# Bound for the sum of the divisors of a number

Let us denote by $s(n) = \sum_{d|n} d$ the sum of divisors of a natural number $n$ ($1$ and $n$ included). If $n$ has at most $5$ distinct prime divisors, prove that $s(n) < \dfrac{77}{16}n$. Also prove that there exists a natural number $n$ for which $s(n) > \dfrac{76}{16}n$ holds.

Attempt:

Let $n = p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}p_4^{\alpha_4}p_5^{\alpha_5}$ where the $p_i$ are primes. The sum of the divisors is $$(1+p_1+\cdots+p_1^{\alpha_1})(1+p_2+\cdots+p_2^{\alpha_2})\cdots(1+p_5+\cdots+p_5^{\alpha_5})=\frac{(p_1^{\alpha_1+1}-1)(p_2^{\alpha_2+1}-1)\cdots(p_5^{\alpha_5+1}-1)}{(p_1-1)(p_2-1)\cdots(p_5-1)}.$$ Then we see that $\dfrac{s(n)}{n}=\prod_{i=1}^5 \dfrac{p_i^{\alpha_i+1}-1}{p_i^{\alpha_i}(p_i-1)}$. But then $$\dfrac{p_i^{\alpha_i+1}-1}{p^\alpha_i(p_i-1)} = \dfrac{p_i-\dfrac{1}{p_i^{\alpha_i}}}{p_i-1}<\dfrac{p_i}{p_i-1}=1+\dfrac{1}{p_i-1} = \dfrac{p_i}{p_i-1}.$$ Since the last expression is decreasing for all $p_i$, we have

$$\prod_{i=1}^5 \frac{p_i^{\alpha_i+1}}{p_i^{\alpha_i}(p_i-1)} < \prod_{i=1}^5 \dfrac{p_i}{p_i-1} \le \dfrac{2}{1}\cdot \dfrac{3}{2}\cdot \dfrac{5}{3}\cdot \dfrac{7}{6}\cdot \dfrac{11}{10}=\dfrac{77}{16}.$$

If we have less than $5$ primes, notice that by repeating the same process we arrive at smaller bounds. If $n = 1$ then clearly our bound holds.

How do we find an $n$ such that $s(n) > \dfrac{76}{16}n$?

• How do we find an n such that s(n)>76n/16? ... By allowing more than 5 prime factors? Jun 10, 2016 at 20:13
• Can't we make 1/p^a_i arbitrarily close to 0 and conclude we can find a_i high enough for numbers with 6 prime factors we can get s(n)/n arebitrarily close to 77*13/16*12? Jun 10, 2016 at 20:56

This is from https://oeis.org/A004394 and https://oeis.org/A004394/b004394.txt and, for locating the first ratio larger than $4.75,$ the zipped file where it says:

T. D. Noe, First 1000000 superabundant numbers (21 MB, zipped)

which gave the line

    38   4.7788655788655789         7.0338578330 S {13,5,3,2}


where the 7.03 is logarithm base ten of the number, meaning a little over ten million, and the final {} is a very brief summary of the factorization, adapted for numbers that are (must be) products of primorials.

and the first text file by Noe gave

38 10810800

Then I checked it myself, and it works as advertised,

 10810800 = 2^4 3^3 5^2 7 11 13
51663360 = 2^9 3 5 7 31^2
ratio  4.778865578865579


The largest item from the Noe list that is small enough to be accepteb by my computer without extra work, i.e. less than $2^{31},$ is

    51   5.2376106865270952         9.1451203466 S {19,0,3,0,2}

1396755360 2^5 3^3 5 7 11 13 17 19
7315660800 2^13 3^6 5^2 7^2
ratio  5.237610686527095
log ten of the number  9.145120346625335


Hmm. Now that I think of it, there is no real difficulty inputting on of Noe's numbers using his factoring summary, the largest prime he ever uses is 237173. Which is big, but still

