Factorial Number Base

Today I turned 25199 days old and one of the entries was OEIS A200748: smallest number requiring \(n\) terms to be expressed as a sum of factorials. It took me some time to work out what this really meant and along the way I found out how to represent decimal numbers in factorial base form. As Wikipedia explains:

The factorial number system is a mixed radix numeral system: the \(i\)-th digit from the right has base \(i\), which means that the digit must be strictly less than \(i\), and that (taking into account the bases of the less significant digits) its value to be multiplied by \((i − 1)\)! (its place value).

So the representation of 25199 in this system is:

4*7! + 6*6! + 5*5! + 4*4! + 3*3! + 2*2! + 1*1! +0*0!

This is shown below as it appears in my Google worksheet:

As can be seen, 4+6+5+4+3+2+1+0 = 25 is the number of terms that OEIS A200748 is referring to and 25199 is the smallest number that can be represented using this system. The initial terms of the sequence are:

0, 1, 3, 5, 11, 17, 23, 47, 71, 95, 119, 239, 359, 479, 599, 719, 1439, 2159, 2879, 3599, 4319, 5039, 10079, 15119, 20159, 25199, 30239, 35279, 40319, 80639, 120959, 161279, 201599, 241919, 282239, 322559, 362879, 725759, 1088639, 1451519, 1814399, 2177279

A modification of the worksheet shows that the smallest number that can be represented using 24 terms is 20159:

Similarly, the smallest number that can be represented using 26 terms is 30239:

The worksheet can be used to confirm all the terms in the OEIS sequence by adding additional rows e.g. the smallest number that can be represented using 41 terms is 2177279:

Repeating Digits in the Squares of Integers

Today I turned 25197 days old and none of the entries in the Online Encyclopaedia of Integer Sequences (OEIS) made much sense to me. I searched about for something of interest about this number and, after some dead-ends, I thought I'd square the number. The result was 634888809 which immediately caught my attention because of the repetition of the digit 8. I thought this result would be relatively rare and something that distinguishes 25197 from most other numbers. My question was how rare? I set about investigating using a Google spreadsheet and had soon found all the numbers between 1 and 26000 whose squares contain at least four consecutive 8's. It turns out that there is a total of only 14 such numbers.

Here are the numbers with their associated squares

Runs of four or more 8's (there are 14 of these):

1609	2588881
6992	48888064
9428	88887184
10094	101888836
12202	148888804
16090	258888100
16667	277788889
16849	283888801
20221	408888841
20359	414488881
21187	448888969
22917	525188889
24267	588887289
25197	634888809

It can be seen that the squares of 12202 and 20221 contain a run of five 8's. Naturally, I was curious about runs of the other digits 1, 2, 3, 4, 5, 6, 7 and 9 and once again used the spreadsheet to investigate the matter. The results are detailed below:

Run of four or more 1's (there are 11 such numbers:

2848	8111104
3334	11115556
10541	111112681
10542	111133764
10543	111154849
10544	111175936
10545	111197025
17062	291111844
20276	411116176
20521	421111441
23877	570111129

What's interesting above is the cluster of numbers 10541, 10542, 10543, 10544 and 10545 that all produce at least four leading 1's (10541 produces five leading 1's).

Runs of four or more 2's (there are 18 of these):

3335	11122225
3496	12222016
4585	21022225
4714	22221796
5415	29322225
6101	37222201
8332	69422224
8557	73222249
13335	177822225
14585	212722225
15415	237622225
16665	277722225
16668	277822224
20548	422220304
20838	434222244
23335	544522225
24585	604422225
25415	645922225

Runs of four or more 3's (there are eight of these):

5774	33339076
7303	53333809
11547	133333209
18037	325333369
18257	333318049
18258	333354564
18259	333391081
23094	533332836

There is another cluster here consisting of 18257, 18258 and 18259, all producing a run of four leading 3's when squared. There is also a run of five 3's produced by 11547 when it is squared.

Runs of four or more 4's (there are 12 of these):

5696	32444416
6667	44448889
11595	134444025
15393	236944449
15857	251444449
21081	444408561
21082	444450724
21083	444492889
24229	587044441
24788	614444944
25386	644448996
25771	664144441

Here again there is a cluster of 21081, 21082 and 21083, all producing a run of four 4's when squared. One number 15857 produces a run of five 4's when squared. The square of 24788 contains six 4's though only four of them are consecutive.

Runs of four or more 5's (there are 14 of these):

2357	5555449
3944	15555136
4166	17355556
8334	69455556
10274	105555076
10419	108555561
12472	155550784
15986	255552196
16666	277755556
20629	425555641
20834	434055556
22925	525555625
23570	555544900
23571	555592041

Here we see that the square of 22925 contains six 5's, although only four of them are consecutive.

