Abundant But Not Zumkeller

A property of the number \( \textbf{28062}\) associated with my diurnal age today prompted me to look more closely at abundant numbers that are not Zumkeller. 28062 has the following properties:$$ \begin{align} 28062 &=2 \times 3^2 \times 1559\\ \text{ divisors }  &\rightarrow 1, 2, 3, 6, 9, 18, 1559, 3118, 4677, 9354, 14031, 28062 \end{align}$$Normally, for abundant numbers, the set of divisors can be divided into two mutually exclusive sets whose elements sum to the same number. However, this simply can't be done with 28062.

Now up to 40000, there are 718 abundant numbers that are abundant but not Zumkeller. 146 of these have an odd sum of divisors and an even split is not possible. The other 572 have an even sum of divisors and thus have the potential for an even split but it proves impossible to find one. Here are the details:

There are 572 abundant numbers with an \( \textbf{even}\) sum of divisors. They are (up to 40000):

[738, 748, 774, 846, 954, 1062, 1098, 1206, 1278, 1314, 1422, 1494, 1602, 1746, 1818, 1854, 1926, 1962, 2034, 2286, 2358, 2466, 2502, 2682, 2718, 2826, 2934, 3006, 3114, 3222, 3258, 3438, 3474, 3492, 3546, 3582, 3636, 3708, 3798, 3852, 3924, 4014, 4068, 4086, 4122, 4194, 4302, 4338, 4518, 4572, 4626, 4716, 4734, 4842, 4878, 4932, 4986, 5004, 5058, 5094, 5274, 5364, 5436, 5526, 5598, 5634, 5652, 5706, 5868, 5958, 6012, 6066, 6228, 6246, 6282, 6354, 6444, 6462, 6516, 6606, 6714, 6822, 6876, 6894, 6948, 7002, 7092, 7146, 7164, 7218, 7362, 7542, 7544, 7578, 7596, 7758, 7794, 7902, 7974, 8028, 8082, 8172, 8226, 8244, 8298, 8334, 8388, 8406, 8604, 8622, 8676, 8766, 8838, 8982, 9036, 9054, 9162, 9252, 9378, 9414, 9468, 9684, 9738, 9756, 9846, 9972, 10026, 10116, 10134, 10184, 10188, 10242, 10278, 10386, 10548, 10566, 10674, 10782, 10818, 10926, 11034, 11052, 11106, 11142, 11196, 11268, 11358, 11412, 11538, 11574, 11646, 11754, 11862, 11898, 11916, 12114, 12132, 12186, 12294, 12438, 12492, 12564, 12618, 12708, 12762, 12924, 12942, 13086, 13194, 13212, 13302, 13374, 13428, 13518, 13626, 13644, 13698, 13788, 13842, 13914, 14004, 14166, 14184, 14292, 14328, 14346, 14436, 14562, 14598, 14724, 14778, 14814, 14886, 14922, 15084, 15102, 15156, 15192, 15354, 15426, 15462, 15516, 15534, 15588, 15786, 15804, 15858, 15894, 15948, 15966, 16056, 16164, 16326, 16344, 16398, 16452, 16488, 16542, 16596, 16668, 16722, 16776, 16812, 16866, 16938, 17046, 17154, 17208, 17244, 17352, 17406, 17478, 17532, 17586, 17676, 17694, 17838, 17946, 17964, 18072, 18108, 18162, 18234, 18324, 18342, 18378, 18504, 18558, 18594, 18702, 18756, 18828, 18882, 18918, 18936, 19098, 19134, 19242, 19368, 19476, 19512, 19566, 19638, 19674, 19692, 19746, 19854, 19944, 19962, 20052, 20106, 20214, 20232, 20268, 20322, 20376, 20484, 20556, 20718, 20754, 20772, 20934, 21078, 21096, 21132, 21258, 21348, 21366, 21474, 21564, 21618, 21636, 21834, 21852, 21906, 22014, 22068, 22104, 22122, 22158, 22212, 22266, 22284, 22300, 22392, 22482, 22536, 22662, 22700, 22716, 22824, 22900, 22986, 23022, 23076, 23094, 23148, 23202, 23238, 23292, 23300, 23346, 23418, 23454, 23508, 23526, 23724, 23742, 23778, 23796, 23832, 23886, 23900, 24100, 24228, 24264, 24372, 24498, 24588, 24606, 24714, 24858, 24876, 24984, 25100, 25128, 25182, 25236, 25362, 25416, 25524, 25614, 25686, 25700, 25722, 25794, 25848, 25884, 25902, 26046, 26118, 26154, 26172, 26262, 26300, 26388, 26424, 26478, 26604, 26658, 26694, 26748, 26766, 26802, 26856, 26874, 26900, 26982, 27036, 27100, 27198, 27252, 27288, 27396, 27414, 27558, 27576, 27684, 27700, 27774, 27828, 27882, 27954, 28008, 28062, 28100, 28206, 28278, 28300, 28332, 28422, 28494, 28584, 28692, 28746, 28818, 28872, 28926, 28962, 29034, 29124, 29142, 29178, 29196, 29286, 29300, 29448, 29466, 29556, 29628, 29772, 29826, 29844, 29934, 30006, 30042, 30168, 30204, 30312, 30474, 30546, 30582, 30700, 30708, 30762, 30852, 30924, 30978, 31014, 31032, 31068, 31100, 31176, 31194, 31300, 31338, 31446, 31554, 31572, 31608, 31662, 31700, 31716, 31788, 31896, 31932, 31986, 32094, 32166, 32202, 32328, 32418, 32598, 32652, 32796, 32814, 32904, 32958, 33084, 33100, 33192, 33246, 33336, 33444, 33498, 33606, 33624, 33678, 33700, 33714, 33732, 33786, 33822, 33876, 34002, 34092, 34218, 34308, 34326, 34434, 34488, 34700, 34758, 34794, 34812, 34900, 34956, 35064, 35082, 35118, 35172, 35300, 35352, 35388, 35514, 35622, 35676, 35766, 35874, 35892, 35900, 35928, 35946, 35982, 36054, 36198, 36216, 36306, 36324, 36468, 36486, 36522, 36648, 36684, 36700, 36702, 36756, 36954, 37116, 37134, 37188, 37242, 37300, 37404, 37458, 37494, 37512, 37566, 37602, 37656, 37764, 37782, 37836, 37900, 37998, 38034, 38196, 38268, 38300, 38322, 38358, 38466, 38484, 38538, 38574, 38754, 38898, 38900, 38952, 39132, 39222, 39276, 39348, 39384, 39492, 39654, 39700, 39708, 39726, 39834, 39924, 39978]