Why not? Here is the beginning of the giant Noe file

// This file contains information about the first 1000000 superabundant
// numbers (SA numbers).  For each SA number n, we give its abundance,
// which is Sigma(n)/n, its base-10 logarithm, and its factorization.
// In column "*", a "C" indicates that the number is also a colossally
// abundant number  The factorization is given in a very compact form.
// For example, {13,5,0,2} means 13 * 11 * 7 * 5^2 * 3^2 * 2^4.
//
// Created by Tony D. Noe, [email protected] on 15-Oct-2009.
// Algorithm developed with help from Devin Kilminster.
//
// Corrected 30-Oct-2009: some SA numbers were erroneously marked with a "C".
//
// position       abundance                log10     *   factorization
//
1   1.0000000000000000         0.0000000000 S {0}
2   1.5000000000000000         0.3010299957 C {2}
3   1.7500000000000000         0.6020599913 S {0,2}
4   2.0000000000000000         0.7781512504 C {3}
5   2.3333333333333333         1.0791812460 C {3,2}
6   2.5000000000000000         1.3802112417 S {3,0,2}
7   2.5277777777777778         1.5563025008 S {0,3}
8   2.5833333333333333         1.6812412374 S {3,0,0,2}
9   2.8000000000000000         1.7781512504 C {5,2}
10   3.0000000000000000         2.0791812460 C {5,0,2}
11   3.0333333333333333         2.2552725051 S {5,3}
12   3.1000000000000000         2.3802112417 S {5,0,0,2}
13   3.2500000000000000         2.5563025008 C {5,3,2}
14   3.3583333333333333         2.8573324964 S {5,3,0,2}
15   3.4285714285714286         2.9242792861 S {7,0,2}
16   3.4666666666666667         3.1003705451 S {7,3}
17   3.5428571428571429         3.2253092817 S {7,0,0,2}
18   3.7142857142857143         3.4014005408 C {7,3,2}
19   3.8380952380952381         3.7024305364 C {7,3,0,2}
20   3.9000000000000000         4.0034605321 S {7,3,0,0,2}
21   3.9365079365079365         4.1795517912 S {7,0,3,2}
22   3.9660317460317460         4.4014005408 S {7,5,0,2}
23   4.0519480519480520         4.4427932259 S {11,3,2}
24   4.1870129870129870         4.7438232216 C {11,3,0,2}
25   4.2545454545454546         5.0448532173 S {11,3,0,0,2}
26   4.2943722943722944         5.2209444763 S {11,0,3,2}
27   4.3265800865800866         5.4427932259 S {11,5,0,2}
28   4.3636363636363636         5.5219744720 S {11,0,3,0,2}
29   4.3963636363636364         5.7438232216 S {11,5,0,0,2}
30   4.3982683982683983         5.8230044677 S {11,0,3,0,0,2}
31   4.5090909090909091         5.8577665739 C {13,3,0,2}
32   4.5818181818181818         6.1587965696 C {13,3,0,0,2}
33   4.6247086247086247         6.3348878286 S {13,0,3,2}
34   4.6593939393939394         6.5567365782 S {13,5,0,2}
35   4.6993006993006993         6.6359178243 C {13,0,3,0,2}
36   4.7345454545454546         6.8577665739 S {13,5,0,0,2}
37   4.7365967365967366         6.9369478200 S {13,0,3,0,0,2}
38   4.7788655788655789         7.0338578330 S {13,5,3,2}
39   4.8559440559440559         7.3348878286 C {13,5,3,0,2}
40   4.8967503085150144         7.5653367500 S {17,0,3,2}
41   4.9334759358288770         7.7871854996 S {17,5,0,2}
42   4.9757301522007404         7.8663667457 S {17,0,3,0,2}
43   5.0130481283422460         8.0882154953 S {17,5,0,0,2}
44   5.0152200740436035         8.1673967413 S {17,0,3,0,0,2}
45   5.0599753187988482         8.2643067543 S {17,5,3,2}
46   5.1415878239407651         8.5653367500 C {17,5,3,0,2}
47   5.1544740089631731         8.8440903510 S {19,0,3,2}
48   5.1823940765117236         8.8663667457 S {17,5,3,0,0,2}
49   5.1844343891402715         9.0424580047 S {17,5,0,3,2}
50   5.1931325640303969         9.0659391006 S {19,5,0,2}
51   5.2376106865270952         9.1451203466 S {19,0,3,0,2}
52   5.2768927666760484         9.3669690962 S {19,5,0,0,2}
53   5.2791790253090563         9.4461503423 S {19,0,3,0,0,2}
54   5.3262898092619455         9.5430603553 S {19,5,3,2}
55   5.4121977094113317         9.8440903510 C {19,5,3,0,2}
56   5.4551516594860248        10.1451203466 S {19,5,3,0,0,2}
57   5.4572993569897595        10.3212116057 S {19,5,0,3,2}
58   5.4766286345233714        10.4461503423 S {19,5,3,0,0,0,2}
59   5.5006112566484084        10.6222416013 S {19,5,0,3,0,2}
60   5.5088440970793912        10.6891883910 S {19,7,3,0,2}
61   5.5578676270559431        10.9047881913 S {23,5,3,2}
62   5.6475106532987809        11.2058181870 C {23,5,3,0,2}
63   5.6923321664201998        11.5068481826 C {23,5,3,0,0,2}
64   5.6945732420762708        11.6829394417 S {23,5,0,3,2}
65   5.7147429229809093        11.8078781783 S {23,5,3,0,0,0,2}
66   5.7397682678070348        11.9839694374 S {23,5,0,3,0,2}
67   5.7483590578219734        12.0509162270 S {23,7,3,0,2}