Runs of four or more 6's (there are ten of these):

1291	1666681
1633	2666689
2582	6666724
5164	26666896
8165	66667225
12910	166668100
16330	266668900
17224	296666176
25819	666620761
25820	666672400

Here there are a pair of numbers, 25819 and 25820, that both produce a run of four leading 6's when squared.

Runs of four or more 7's (there are seven of these):

8819	77774761
11076	122677776
13924	193877776
16964	287777296
18915	357777225
21858	477772164
24037	577777369

Runs of four or more 9's (there are only two of these):

7071	49999041
14142	199996164

It's surprising that there are only two numbers between 1 and 26000 that produce four sequential 9's when squared and the one is double the other. I'll need to investigate the reason for this paucity.

I've proposed the sequence of numbers that produce four or more 8's when squared as a candidate for the OEIS. If it's approved, I'll propose the other runs of four or more digits are candidate sequences as well.

Generating Functions

On March 22nd 2018, my number of the day was 25190 described as a member of OEIS A014334 whose terms arise from the exponential convolution of Fibonacci numbers with themselves. This meant the term arose from a generating function of a particular type, namely exponential. The term convolution means that the generating function is multiplied by itself to generate a new function. The first eleven terms of OEIS A014334 are 0, 0, 2, 6, 22, 70, 230, 742, 2406, 7782 and 25190. So how are these terms generated?

Firstly, we need to establish what the exponential generating function is for the Fibonacci numbers. It turns out to be: $$ \sum_{n=0}^{\infty} F_n \frac{t^n}{n!}$$ $$\text{where } F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta} \text{ and } \alpha=\frac{1+\sqrt{5}}{2} \text{ and } \beta=\frac{1-\sqrt{5}}{2}$$
Now the convolution of the generating function with itself yields new terms that can be found from the original Fibonacci numbers. Let's define two exponential generating functions: $$ A(t)=\sum_{n=0}^{\infty} a_n \frac{t^n}{n!} \text{ and } B(t)=\sum_{n=0}^{\infty} b_n \frac{t^n}{n!}$$ $$ \text{then } A(t).B(t)=\sum_{n=0}^{\infty} \bigg(  \sum_{k=0}^{n} \binom{n}{k}.a_k.b_{n-k} \bigg) \frac{t^n}{n!}$$ Let's remember that \(a_n = b_n =F_n \) and so all the coefficients of this generating function can be found by setting \(n=0, 1, 2, 3, ... \). For example, when \( n=4 \) the coefficient is $$ \binom{4}{0}.F_0.F_4+\binom{4}{1}.F_1.F_3+\binom{4}{2}.F_2.F_2+\binom{4}{3}.F_3.F_1+\binom{4}{4}.F_4.F_0$$Remember that \( F_0=0, F_1=1, F_2=1, F_3=2, F_4=3 \) and so substituting these values, the coefficient becomes \( 1*0*3+4*1*2+6*1*1+4*2*1+1*3*0=22 \). The magnitude of these coefficients rises quickly as sequence OEIS A014334 shows. So this is how the number 25190 can be calculated (by setting n=10).

REFERENCE: https://www.mathstat.dal.ca/FQ/Scanned/11-3/church.pdf

The 13x+1 and 11x+ 1 Maps

13x+1

Having looked at the 17x+1 map is my previous post, it was seen at 43 and 61 were special in the sense that they returned to their starting point after 84 steps, whereas every other prime from 17 to 97 terminated in at most 56 steps. Working again in Google Sheets, I applied the 13x+1 algorithm to the primes between 17 and 109. For any integer N>11, the algorithm is as follows:

• IF 2|N THEN N --> N/2 
• IF 3|N THEN N --> N/3 
• IF 5|N THEN N --> N/5 
• IF 7|N THEN N --> N/7 
• IF 11|N THEN N --> N/11 
• ELSE N --> 13*N+1

Here are the results and, for 13x+1, it appears that 19 and 101 are special in that they are the only primes so far that I've found to return to their starting point (151 joins the trajectory of 19 after 10 steps):

• 13 terminates after 25 steps
• 17 terminates after 23 steps
• 19 repeats after 15 steps
• 23 terminates after 6 steps
• 29 terminates after 6 steps
• 31 repeats after 15 steps (overlaps with trajectory of 19)
• 37 terminates after 20 steps
• 41 terminates after 40 steps
• 43 terminates after 7 steps
• 47 terminates after 28 steps
• 53 terminates after 10 steps
• 59 terminates after 10 steps
• 61 terminates after 10 steps
• 67 terminates after 60 steps
• 71 terminates after 6 steps
• 73 repeats after 15 steps (overlaps with trajectories of 19 and 31)
• 79 terminates after 43 steps
• 83 terminates after 8 steps
• 89 terminates after 37 steps
• 97 terminates after 24 steps
• 101 repeats after 15 steps (overlaps with trajectories of 19, 31 and 73)
• 103 terminates after 64 steps
• 107 terminates after 12 steps
• 109 terminates after 56 steps

The 19, 31, 73 and 101 sequences overlap as the diagram below shows (progression is anticlockwise):

Further investigation has shown that 151 joins the trajectory of 19 after 10 steps. Primes 113, 127, 131, 137, 139 and 149 all terminate. It will be interesting to investigate whether other non-terminating primes join this above circle.