There are 146 abundant numbers with an \( \textbf{odd}\) sum of divisors. These are up to 40000:

[18, 36, 72, 100, 144, 162, 196, 200, 288, 324, 392, 400, 450, 576, 648, 784, 800, 882, 900, 968, 1152, 1296, 1352, 1458, 1568, 1600, 1764, 1800, 1936, 2178, 2304, 2450, 2500, 2592, 2704, 2916, 3042, 3136, 3200, 3528, 3600, 3872, 4050, 4356, 4608, 4624, 4900, 5000, 5184, 5202, 5408, 5776, 5832, 6050, 6084, 6272, 6400, 6498, 7056, 7200, 7744, 7938, 8100, 8450, 8464, 8712, 9216, 9248, 9522, 9604, 9800, 10000, 10368, 10404, 10816, 11025, 11250, 11552, 11664, 12100, 12168, 12544, 12800, 12996, 13122, 13456, 14112, 14400, 15138, 15376, 15488, 15876, 16200, 16900, 16928, 17298, 17424, 18432, 18496, 19044, 19208, 19600, 19602, 20000, 20736, 20808, 21632, 22050, 22500, 23104, 23328, 23716, 24200, 24336, 24642, 25088, 25600, 25992, 26244, 26912, 27378, 28224, 28800, 28900, 30258, 30276, 30752, 30976, 31752, 32400, 33124, 33282, 33800, 33856, 34596, 34848, 36100, 36450, 36864, 36992, 38088, 38416, 39200, 39204, 39762, 40000]

The total number is 718

Note how I've marked 11025 in red. This is because it is the only odd number with an odd number of divisors. We have:$$11025 = 3^2 \times 5^2 \times 7^2 = 105^2 \text{ with 27 divisors}$$The next such number is 99225 where:$$99225=3^4 \times 5^2 \times 7^2 = 315^2 \text{ with 243 divisors}$$

Let's not confuse oddness and evenness of the divisor sums with the oddness and evenness of the numbers themselves. The majority of Zumkeller numbers are even. To quote from an earlier blog of mine:

In the range up to 100,000 there are 24362 even Zumkeller numbers comprising 24.362% of the range. However, there are only 208 odd Zumkeller numbers in that range, comprising 0.208%.

Number's Divisors to Sequence Algorithm

A logical progression to looking at the sequence produced by considering the factors of a number (see Number's Factors to Sequence Algorithm 1 and Number's Factors to Sequence Algorithm 2) is to consider the sequence formed by its divisors. The rules this time around are very similar to that for factors except that we are dealing now with the number's divisors.

 Suppose we take any positive integer \(n \gt 1\) 

• if prime, double it and add 1: \(n \rightarrow 2n+1\)
  
• if composite, determine its number of divisors \(d\)
  
• if \( n \pmod d \equiv 0\) then \(n \rightarrow \dfrac{n}{d} \)
  
• if \( n \pmod d \not\equiv 0 \) then \(n \rightarrow n \times d\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

Let's use 28059 as an example. Applying the above rules leads to the following sequence:

28059, 224472, 7183104, 517183488, 2693664, 37412, 448944, 17957760, 140295, 2244720, 28059

We end up right where we started. Here the details with number of divisors shown (permalink):

28059 --> 8

224472 --> 32

7183104 --> 72

517183488 --> 192

2693664 --> 72

37412 --> 12

448944 --> 40

17957760 --> 128

140295 --> 16

2244720 --> 80

28059 --> 8

As before we are interested in record lengths and my algorithm was not up to the job of determing these so I had to call on Gemini for help. It came up with the following record breaking numbers in the range up to one million (permalink). This algorithm proved to be faulty. For the correction see blog post titled "A Correction".

2, 3, 6, 11, 22, 44, 50, 99, 125, 206, 350, 463, 487, 974, 1375, 1573, 1625, 5200, 14157, 16879, 18747, 39325, 89237, 151911, 563553, 803133

Here are the details of the lengths:

Number     | Length     | Status
-----------------------------------
2          | 5          | New Record!     
3          | 8          | New Record!     
6          | 10         | New Record!     
11         | 21         | New Record!     
22         | 23         | New Record!     
44         | 25         | New Record!     
50         | 28         | New Record!     
99         | 32         | New Record!     
125        | 33         | New Record!     
206        | 34         | New Record!     
350        | 37         | New Record!     
463        | 44         | New Record!     
487        | 46         | New Record!     
974        | 48         | New Record!     
1375       | 51         | New Record!     
1573       | 52         | New Record!     
1625       | 60         | New Record!     
5200       | 62         | New Record!     
14157      | 63         | New Record!     
16879      | 64         | New Record!     
18747      | 67         | New Record!     
39325      | 70         | New Record!     
89237      | 71         | New Record!     
151911     | 75         | New Record!     
563553     | 77         | New Record!     
803133     | 82         | New Record!  