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

with the corresponding numbers in decimal

# The comments at the end of this file extend the table to all superabundant
# numbers less than 10^1200.  For compactness, these are written in terms of
# factorials (A000142), k!, and primorials (A034386), p#.
1 1
2 2
3 4
4 6
5 12
6 24
7 36
8 48
9 60
10 120
11 180
12 240
13 360
14 720
15 840
16 1260
17 1680
18 2520
19 5040
20 10080
21 15120
22 25200
23 27720
24 55440
25 110880
26 166320
27 277200
28 332640
29 554400
30 665280
31 720720
32 1441440
33 2162160
34 3603600
35 4324320
36 7207200
37 8648640
38 10810800
39 21621600
40 36756720
41 61261200
42 73513440
43 122522400
44 147026880
45 183783600
46 367567200
47 698377680
48 735134400
49 1102701600
50 1163962800
51 1396755360
52 2327925600
53 2793510720
54 3491888400
55 6983776800
56 13967553600
57 20951330400
58 27935107200
59 41902660800
60 48886437600
61 80313433200
62 160626866400
63 321253732800
64 481880599200
65 642507465600
66 963761198400
67 1124388064800


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

Curious how it works out using just primorials as in the other answer,

2 2
3 3
ratio  1.5
log ten of the number  0.3010299956639812
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
6 2 3
12 2^2 3
ratio  2
log ten of the number  0.7781512503836437
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
30 2 3 5
72 2^3 3^2
ratio  2.4
log ten of the number  1.477121254719663
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
210 2 3 5 7
576 2^6 3^2
ratio  2.742857142857143
log ten of the number  2.32221929473392
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
2310 2 3 5 7 11
6912 2^8 3^3
ratio  2.992207792207792
log ten of the number  3.363611979892144
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
.....omit several before ratio 4.75
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
557940830126698960967415390 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
2660857705190196806418432000 2^38 3^15 5^3 7^2 11 17 19 31
ratio  4.769067903824068
log ten of the number  26.74658814425655
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=


An easy program, as long as you can use oversize integers: just take $LCM(1,2,3,...,n)$ as your sequence of numbers. This only increases when $n$ is prime or a prime power. The numbers have a similar balance of factors to either the Superior Highly Composite Numbers or the Colossally Abundant Numbers. As a result, we get ratio exceeding 4.75 quite early, very little effort

