(s \right)\Gamma \left( s+\frac{1}{d} \right)\cdots\Gamma \left( s+\frac{d-1}{d} \right) \right)=\left(\sqrt{2\pi} \right)^{\sigma(n)-\sigma_0(n)}n^{\sigma_0(n)/4}.$$

I believe that an application of Gronwall's Theorem provide us the following
$$\limsup_{n\to\infty}\left( \left(\sqrt{2\pi} \right)^{\sigma(n)-\sigma_0(n)}n^{\sigma_0(n)/4} \right) ^{\frac{1}{n\log\log}}=(\sqrt{2\pi})^{e^\gamma},$$ where $\gamma $ is the Euler-Mascheroni constant.

Question. Notice that LHS seems a function of $s$. Can you deduce the same superior limit from LHS? I say that you can do simplifications, use Gronwall's Theorem, Stirling approximation... to deduce, if it is right that $$\limsup_{n\to\infty} \left( \frac{\prod_{d\mid n} d^{ds} }{\prod_{d\mid n}\Gamma(ds)} \left( \prod_{d\mid n}\Gamma \left(s \right)\Gamma \left( s+\frac{1}{d} \right)\cdots\Gamma \left( s+\frac{d-1}{d} \right) \right)\right) ^{\frac{1}{n\log\log n}}=(\sqrt{2\pi})^{e^\gamma}.$$ Thanks in advance.

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  • $\begingroup$ why don't you write your reasoning rigorously ? honestly, instead of pressing the "POST" button, once you wrote your question, the points that are unclear are obvious, and you should work on those. $\endgroup$ – reuns Jul 13 '16 at 21:31
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Believable that there is a finite, nonzero, limsup. However, it appears to be a little smaller than $5,$ rather than a little bigger. What needs to be done is Ramanujan's procedure, as used in the Superior Highly Composite numbers. You need to identify, by factorization, what numbers make your quantity especially large. Worth the effort to learn how to do that.

One method, good for computer experiments, is to take the sequence of numbers $L_n = \operatorname{lcm} \{1,2,3, \ldots, n \}.$ As $n$ increases, this increases only when $n$ is a prime or prime power. This is closer to Ramanujan's sequences than the factorials, another sequence that is fairly easy to program. Much closer than the primorials, although worth doing the primorials as well to see what happens.