Let's look at the sequence for the last number in the above list, 803133 (permalink):

803133, 4818798, 77100768, 1606266, 19275192, 616806144, 8566752, 356948, 2141688, 34267008, 1070844, 89237, 178475, 3212550, 346955400, 149884732800, 115651800, 321255, 11565180, 64251, 1156518, 69391080, 19984631040, 23130360, 5551286400, 6425100, 1040866200, 524596564800, 231303600, 514008, 7139, 42834, 1028016, 92521440, 257004, 13878216, 1998463104, 5204331, 218581902, 41967725184, 48573756, 10491931296, 16191252, 2914425360, 3469554, 249807888, 59953893120, 61680960, 20724802560, 15700608, 4019355648, 7850304, 35046, 1121472, 125604864, 356832, 3717, 44604, 2140992, 299738880, 555072, 6608, 132160, 2360, 37760, 1180, 14160, 354, 2832, 56640, 3171840, 19824, 792960, 6195, 99120, 1239, 9912, 317184, 22837248, 118944, 1652, 19824

Like the previous sequences using the number of factors, this sequence using the number of divisors is also \( \textbf{base independent}\).

Number's Factors to Sequence Algorithm 2

 Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if prime, double it and add 1: \(n \rightarrow 2n+1\)
  
• if composite, determine its number of factors \(f\) counted \( \textbf{without} \) multiplicity
  
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
  
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. This process is exactly the same as in my previous post except that the number of factors is counted without multiplicity. Let's apply this algorithm to 28059. The result is the sequence 28059, 9353, 18706, 56118, 224472, 56118. Here are the details (permalink):

• \(28059 = 3 \times 47 \times 199\) with three factors
  3 divides 28059 to give 9353
  
  
• \(9353 = 47 \times 199\) with two factors but 2 doesn't divide 9353
  multiplying by 2 gives 18706
  
  
• \(18706 = 2 \times 47 \times 199\) with three factors but 3 doesn't divide 18706
  multiplying by 3 gives 56118
  
  
• \(56118 = 2 \times 3 \times 47 \times 199\) with four factors but 4 doesn't divide 56118
  multiplying by 4 gives 224472
  
  
• \(224472 = 2^3 \times 3 \times 47 \times 199\) with four distinct prime factors
  4 divided into 224472 gives 56118
  
  
• \(56118\) occurred earlier in the sequence and so we have a loop

The fluctuations between terms are less extreme and the record breaking numbers this time around are 2, 3, 6, 12, 24, 31, 62, 93, 139, 278, 417, 1251, 3753, 8896, 17792, 18433, 36866, 55299, 165897, 248851, 497702, 746553.

Here are the sequence lengths of these record breakers (permalink) up to one million:

2 --> 10

3 --> 14

6 --> 15

12 --> 16

24 --> 17

31 --> 18

62 --> 19

93 --> 21

139 --> 23

278 --> 24

417 --> 26

1251 --> 27

3753 --> 28

8896 --> 29

17792 --> 30

18433 --> 31

36866 --> 32

55299 --> 34

165897 --> 35

248851 --> 37

497702 --> 38

746553 --> 40

When the algorithm ran with multiplicity of factors being counted, the maximum sequence length up to one million was 155. The trajectory of the last number in the previous list (746553) is as follows:

746553, 1493106, 497702, 248851, 497703, 995406, 331802, 165901, 331803, 663606, 221202, 73734, 24578, 12289, 24579, 49158, 16386, 5462, 2731, 5463, 10926, 3642, 1214, 607, 1215, 2430, 810, 270, 90, 30, 10, 5, 11, 23, 47, 95, 190, 570, 2280, 570

Figure 1 shows a graph of the trajectory of 746553 using a logarithmic \(y\) scale.

Figure 1

Number's Factors to Sequence Algorithm 1

Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if prime, double it and add 1: \(n \rightarrow 2n+1\)
  
• if composite, determine its number of factors \(f\) counted with multiplicity
  
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. Let's use 28058 as an example. The sequence generated is 28058, 14029, 28059, 9353, 18706, 56118, 224472, 37412, 9353 and the details are as follows:

• \(28058 = 2 \times 14029\) and there are two factors
  2 divides 28056 to give 14209
  
  
• \(14029\) is prime
  multiplying by 2 and adding 1 we get 28059
  
  
• \(28059 = 3 \times 47 \times 199\) and there are three factors
  3 divides 28059 to give 9353
  
  
• \(9353 = 47 \times 199\) and there are two factors but 2 doesn't divide 9353
  multiplying by 2 gives 18706
  
  
• \(18706 = 2 \times 47 \times 199\) and there are three factors but 3 doesn't divide 18706
  multiplying by 3 gives 56118
  
  
• \(56118 = 2 \times 3 \times 47 \times 199\) and there are four factors but 4 doesn't divide 56118
  multiplying by 4 gives 224472
  
  
• \(224472 = 2^3 \times 3 \times 47 \times 199\) and there are six factors (with multiplicity)
  6 divides 224472 to give 37412
  
  
• \(37412 = 2^2 \times 47 \times 199\) and there are four factors (with multiplicity)
  4 divides 37412 to give 9353
  
  
• \(9353\) occurred earlier in the sequence and so we have a loop

Figure 1 shows the trajectory.

Figure 1

Some numbers return to their starting points. 27056 is one such number. It's sequence is 28056, 4676, 1169, 2338, 7014, 28056. What appeals to me about this sequence is that it is \( \textbf{base independent}\). Here is permalink to generate the sequence of any number entered into it.