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
2 2
3 3
ratio  1.5
log ten of the number  0.3010299956639812
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
6 2 3
12 2^2 3
ratio  2
log ten of the number  0.7781512503836437
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
12 2^2 3
28 2^2 7
ratio  2.333333333333333
log ten of the number  1.079181246047625
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
60 2^2 3 5
168 2^3 3 7
ratio  2.8
log ten of the number  1.778151250383644
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
420 2^2 3 5 7
1344 2^6 3 7
ratio  3.2
log ten of the number  2.623249290397901
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
840 2^3 3 5 7
2880 2^6 3^2 5
ratio  3.428571428571428
log ten of the number  2.924279286061882
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
2520 2^3 3^2 5 7
9360 2^4 3^2 5 13
ratio  3.714285714285714
log ten of the number  3.401400540781545
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
27720 2^3 3^2 5 7 11
112320 2^6 3^3 5 13
ratio  4.051948051948052
log ten of the number  4.44279322593977
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
360360 2^3 3^2 5 7 11 13
1572480 2^7 3^3 5 7 13
ratio  4.363636363636363
log ten of the number  5.556736578246606
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
720720 2^4 3^2 5 7 11 13
3249792 2^7 3^2 7 13 31
ratio  4.509090909090909
log ten of the number  5.857766573910587
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
12252240 2^4 3^2 5 7 11 13 17
58496256 2^8 3^4 7 13 31
ratio  4.774331550802139
log ten of the number  7.088215495288862
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
232792560 2^4 3^2 5 7 11 13 17 19
1169925120 2^10 3^4 5 7 13 31
ratio  5.025612158739094
log ten of the number  8.366969096241691
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=


You have proven that $n$ has to have at least $6$ distinct prime divisors. From your work it should be clear that for prime $p$, $\frac {s(p^n)}{p^n}$ is decreasing with $n$, so you want $n$ to be the product of the first powers of primes. The natural choice is to keep multiplying primes $p_i$ until $\prod \frac {p_i+1}{p_i} \gt \frac {76}{16}$ Alpha finds the product up to $i=20$ to be generous: $\prod_{i=1}^{20} \frac {p_i+1}{p_i} \gt 4.769\gt \frac {76}{16}$

• $s(p^n)/p^n = 1 + 1/p + \ldots + 1/p^n$ is increasing with $n$, but it doesn't increase very much, so it's still reasonable to use a product of distinct primes if you don't need the smallest possible solution. Jun 10, 2016 at 19:27
• @RobertIsrael the smallest is in an OEIS file by T. D. Noe, I put all in an answer, and just added the best primorial Jun 10, 2016 at 19:30
• And it should be $\prod_i \frac{p_i+1}{p_i}$, not $\prod_i \frac{p_i}{p_i-1}$. This is indeed first $> 76/16$ for the product up to $i=20$. Jun 10, 2016 at 19:36
• @RobertIsrael Where are you getting $\prod_{i} \dfrac{p_i+1}{p_i}$ from? Jun 10, 2016 at 19:49
• $s(p) = p+1$ if $p$ is prime. Jun 10, 2016 at 20:01

Let $n = \prod_i^6 p_i^k$.

You've pointed out that $s(n)/n = \prod_i^6 \frac{p_i - \frac 1{p_i^k}}{p_i -1} < \prod_i^6 \frac{p_i }{p_i -1}=77*13/16*12$.

But we can make $k$ arbitrarily large so for any $\epsilon > 0$ in particular any epsilon $77/16 < 77*13/16*12 - \epsilon$ we can find $k$ so that $77*13/16*12 >s(n)/n = \prod_i^6 \frac{p_i - \frac 1{p_i^k}}{p_i -1} > 77*13/16*12 - \epsilon$.

I figure, but haven't varified, that as

$\prod_i^6\frac{p_i}{p_i - 1} = 77*13/16*12 = 1001/192=5.213$

$\prod_i^6\frac{1 + p_i}{p_i} = 32256/30030= 1.0741$

$\log_{1.0741} 5.213 = 23.01$

that $k = 24$ ought to be large enough the $n = \prod_i^6 p_i^{24}$ will satisfy.

Not sure if my reasoning for $k =24$ is valid but if not some larger $k$ will do.