16 = 2^4    5.34796
18 = 2 * 3^2    6.13628
20 = 2^2 * 5    5.54196
24 = 2^3 * 3    7.03566
30 = 2 * 3 * 5    5.96972
36 = 2^2 * 3^2    6.14368
40 = 2^3 * 5    4.87679
42 = 2 * 3 * 7    4.92994
48 = 2^4 * 3    5.8203
60 = 2^2 * 3 * 5    6.29779
72 = 2^3 * 3^2    5.64003
84 = 2^2 * 3 * 7    5.28204
90 = 2 * 3^2 * 5    4.98691
96 = 2^5 * 3    4.9882
108 = 2^2 * 3^3    4.7654
120 = 2^3 * 3 * 5    5.95407
144 = 2^4 * 3^2    5.07828
168 = 2^3 * 3 * 7    5.09363
180 = 2^2 * 3^2 * 5    5.55753
240 = 2^4 * 3 * 5    5.45543
252 = 2^2 * 3^2 * 7    4.81503
336 = 2^4 * 3 * 7    4.7536
360 = 2^3 * 3^2 * 5    5.50485
420 = 2^2 * 3 * 5 * 7    5.22713
480 = 2^5 * 3 * 5    4.98958
504 = 2^3 * 3^2 * 7    4.81824
540 = 2^2 * 3^3 * 5    4.80776
720 = 2^4 * 3^2 * 5    5.2288
840 = 2^3 * 3 * 5 * 7    5.2981
1080 = 2^3 * 3^3 * 5    4.89648
1260 = 2^2 * 3^2 * 5 * 7    5.12088
1440 = 2^5 * 3^2 * 5    4.9128
1680 = 2^4 * 3 * 5 * 7    5.12874
2160 = 2^4 * 3^3 * 5    4.76805
2520 = 2^3 * 3^2 * 5 * 7    5.30125
3360 = 2^5 * 3 * 5 * 7    4.89001
3780 = 2^2 * 3^3 * 5 * 7    4.74138
5040 = 2^4 * 3^2 * 5 * 7    5.22017
7560 = 2^3 * 3^3 * 5 * 7    4.97266
9240 = 2^3 * 3 * 5 * 7 * 11    4.75076
10080 = 2^5 * 3^2 * 5 * 7    5.0427
12600 = 2^3 * 3^2 * 5^2 * 7    4.82894
15120 = 2^4 * 3^3 * 5 * 7    4.9585
18480 = 2^4 * 3 * 5 * 7 * 11    4.74611
20160 = 2^6 * 3^2 * 5 * 7    4.84389
25200 = 2^4 * 3^2 * 5^2 * 7    4.83633
27720 = 2^3 * 3^2 * 5 * 7 * 11    4.97151
30240 = 2^5 * 3^3 * 5 * 7    4.84175
32760 = 2^3 * 3^2 * 5 * 7 * 13    4.8159
50400 = 2^5 * 3^2 * 5^2 * 7    4.74099
55440 = 2^4 * 3^2 * 5 * 7 * 11    5.00746
65520 = 2^4 * 3^2 * 5 * 7 * 13    4.85518
83160 = 2^3 * 3^3 * 5 * 7 * 11    4.8285
110880 = 2^5 * 3^2 * 5 * 7 * 11    4.92966
131040 = 2^5 * 3^2 * 5 * 7 * 13    4.78542
138600 = 2^3 * 3^2 * 5^2 * 7 * 11    4.74786
166320 = 2^4 * 3^3 * 5 * 7 * 11    4.89272
196560 = 2^4 * 3^3 * 5 * 7 * 13    4.75237
221760 = 2^6 * 3^2 * 5 * 7 * 11    4.80869
277200 = 2^4 * 3^2 * 5^2 * 7 * 11    4.82156
332640 = 2^5 * 3^3 * 5 * 7 * 11    4.84284
360360 = 2^3 * 3^2 * 5 * 7 * 11 * 13    4.8239
554400 = 2^5 * 3^2 * 5^2 * 7 * 11    4.78238
665280 = 2^6 * 3^3 * 5 * 7 * 11    4.74607
720720 = 2^4 * 3^2 * 5 * 7 * 11 * 13    4.91796
831600 = 2^4 * 3^3 * 5^2 * 7 * 11    4.76475
942480 = 2^4 * 3^2 * 5 * 7 * 11 * 17    4.73175
1081080 = 2^3 * 3^3 * 5 * 7 * 11 * 13    4.7739
1441440 = 2^5 * 3^2 * 5 * 7 * 11 * 13    4.89339
1663200 = 2^5 * 3^3 * 5^2 * 7 * 11    4.74345
1801800 = 2^3 * 3^2 * 5^2 * 7 * 11 * 13    4.728
1884960 = 2^5 * 3^2 * 5 * 7 * 11 * 17    4.71243
2162160 = 2^4 * 3^3 * 5 * 7 * 11 * 13    4.88341
2882880 = 2^6 * 3^2 * 5 * 7 * 11 * 13    4.81685
3603600 = 2^4 * 3^2 * 5^2 * 7 * 11 * 13    4.84251
4324320 = 2^5 * 3^3 * 5 * 7 * 11 * 13    4.87406
5405400 = 2^3 * 3^3 * 5^2 * 7 * 11 * 13    4.71395
5765760 = 2^7 * 3^2 * 5 * 7 * 11 * 13    4.72126
6486480 = 2^4 * 3^4 * 5 * 7 * 11 * 13    4.7436
7207200 = 2^5 * 3^2 * 5^2 * 7 * 11 * 13    4.83926
8648640 = 2^6 * 3^3 * 5 * 7 * 11 * 13    4.81098
10810800 = 2^4 * 3^3 * 5^2 * 7 * 11 * 13    4.84049

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$L_n = \operatorname{lcm} \{1,2,3, \ldots, n \}:$