An investigation into what numbers produced sequences of record lengths returned the following number in the range up to one million:

2, 3, 6, 8, 13, 19, 38, 57, 76, 304, 1024, 1579, 2401, 3584, 10331, 12119, 12500, 15379, 24251, 30689, 48661, 57122, 66749, 116603, 155201, 232801, 465602, 698403, 931204

The final number in the list (931204) produces a sequence of length 155. Here are the full details for all the numbers in the list (permalink):

2 --> 11

3 --> 15

6 --> 16

8 --> 18

13 --> 20

19 --> 24

38 --> 25

57 --> 27

76 --> 29

304 --> 32

1024 --> 34

1579 --> 40

2401 --> 43

3584 --> 49

10331 --> 51

12119 --> 53

12500 --> 61

15379 --> 64

24251 --> 65

30689 --> 66

48661 --> 69

57122 --> 92

66749 --> 105

116603 --> 145

155201 --> 146

232801 --> 150

465602 --> 151

698403 --> 153

931204 --> 155

The sequence for 931204 is as follows: 

931204, 2793612, 698403, 1396806, 465602, 232801, 465603, 931206, 310402, 155201, 310403, 1241612, 7449672, 931209, 4656045, 27936270, 223490160, 2458391760, 204865980, 1843793820, 167617620, 16761762, 134094096, 1475035056, 122919588, 13657732, 95604124, 764832992, 69530272, 695302720, 8343632640, 556242176, 7231148288, 516510592, 6198127104, 92971906560, 5468935680, 341808480, 28484040, 256356360, 2819919960, 234993330, 26110370, 182772590, 1462180720, 132925520, 13292552, 1661569, 8307845, 49847070, 398776560, 4386542160, 365545180, 3289906620, 299082420, 29908242, 239265936, 2631925296, 219327108, 1973943972, 179449452, 1794494520, 149541210, 16615690, 2373670, 14242020, 113936160, 1253297760, 104441480, 939973320, 85452120, 8545212, 68361696, 751978656, 62664888, 563983992, 51271272, 512712720, 42726060, 4747340, 33231380, 265851040, 2924361440, 35092337280, 2339489152, 179960704, 2159528448, 32392926720, 550679754240, 30593319680, 458899795200, 26994105600, 1687131600, 140594300, 1265348700, 115031700, 11503170, 92025360, 1012278960, 84356580, 759209220, 69019020, 6901902, 55215216, 607367376, 50613948, 5623772, 803396, 4820376, 602547, 3012735, 18076410, 144611280, 13146480, 1314648, 164331, 821655, 4929930, 39439440, 433833840, 36152820, 4016980, 28118860, 224950880, 20450080, 2045008, 255626, 1278130, 7668780, 61350240, 674852640, 56237720, 506139480, 46012680, 4601268, 36810144, 404911584, 33742632, 303683688, 27607608, 276076080, 23006340, 2556260, 365180, 2191080, 273885, 54777, 219108, 36518, 146072, 876432, 109554, 547770, 91295, 365180

The range of values in this sequence is extreme, ranging from a minimum of 36,518 to a maximum of 550,679,754,240. Figure 2 shows the trajectory with a log scale being necessary for the \(y\) axis.

Figure 2

A Little Maths Puzzle

This puzzle appeared on PUZZLE A DAY and asks:

What is the only positive number that is equal to the average of its digits? Trailing zeros are not permitted. The trivial single digit answers don’t count as the question specifies ‘digits’. The clue provided is that the number consists of two digits. 

I made the mistake of thinking that the number is a two digit integer and reached an impasse. Let's represent the number with the digits \(x\) and \(y\). Then we have:$$ \begin{align} 10x + y &=\frac{x+y}{2} \\ 20x + 2y &=x+y \\ 19x + y &= 0 \end{align}$$Clearly this yields no solutions. However. Let's try the number \(x.y\) where the dot represents the decimal place. Then we have:$$ \begin{align} x + \frac{y}{10} &= \frac{x+y}{2} \\10x+y &= 5x+ 5y \\5x-4y &= 0 \end{align}$$This does yield a solution when \(x=4\) and \(y=5\). So the number must be 4.5 which indeed is the average of 4 and 5.

It's just a simple little problem but once you lock into that mindset that the number is an integer a solution seems impossible. I asked Gemini if there are positive numbers that satisfy for the geometric and harmonic means and the short answer is NO. 

Cyclotomic Polynomials Revisited

It was in November of 2020 that I made my first and only post about cyclotomic polynomials. Here is a link to that post. Today I was reminded of this category of polynomials because the number associated with my diurnal age today, 28057, is connected with the third cyclotomic polynomial, namely \(x^2 + x + 1\) because the number is a member of OEIS A053183:

  OEIS A053183: primes of the form \(p^2 + p + 1\) when \(p\) is prime.

In the case of 28057, the value of \(p\) is 167 and thus:$$28057 = 167^2 + 167 + 1$$What follows is what Gemini had to say about this particular polynomial.

The Complex Roots of $x^2 + x + 1 = 0$

The roots of the polynomial $x^2 + x + 1$ are the two primitive cube roots of unity. In mathematical literature, these are commonly denoted by the Greek letters ω (omega) and ω².

1. Numerical Values

Using the quadratic formula, we find the exact complex coordinates:

• x₁: $-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ (approx. $-0.5 + 0.866i$)
  
  
• x₂: $-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$ (approx. $-0.5 - 0.866i$)

2. Geometric Representation

When plotted on the Argand plane (complex plane), these roots exhibit specific geometric properties:

• The Unit Circle: Both roots have a magnitude of 1, meaning they sit exactly on the circumference of a circle with radius 1 centered at the origin.
  
  
• Conjugate Symmetry: The roots are reflections of each other across the real (horizontal) axis.
  
  
• The Equilateral Triangle: These two roots, combined with the third cube root ($x = 1$), form the vertices of a perfect equilateral triangle.

For further exploration of complex numbers and their visualization, you can use tools like the GeoGebra Complex Number Tool or check detailed proofs on Wolfram MathWorld.

All numbers generated by \(x^2+x+1\) for positive integer values of \(x\) are one more than the pronic numbers because the polynomial can be written as \(x(x+1) + 1 \).

An Interesting Limit

Here is a report generated by Gemini Pro based on the content of this video. I found the video moved a little too quickly at times and the author read out every mathematical expression symbol by symbol which wasn't really necessary and it didn't aid comprehension. It's useful to be able to take the content of the video and put into a report that can be read at your leisure. If you want to tweak the LaTeX code, you can do so easily and I've done that below for a few of expressions.