• As you are reading the question, the limit is five prime factors, so the upper limit for $i$ should be $5$, not $6$. Your reasoning is correct that we can get arbitrarily close to $\frac {77}{16}$ with a high enough power of $11$ primorial $=25410$. I haven't checked $k=24$ Jun 11, 2016 at 0:28
• With five you can not get higher than 77/16 as was proven in in the OP. So to get more than 77/16 you do need at least six. By the same argument as the the limit for six factors would be 77*13/12*16 > 77*13/12*16. The OP'S question was how can we know there is any number where the result is more than 77/16, the upper limit of 5 factor. My point is that as the upper limit for 6 is greater than the upper limit for 5 is enough to conclude it does as we can get arbitrarily case to the limit. And the limits get larger with more factors. Jun 11, 2016 at 2:32
• No, the upper limit has to be more than five as s (n)/n is bounded above by 77/16 for 5 factors. To get above 77/16 we need at least 6 factors. Then the upper limit 77*13/12*16. The question was can we find any n where s (n)/n > 77*13/12*16. My post is to say they we can get arbitrarily close to any upper limit by higher powers. As the upper limits increase with factors. The answer is yes, we can find s(n)/n > 77/16 if there are 6 factors. Likewise we can find > 77.13/12.16 if there are more than six factors and so on. Jun 11, 2016 at 2:41
• I'm not trying to get arbitrarily close to 17/16. I'm trying to get distinctly higher than 17/16. And to get higher than 77/16, I must have at least 6 factors. With six factors, I can get arbitrarily close to (but not higher than) 1001/192. But arbitrarily close to 1001/192 is higher than 77/16 so that is enough. Jun 11, 2016 at 2:57

Worth a separate answer; it follows from Ramanujan's work that when $$n = \operatorname{lcm} \{1,2,3, \ldots, N \}$$ then $s(n)/n$ is quite large; this is a very efficient way to find large values of the ratio, and so on. Note that the Prime Number Theorem says that $$n \approx e^N.$$ I had the computer print out the logarithm base ten. Just multiply that by 2.302585 and compare with $N.$ Meanwhile, this produces something new every time $N$ is a prime or prime power. To keep to a reasonable printout, I told it to print out only when the ratio $s(n)/n$ passes an integer.

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 4 2^2
logten 1.07918
ratio decimal  2.333333333333333
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 7 7
logten 2.6232492903979
ratio decimal  3.2
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 11 11
logten 4.442793225939769
ratio decimal  4.051948051948052
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 19 19
logten 8.366969096241691
ratio decimal  5.025612158739094
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 32 2^5
logten 14.15957787871218
ratio decimal  6.030712154801813
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 61 61
logten 26.77168552928217
ratio decimal  7.107376448297107
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 103 103
logten 44.86051831427165
ratio decimal  8.046547557699006
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 169 13^2
logten 73.61952039853439
ratio decimal  9.002387229747514
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 289 17^2
logten 127.4880844692016
ratio decimal  10.00473574124096
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 509 509
logten 223.2730318634646
ratio decimal  11.01536677911805
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 881 881
logten 383.7686828082957
ratio decimal  12.01079052806564
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 1531 1531
logten 665.0185935210201
ratio decimal  13.00272756436563
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 2671 2671
logten 1153.297364165875
ratio decimal  14.00190242107617
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 4639 4639
logten 2010.976688476967
ratio decimal  15.00184149854144
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 8089 8089
logten 3507.89371950155
ratio decimal  16.00131709994758
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 14107 14107
logten 6130.602439140246
ratio decimal  17.00023340649911
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 24649 157^2
logten 10714.01752999065
ratio decimal  18.0002306740445
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 43063 43063
logten 18753.79867942667
ratio decimal  19.00021460572149
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 75617 75617
logten 32840.3415295448
ratio decimal  20.00025815915193
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 132437 132437
logten 57505.30935799197
ratio decimal  21.00007191088712
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 231719 231719
logten 100737.0004955105
ratio decimal  22.00005484738577
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 406633 406633
logten 176517.1636552099
ratio decimal  23.00005603637724
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 712169 712169
logten 309347.9227482136
ratio decimal  24.00000712886767
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
N 1248407 1248407
logten 542185.3095075123
ratio decimal  25.00000283110379
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=