12 =  2^2 3    8.95407
60 =  2^2 3 5    6.29779
420 =  2^2 3 5 7    5.22713
840 =  2^3 3 5 7    5.2981
2520 =  2^3 3^2 5 7    5.30125
27720 =  2^3 3^2 5 7 11    4.97151
360360 =  2^3 3^2 5 7 11 13    4.8239
720720 =  2^4 3^2 5 7 11 13    4.91796
12252240 =  2^4 3^2 5 7 11 13 17    4.81228
232792560 =  2^4 3^2 5 7 11 13 17 19    4.76407
5354228880 =  2^4 3^2 5 7 11 13 17 19 23    4.71132
26771144400 =  2^4 3^2 5^2 7 11 13 17 19 23    4.79067
80313433200 =  2^4 3^3 5^2 7 11 13 17 19 23    4.87704
2329089562800 =  2^4 3^3 5^2 7 11 13 17 19 23 29    4.84324
72201776446800 =  2^4 3^3 5^2 7 11 13 17 19 23 29 31    4.83041
144403552893600 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31    4.90603
5342931457063200 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37    4.88247
219060189739591200 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41    4.86137
9419588158802421600 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41 43    4.85267
442720643463713815200 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41 43 47    4.84455
3099044504245996706400 =  2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47    4.90139
164249358725037825439200 =  2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53    4.89442
9690712164777231700912800 =  2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53 59    4.88279
591133442051411133755680800 =  2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53 59 61    4.87794
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  • $\begingroup$ Very thanks much for your approach and ideas. This night I will try understand your answer and calculations. Thanks for encourage to to study by means of experiments also, if those were your words. $\endgroup$ – user243301 Jul 14 '16 at 5:57
  • $\begingroup$ Very thanks much for your trick, your claim ...it appears to be a little smaller than 5... agree with the numeric value for $\sqrt{2\pi}^{e^{\gamma}}$ Please: 0) Can you explain if there is a typo, I don't understand, why you've labed two sequences as $lcm(1,2\ldots,n)$, it is a typo? I ask if one of those is for factorials; 1) the article that you say is Ramanujan, Highly composite numbers published in 1915 isn't? 2) The trick is the same that the effective construction explained in Wikipedia? $\endgroup$ – user243301 Jul 14 '16 at 19:08
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    $\begingroup$ @user243301 yes, 1915. The problem is that your quantity is not number-theoretic multiplicative, so I would say there is no easy way to duplicate Ramanujan. In attempting to do so, we would consider some real $\delta > 0,$ and replace your exponent denominator, currently $n \log \log n,$ by $n^{1 + \delta}.$ After proving that this quantity takes a maximum, we identify the integer $N_\delta$ where the maximum is achieved. In your case, i do not see how to predict the separate exponents for the prime factors of $N_\delta$ $\endgroup$ – Will Jagy Jul 14 '16 at 19:49
  • $\begingroup$ Currently, it is sufficient to me. You was very kind to help with the question. Very thanks much. $\endgroup$ – user243301 Jul 14 '16 at 19:56
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I ran a better program; it may be difficult to prove all aspects of a Ramanujan type characterization here. However, it does appear, if the largest prime that divides a number is $p,$ then the maximum your quantity can be occurs when the number is a product of primorials. That means $$ 2^{e_2} 3^{e_3} 5^{e_5} 7^{e_7} \cdots p^{e_p}, $$ where $$ e_2 \geq e_3 \geq e_5 \geq \cdots \geq e_p, $$ and $e_p = 1.$

Here are the best i could find for $p = 7, 11, \dots, 31.$ It seems to me that the ratios of exponents are sticking to a pattern, but I am not sure what that is. In some $n!,$ the exponent of a prime factor $p$ is proportional to $1/p.$ In Ramanujan's numbers, the exponent of a prime factor $p$ is proportional to $1/ \log p.$ I cannot tell what types of ratios we are seeing here.

2520 =  2^3 3^2 5 7     5.30125     
p1 210  p2 6  p3 2  p4 1  p5 1  p6 1  p7 1

55440 =  2^4 3^2 5 7 11     5.00746     
p1 2310  p2 6  p3 2  p4 2  p5 1  p6 1  p7 1

720720 =  2^4 3^2 5 7 11 13     4.91796     
p1 30030  p2 6  p3 2  p4 2  p5 1  p6 1  p7 1

367567200 =  2^5 3^3 5^2 7 11 13 17     4.87722     
p1 510510  p2 30  p3 6  p4 2  p5 2  p6 1  p7 1

6983776800 =  2^5 3^3 5^2 7 11 13 17 19     4.92144   
p1 9699690  p2 30  p3 6  p4 2  p5 2  p6 1  p7 1

160626866400 =  2^5 3^3 5^2 7 11 13 17 19 23     4.93625    
p1 223092870  p2 30  p3 6  p4 2  p5 2  p6 1  p7 1

9316358251200 =  2^6 3^3 5^2 7 11 13 17 19 23 29     4.91917     
p1 6469693230  p2 30  p3 6  p4 2  p5 2  p6 2  p7 1

2021649740510400 =  2^6 3^3 5^2 7^2 11 13 17 19 23 29 31     4.93356     
p1 200560490130  p2 210  p3 6  p4 2  p5 2  p6 2  p7 1
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  • $\begingroup$ Very thanks much for your new contribution. $\endgroup$ – user243301 Jul 15 '16 at 5:21

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