The Limit of the nth Root: An Exploration of Euler's Sequence

At first glance, the limit of the sequence involving factorials and powers can seem intuitive, but as with many problems in analysis, the "obvious" answer can be misleading. This report dismantles the problem of finding the limit of the nth root of n factorial divided by n to the power of n. Following the path likely walked by Leonhard Euler, we move from discrete sequences to continuous calculus to find a beautiful truth connecting factorials, limits, and Euler’s number, e.

1. The Sequence and Intuition

We begin by defining our sequence \( a_n \):

$$ a_n = \frac{n!}{n^n} $$

Intuitively, the denominator \( n^n \) grows much faster than the numerator \( n! \), suggesting that the terms drop toward zero. To prove this rigorously, we employ the Ratio Test. We investigate the ratio of consecutive terms:

$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} $$

Simplifying the factorials (\( (n+1)! = (n+1)n! \)) and regrouping terms gives:

$$ \frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left( \frac{n}{n+1} \right)^n $$

This expression can be rewritten to reveal a fundamental definition of Euler's number:

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{\left( \frac{n+1}{n} \right)^n} = \lim_{n \to \infty} \frac{1}{\left( 1 + \frac{1}{n} \right)^n} = \frac{1}{e} $$

Since \( \dfrac{1}{e} \approx 0.3679 \) is strictly less than 1, the Ratio Test confirms that the sequence converges to 0.

2. The Main Problem: The nth Root

Establishing that the base sequence goes to 0 is just the preamble. The true challenge is finding the limit of the nth root of this sequence:

$$ L = \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} $$

Since the base approaches 0 and the exponent \( \frac{1}{n} \) also approaches 0, we are facing the indeterminate form \( 0^0 \). To resolve this, we use the natural logarithm to transform the product into a sum, bridging the gap between discrete algebra and integral calculus.

3. From Logarithms to Riemann Sums

Let \( y = \sqrt[ \uproot{4} n]{a_n} \). Taking the natural log of the limit allows us to work with sums:

$$ \ln(L) = \lim_{n \to \infty} \ln \left( \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} \right) = \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) $$

Using logarithm rules, we expand the term inside:

$$ \ln \left( \frac{n!}{n^n} \right) = \ln(n!) - \ln(n^n) = \sum_{k=1}^{n} \ln(k) - n \ln(n) $$

We can cleverly rewrite \( n \ln(n) \) as a sum of \( n \) identical terms: \( \displaystyle \sum_{k=1}^{n} \ln(n) \). Substituting this back into our limit equation gives:

$$ \ln(L) = \lim_{n \to \infty} \frac{1}{n} \left( \sum_{k=1}^{n} \ln(k) - \sum_{k=1}^{n} \ln(n) \right) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( \ln(k) - \ln(n) \right) $$

Combining the logs, we arrive at a recognizable form:

$$ \ln(L) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} \right) $$

This structure—a sum of function values multiplied by a small width \( \frac{1}{n} \) is precisely the definition of a Riemann Sum on the interval \([0, 1]\). As \( n \to \infty \), the sum converges to a definite integral:

$$ \ln(L) = \int_{0}^{1} \ln(x) \, dx $$

4. Evaluating the Integral

We now evaluate the improper integral of \( \ln(x) \) from 0 to 1. Using integration by parts (where \( u = \ln(x) \) and \( dv = dx \)), we get:

$$ \int \ln(x) \, dx = x \ln(x) - x $$

We evaluate the definite integral by taking the limit as the lower bound approaches 0:

$$ \Bigl[ x \ln(x) - x \Bigr]_0^1 = (1 \ln(1) - 1) - \lim_{a \to 0^+} (a \ln(a) - a) $$

We know \( \ln(1) = 0 \), so the first term is \( -1 \). For the limit \( \lim_{a \to 0^+} a \ln(a) \), we use L'Hôpital's rule, which shows that this term vanishes to 0. Thus:

$$ \ln(L) = -1 - 0 = -1 $$

Conclusion

We have found that the natural logarithm of the limit is \( -1 \). To find the original limit \( L \), we simply exponentiate both sides:

$$ L = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}}=e^{-1} = \frac{1}{e} $$

We have proven that while the sequence \( \dfrac{n!}{n^n} \) collapses to zero, its geometric mean—the nth root—stabilizes at exactly 1 over e. This problem serves as an elegant bridge connecting basic sequence properties, the definition of e, and the power of integral calculus.

Newton's Approximation For Pi

I watched an interesting YouTube video about how Newton approximated \( \pi \) and I thought it would be an interesting exercise to get Gemini to access the video and recreate the steps outlined. Once Gemini had created its response, I then got it to create a HTML version of the information for easy pasting into Blogger. Here is what it came up with and it's really quite impressive:

Based on the video provided, specifically the section starting around [00:06:50], here is the step-by-step derivation of Newton’s approximation of $\pi$.

Newton's method combines two different ways of calculating the area of a specific portion of a circle: using calculus (integration of an infinite series) and geometry. By equating these two results, he solved for $\pi$.