My conjecture is that all numbers, under the 13x+1 algorithm, will either enter this loop or else get reduced to 1.

11x+1

For the 11x+1 map, the algorithm to be applied is that for any integer N>7:

• IF 2|N THEN N --> N/2 
• IF 3|N THEN N --> N/3 
• IF 5|N THEN N --> N/5 
• IF 7|N THEN N --> N/7  
• ELSE N --> 11*N+1

Once again, most numbers terminate but there is a cycle as shown below, starting at 17 and proceeding in an anticlockwise direction:

My conjecture is that all numbers, under the 11x+1 algorithm, will either enter this loop or else get reduced to 1.

What I am seeing thus far with the Px+1 map is that values of P=17, 13 and 11 will reduce most numbers to 1 but in each case there is a set of prime numbers that form part of a cycle. I haven't tested for P=7 or 5 yet. However, we know that for P=3 there is no known cycle of this nature.

The 17x+1 Map

Having recently written yet again about the Collatz trajectory, I was pleasantly surprised today to come upon a more generalised version of it. It goes by the name of the Px+1 map of which the Collatz trajectory is a specific example in which P=3. The Px+1 trajectory or map is an algorithm that states if x is divisible by any prime < P then divide out these primes one at a time starting with the smallest; otherwise multiply x by P and add 1.

My number for today is 25186 and it appears as an entry in OEIS A057534 that states:

• a(n+1) = a(n)/2 if 2|a(n)
• a(n)/3 if 3|a(n)
• a(n)/5 if 5|a(n)
• a(n)/7 if 7|a(n)
• a(n)/11 if 11|a(n)
• a(n)/13 if 13|a(n)
• else 17*a(n)+1

This is a particular example of the Px+1 map in which P=17 and this generates a sequence, part of which is shown below:

61, 1038, 519, 173, 2942, 1471, 25008, 12504, 6252, 3126, 1563, 521, 8858, 4429, 75294, 37647, 12549, 4183, 71112, 35556, 17778, 8889, 2963, 50372, 25186, 12593, 1799, 257, 4370, 2185, 437, 7430, 3715, 743, 12632, 6316, 3158, 1579, 26844, 13422, ...

I pondered long and hard over why 61 was chosen as the starting point for this sequence. Once a(0)=63 then the other terms follow according to the algorithm but why start there? I used Google Sheets to investigate the matter. As shown below, the sequence repeats after 84 steps. I'll leave the numbers vertical as it's easier to scan them.

61 prime
1038
519
173 prime
2942
1471 prime
25008
12504
6252
3126
1563
521 prime
8858
4429 (43 x 103)
75294
37647
12549
4183 (47 x 89)
71112
35556
17778
8889
2963 prime
50372
25186
12593
1799
257 prime
4370
2185
437 (19 x 23)
7430
3715
743 prime
12632
6316
3158
1579 prime
26844
13422
6711
2237 prime
38030
19015
3803 prime
64652
32326
16163
2309 prime
39254
19627 (19 x 1033)
333660
166830
83415
27805
5561 (67 x 83)
94538
47269 prime
803574
401787
133929
44643
14881 (23 x 647)
252978
126489
42163
3833 prime
65162
32581 (31 x 1051)
553878
276939
92313
30771
10257
3419
263 prime
4472
2236
1118
559
43 prime
732
366
183
61 prime

This still didn't explain why 61 was chosen as the starting point so I decided to investigate all the primes from 17 to 59 and from 67 up to 97. Here is a summary of my results:

17 terminates after 13 steps
19 terminates after 7 steps
23 terminates after 6 steps
29 terminates after 10 steps
31 terminates after 7 steps
37 terminates after 6 steps
41 terminates after 53 steps
43 repeats after 84 steps (joins 61 trajectory after 4 steps)
47 terminates after 8 steps
53 terminates after 56 steps (joins the 41 trajectory after 3 steps)
59 terminates after 13 steps
61 repeats after 84 steps
67 terminates after 12 steps
71 terminates after 15 steps
73 terminates after 11 steps
79 terminates after 9 steps
83 terminates after 36 steps
89 terminates after 9 steps
97 terminates after 6 steps

I thought it odd that 43 and 61 both repeated after 84 steps and closer investigation revealed that 43 is actually on the trajectory of 61 because of 43, 732, 366, 183, 61. So I guess 61 was chosen because it's the first sequence that loops instead of terminating. 43 also loops but actually joins the trajectory of 61 after 4 steps.