Step 1: The Equation of the Circle

Newton started with the equation of a unit circle centered at the origin:

$$x^2 + y^2 = 1$$

Solving for $y$ gives the function for the upper semicircle:

$$y = \sqrt{1 - x^2} = (1 - x^2)^{\frac{1}{2}}$$

Newton wanted to find the area under this curve from $x = 0$ to $x = \frac{1}{2}$. This can be expressed as the definite integral:

$$\text{Area} = \int_{0}^{\frac{1}{2}} (1 - x^2)^{\frac{1}{2}} \, dx$$

Step 2: Binomial Expansion

Since calculus was in its infancy, Newton applied his generalized binomial theorem to expand $(1 - x^2)^{\frac{1}{2}}$ into an infinite series. The general form for $(1+u)^n$ is:

$$(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots$$

By substituting $u = -x^2$ and $n = \frac{1}{2}$, he derived the series for the circle's equation:

$$(1 - x^2)^{\frac{1}{2}} = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{16}x^6 - \frac{5}{128}x^8 - \dots$$

Step 3: Term-by-Term Integration

Newton then found the area (fluxion) by integrating the series term by term:

$$\int \left( 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 - \frac{1}{16}x^6 - \dots \right) \, dx$$

This yields the algebraic formula for the area:

$$\text{Area}(x) = x - \frac{1}{6}x^3 - \frac{1}{40}x^5 - \frac{1}{112}x^7 - \frac{5}{1152}x^9 - \dots$$

Step 4: Calculating the Area for $x = \frac{1}{2}$

Newton improved efficiency by choosing the limit $x = \frac{1}{2}$ (instead of 1), which makes the powers of $x$ shrink very rapidly. Substituting $x = \frac{1}{2}$ into the series:

$$\text{Area} \approx \frac{1}{2} - \frac{1}{6}\left(\frac{1}{2}\right)^3 - \frac{1}{40}\left(\frac{1}{2}\right)^5 - \frac{1}{112}\left(\frac{1}{2}\right)^7 - \dots$$

Simplifying the fractions:

$$\text{Area} \approx \frac{1}{2} - \frac{1}{48} - \frac{1}{1280} - \frac{1}{14336} - \dots$$

This sum gives a numerical value for the area.

Step 5: Equating to Geometry

Screenshot from video

Geometrically, the area under the unit circle from $x=0$ to $x=\frac{1}{2}$ corresponds to a slice of the circle. As described in the video, this region consists of two shapes:

1. A triangle with base $\frac{1}{2}$ and height $\frac{\sqrt{3}}{2}$ (since $y = \sqrt{1 - (0.5)^2} = \sqrt{0.75} = \frac{\sqrt{3}}{2}$).
2. A circular sector of $30^{\circ}$ (or $\frac{\pi}{6}$ radians).

The total geometric area is:

$$\text{Area} = \text{Area}_{\text{triangle}} + \text{Area}_{\text{sector}}$$ $$\text{Area} = \left( \frac{1}{2} \cdot \text{base} \cdot \text{height} \right) + \left( \frac{\theta}{2\pi} \cdot \pi r^2 \right)$$ $$\text{Area} = \left( \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \right) + \left( \frac{\pi/6}{2\pi} \cdot \pi (1)^2 \right)$$ $$\text{Area} = \frac{\sqrt{3}}{8} + \frac{\pi}{12}$$

Step 6: Solving for $\pi$

Finally, Newton equated the Algebraic Area (from Step 4) with the Geometric Area (from Step 5) and solved for $\pi$:

$$\frac{\sqrt{3}}{8} + \frac{\pi}{12} = \text{Series Sum}$$ $$\frac{\pi}{12} = \text{Series Sum} - \frac{\sqrt{3}}{8}$$ $$\pi = 12 \left( \text{Series Sum} - \frac{\sqrt{3}}{8} \right)$$ $$\pi = 12 \left[ \left( \frac{1}{2} - \frac{1}{48} - \frac{1}{1280} - \dots \right) - \frac{\sqrt{3}}{8} \right]$$

Using just the first few terms of the series, Newton achieved an approximation of $\pi \approx 3.14161$.

Fibonacci-Related Numbers

My diurnal age today is \( \textbf{28047} \) and this number has the interesting property that it's part of a Fibonacci sequence with seed numbers of 3 and 9. This leads to the following sequence of numbers:

3, 9, 12, 21, 33, 54, 87, 141, 228, 369, 597, 966, 1563, 2529, 4092, 6621, 10713, 17334, 28047, ...

These initial numbers form part of OEIS A022379. 

The thought occurred to me that given two seed digits between 0 and 9, there ought to be a limited set of numbers from 10 to let's say 40000 that are generated by two single digit seeds. I got Gemini to write a Python program (permalink) to investigate this and it turns out that there are 651 numbers with this property. Here they are:

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 107, 108, 109, 110, 111, 112, 113, 115, 116, 117, 118, 120, 121, 123, 124, 125, 126, 128, 129, 131, 133, 134, 136, 137, 138, 139, 141, 142, 144, 146, 147, 149, 150, 152, 154, 155, 157, 159, 160, 162, 163, 165, 167, 168, 170, 173, 175, 176, 178, 180, 181, 183, 186, 188, 189, 191, 194, 196, 199, 201, 202, 204, 207, 209, 212, 215, 217, 220, 222, 223, 225, 228, 230, 233, 236, 238, 241, 243, 246, 249, 251, 254, 257, 259, 262, 264, 267, 270, 272, 275, 280, 283, 285, 288, 291, 293, 296, 301, 304, 306, 309, 314, 317, 322, 325, 327, 330, 335, 338, 343, 348, 351, 356, 359, 361, 364, 369, 372, 377, 382, 385, 390, 393, 398, 403, 406, 411, 416, 419, 424, 427, 432, 437, 440, 445, 453, 458, 461, 466, 471, 474, 479, 487, 492, 495, 500, 508, 513, 521, 526, 529, 534, 542, 547, 555, 563, 568, 576, 581, 584, 589, 597, 602, 610, 618, 623, 631, 636, 644, 652, 657, 665, 673, 678, 686, 691, 699, 707, 712, 720, 733, 741, 746, 754, 762, 767, 775, 788, 796, 801, 809, 822, 830, 843, 851, 856, 864, 877, 885, 898, 911, 919, 932, 940, 945, 953, 966, 974, 987, 1000, 1008, 1021, 1029, 1042, 1055, 1063, 1076, 1089, 1097, 1110, 1118, 1131, 1144, 1152, 1165, 1186, 1199, 1207, 1220, 1233, 1241, 1254, 1275, 1288, 1296, 1309, 1330, 1343, 1364, 1377, 1385, 1398, 1419, 1432, 1453, 1474, 1487, 1508, 1521, 1529, 1542, 1563, 1576, 1597, 1618, 1631, 1652, 1665, 1686, 1707, 1720, 1741, 1762, 1775, 1796, 1809, 1830, 1851, 1864, 1885, 1919, 1940, 1953, 1974, 1995, 2008, 2029, 2063, 2084, 2097, 2118, 2152, 2173, 2207, 2228, 2241, 2262, 2296, 2317, 2351, 2385, 2406, 2440, 2461, 2474, 2495, 2529, 2550, 2584, 2618, 2639, 2673, 2694, 2728, 2762, 2783, 2817, 2851, 2872, 2906, 2927, 2961, 2995, 3016, 3050, 3105, 3139, 3160, 3194, 3228, 3249, 3283, 3338, 3372, 3393, 3427, 3482, 3516, 3571, 3605, 3626, 3660, 3715, 3749, 3804, 3859, 3893, 3948, 3982, 4003, 4037, 4092, 4126, 4181, 4236, 4270, 4325, 4359, 4414, 4469, 4503, 4558, 4613, 4647, 4702, 4736, 4791, 4846, 4880, 4935, 5024, 5079, 5113, 5168, 5223, 5257, 5312, 5401, 5456, 5490, 5545, 5634, 5689, 5778, 5833, 5867, 5922, 6011, 6066, 6155, 6244, 6299, 6388, 6443, 6477, 6532, 6621, 6676, 6765, 6854, 6909, 6998, 7053, 7142, 7231, 7286, 7375, 7464, 7519, 7608, 7663, 7752, 7841, 7896, 7985, 8129, 8218, 8273, 8362, 8451, 8506, 8595, 8739, 8828, 8883, 8972, 9116, 9205, 9349, 9438, 9493, 9582, 9726, 9815, 9959, 10103, 10192, 10336, 10425, 10480, 10569, 10713, 10802, 10946, 11090, 11179, 11323, 11412, 11556, 11700, 11789, 11933, 12077, 12166, 12310, 12399, 12543, 12687, 12776, 12920, 13153, 13297, 13386, 13530, 13674, 13763, 13907, 14140, 14284, 14373, 14517, 14750, 14894, 15127, 15271, 15360, 15504, 15737, 15881, 16114, 16347, 16491, 16724, 16868, 16957, 17101, 17334, 17478, 17711, 17944, 18088, 18321, 18465, 18698, 18931, 19075, 19308, 19541, 19685, 19918, 20062, 20295, 20528, 20672, 20905, 21282, 21515, 21659, 21892, 22125, 22269, 22502, 22879, 23112, 23256, 23489, 23866, 24099, 24476, 24709, 24853, 25086, 25463, 25696, 26073, 26450, 26683, 27060, 27293, 27437, 27670, 28047, 28280, 28657, 29034, 29267, 29644, 29877, 30254, 30631, 30864, 31241, 31618, 31851, 32228, 32461, 32838, 33215, 33448, 33825, 34435, 34812, 35045, 35422, 35799, 36032, 36409, 37019, 37396, 37629, 38006, 38616, 38993, 39603, 39980

The following code (permalink) can be used to find the seeds of any number in this list. It's not impressive code I'm sure but it seems to do the job. Let's try with the last number in the previous list, 39980.

c=39980 L=[] b=round(c/((1+sqrt(5))/2)) L.append(c);L.append(b) a=0 while b>=a and c>b: a=c-b if a <b: c=b b=a L.append(a) L.reverse() if len(str(L[1]))==2: a=L[1]-L[0] b=L[0] print("Seed numbers for Fibonacci sequence are", a,"and",b) X=[a,b] L.remove(L[0]) print(X+L) else: print("Seed numbers for Fibonacci sequence are", L[0],"and",L[1]) print(L)

Seed numbers for Fibonacci sequence are 9 and 4

[9, 4, 13, 17, 30, 47, 77, 124, 201, 325, 526, 851, 1377, 2228, 3605, 5833, 9438, 15271, 24709, 39980]

Interestingly, 39980 is listed in the OEIS as OEIS A022132 but the seeds are listed as 4 and 13 but this is equivalent of course to seeds of 9 and 4. Zero can be a seed number in the first position but it can't then be followed by another zero and so there are 10 x 9 = 90 distinct pairs of single digit seeds. 

Two different pairs of single digit seeds can lead to the same sequence. For example:

• seeds of 1 and 3 \( \rightarrow\) 1, 3, 4, 7, 11, 18, 29, ...
• seeds of 2 and 1 \( \rightarrow \) 2, 1, 3, 4, 7, 11, 18, 29, ...

Figure 1 shows a plot of these Fibonacci-related numbers and it's interesting that they form a definite exponential curve with the density of points noticeably thinning out as the numbers become larger.

Figure 1

ADDENDUM:

After completing this post, I discovered that I'd already covered this topic in a post from Thursday, 12th June 2025 titled Fibonacci Numbers Derived From Single Digits. However, in that post I only considered seed digits where the first was smaller than the second and so I discounted a number like 59 because it requires seed digits of 7 and 3 such that the sequence 7, 3, 10, 13, 23, 36, 59 is generated.

What's Special About 28046?

Figure 1

Figure 1 shows a screenshot of a post that I uploaded on the 27th September 2022 in which I noted that the number 26840 contained all the even digits exactly once. Today I turned 28046 days old and this number is the next such number after 26840 with this property. There are only 96 such numbers overall and as I also noted in my blog post at the time:

"I'll only see another six such numbers in my lifetime: 28046, 28064, 28406, 28460, 28604, 28640."

THE 28046 MONUMENT: A COLOSSAL ACHIEVEMENT

So, one down and five to go with the next one only 18 days away. After that, there's a gap of slightly less than a year. The property of these numbers is very much base dependent of course but nonetheless interesting from a recreational mathematical perspective.

Bogey Numbers

Video based on blog content, generated by NotebookLM

In a darts match, once you reach a score of between 170 and 2 inclusive, you can defeat your opponent with one, two or three darts provided you score doesn't correspond to one of the so-called "bogey numbers". These numbers are:

• \(169 =13^2 \)
• \(168=2^3 \times 3 \times 7 \)
• \(166=2 \times 83 \)
• \(165=3 \times 5 \times 11\)
• \(163 \text{ is prime} \)
• \(162 = 2 \times 3^4\)
• \(159 = 3 \times 53\)

The maximum of score of 170 can be finished with two consecutive treble 20s followed by 50 (the red of the bullseye equivalent to 2 x 25). With a score of 2, you finish the match off with a single dart directed at double 1.

With my 77th birthday fast approaching, I wanted to consider how many ways there are to finish off a darts match from a score of 77. It turns out that there are 905 different ways (permalink) using two or three darts. Here are the possible two dart finishes:

• 2-Darts: T9, DBull
• 2-Darts: T13, D19
• 2-Darts: T15, D16
• 2-Darts: T17, D13
• 2-Darts: T19, D10

The other 900 ways involve three darts, for example:  S1, D13, DBull. Figure 1 shows a plot of the total numbers of ways to finish for scores between 2 and 170. Click to enlarge. A score of 58 offers the largest number of ways to check out, 1404.

Figure 1: permalink

Of course professional darts players only choose from a very narrow selection out of all these possibilities as Gemini explains.

You have hit on the fundamental difference between the math of darts (which is what the code calculates) and the art of darts (which is what pros play).

While Python might find 50 ways to check out a score, a professional like Michael van Gerwen or Luke Littler might only ever consider one or two of them.

Infographic based on blog content
generated by NotebookLM

Here is the breakdown of the logic top players use to filter those thousands of possibilities down to the "correct" shot.

1. The Principle of "Splitting the Double" (The Safety Net)

This is the single most important rule in professional darts. Pros prioritize doubles that can be halved (split) and still remain a double if they miss into the single segment.

• The "King" Double is D16 (32):

  • If you aim for D16 (32) and hit Single 16, you have 16 left. That is D8.

  • If you aim for D8 and miss into S8, you have 4 left (D2).

  • If you aim for D2 and miss into S2, you have 2 left (D1).

  • Result: You have 4 layers of safety. You can miss your double 3 times and still be on a finish without needing a "setup" dart to get back to an even number.

• The Problem with D20 (Tops):

  • D20 is popular, but less safe. If you miss D20 (40) and hit S20, you have 20 left (D10).

  • If you miss D10 (20) and hit S10, you have 10 left (D5).

  • D5 is an odd number. You cannot finish on 5. You must now waste a dart hitting a Single 1 to leave D2.

The Decision: If a pro has a choice between leaving D20 or D16, they will almost always maneuver to leave D16.

2. "Percentage Darts" (Managing the Miss)

Pros assume they might miss the Triple and hit a Single. They choose routes where a "bad" throw (a single) still leaves a finishable score.

• Example: You have 61 points left with 2 darts.

  • The "Amateur" Math Route: Triple 15 (45) leaves D8 (16).

    • Risk: If you miss T15 and hit S15, you have 46 left. You have 1 dart remaining. You cannot check out 46 with 1 dart. Turn over.

  • The "Pro" Route: Single 25 (Outer Bull) leaves D18 (36).

    • Safety: If you miss the Bull and hit S25, you have 36 left (D18). If you miss widely and hit S11 (a common neighbor to the bull), you have 50 left (Double Bull). You still have a shot at a double.

The Decision: The pro chooses the route where a miss keeps the turn alive.

3. Avoiding the "Bogey" Singles (Odd vs. Even)

The goal is to get to an even number as fast as possible.

• If you are on an odd number (e.g., 17, 33, 45), you must hit an odd segment to get to an even finish.

• If you are on an even number, you must hit an even segment (or a Triple even) to stay even.

Pros hate finishing on odd doubles (D19, D17, D15, etc.) because if they hit the single, they are stuck on an odd number again.

• Bad: Aiming for D19 (38). Hit S19. Left with 19. Now you need S3 to leave D8. It's messy.

• Good: Aiming for D18 (36). Hit S18. Left with 18 (D9). Still messy, but D18 is physically close to D4 and D1, making the recovery easier.

4. Board Management (Staying in Rhythm)

Dart players rely heavily on muscle memory. They prefer routes that require minimal movement around the board.

• Example: 82 left.

  • Option A: Bull (50) + D16 (32). (Requires moving from center to right edge).

  • Option B (Pro Preference): D20 (40) + S10 + D16? No.

  • Option C (The "Top" Route): T14 (42) leaves D20 (40).

  • The "Phil Taylor" Route: Bull (50) leaves D16. (Some pros love the Bull because it fixes "messy" math instantly).

5. The "Matchplay" Factor (Aggressive vs. Conservative)

Finally, the choice depends on what the opponent is doing.

• Scenario A: Opponent is way back (on 300 points).

  • Strategy: Go safe. Don't try a risky T20-T20-D20 finish. Just hit big scores to get down to D16 reliably. Use 3 darts to set up the perfect number.

• Scenario B: Opponent is on a finish (e.g., 32).

  • Strategy: Aggressive. You must check out this turn. If you have 170, you must go T20-T20-Bull. You don't care about safety; you care about possibility.

Summary: The "Pro" Algorithm

If you were to program a "Pro Bot" rather than a "Math Bot," the logic would look like this:

1. Can I finish this turn?

2. If yes: Pick the route where a single (missed triple) leaves a finish for the next dart.

3. If no: Pick the route that leaves 32 (D16) or 40 (D20) for the next turn.

4. Avoid: Any route that leaves an odd number remainder.

Gemini created a program that will calculate the better choices when considering checking out. Figure 2 shows what it suggests for 77 (permalink).

Figure 2