It would interesting be to explore what happens to larger primes under the 17x+1 map and also to experiment with other values of P, all possible of course with a spreadsheet.

My conjecture is that there is just the single 61 loop as shown above and that any number not in that loop must either enter it or else get reduced to 1 sooner or later.

Losing at Lotto

Figure 1

Each week, I play four games in the Saturday Night Gold Lotto and, though I seldom win a prize, I always manage to choose at least one winning number. For each game in this particular lottery, you must choose six numbers out of a total of 45 (numbered from 1 to 45) and, during the draw, six main numbers are chosen as well as two supplementary numbers (so there are eight "winning" numbers). If you select all six main numbers, you share in the first division prize of $4,000,000. The amount is divided among the total numbers of winners which on average is between four and five. Combinations of main numbers and supplementary numbers win lesser prizes. Tonight, I was surprised to see that I'd managed to select none of the eight winning numbers. I've attached a screenshot of my results (see Figure 1). This got me wondering what was the probability of doing this? It had to be fairly low because I couldn't recall it ever happening to me before.

To tackle the problem, let's consider a single game in which there are 37 "losing" numbers along with the eight "winning" numbers. How many ways can these 37 numbers be chosen if I am selecting six numbers at a time, disregarding the order in which they are chosen. There are \( ^{37}C_6 \)= 2516 ways to do this out of a total of \(^{45}C_6\) = 8815 possibilities. So the probability of choosing six losing numbers in a single game is 2516/8815 which is almost 0.29 or 29%. Thus for every 100 games I play, around 29 will result in my missing all of the winning numbers. However, it is highly unlikely that I'll "succeed" in doing this four times in a row because each game is independent of the other and so the probabilities are multiplied. Thus the fraction 2516/8815 is raised to the fourth power and this gives a probability of about 0.0066 or 0.66%. In other words, if I play 10,000 sets of four games each, I'll only miss all of the winning numbers about 66 times. Tonight was one of those times.

Below is shown the WolframAlpha calculation:

For an earlier post on the probability of winning at Oz Lotto, follow this link.

UPDATE: on Saturday March 6th 2021 I also managed to select no numbers out of four games. Thus it is almost three years since this last happened. Figure 2 shows a screenshot of the results.

Figure 2

The Collatz Conjecture Revisited

Some time ago I posted about the Collatz Conjecture. Today's and yesterday's numbers (25183 and 25182 respectively) are connected to this conjecture. In general, these numbers arise because I'm tracking the number of days that I've been alive, numbering the day I was born (April 3rd 1949) as day zero and counting forward from there.

Both numbers appear in the Online Encyclopaedia of Integer Sequences (OEIS) A224303, whose members comprise numbers n for which number of iterations to reach the largest equals number of iterations to reach 1 from the largest in Collatz (3x+1) trajectory of n.

It's easy enough to set up a spreadsheet that calculates the number of steps to reach 1 and also the number of steps to reach the largest number in the trajectory. This is what I've done in Google Sheets and I've included a screenshot below for 25183.

As can be seen, 116 steps are required to get to the largest number (6,810,136) in the trajectory and then the same number of steps to reach 1, making for 232 steps in all. The steps for the previous number 25182 are the same. Here is a graph of the trajectory:

Looking at the sequence of such numbers, it's apparent that they tend to cluster and often appear in groups of two or more. Here is the list as it is shown in OEIS A224303 (with clusters shown in different colours):

1, 6, 120, 334, 335, 804, 1249, 2008, 2010, 2012, 2013, 6556, 6557, 6558, 6801, 6802, 6803, 7496, 7498, 7500, 7501, 7505, 10219, 22633, 25182, 25183, 27074, 27075, 27864, 27866, 27868, 31838, 31839, 32078, 36630, 36633, 36690, 36691, 36914, 39126, 39344

The second member of the sequence, 6, is given as an example: 6 is in the list because the Collatz trajectory of 6 is {6, 3, 10, 5, 16, 8, 4, 2, 1} and four steps are required to reach the largest number number (16) and four steps are required to reach 1 from 16:

6 --> 3 --> 10 --> 5 --> 16 and then 16 --> 8 --> 4 ---> 2 --> 1

Of course, there's a site on the Internet that will calculate the number of steps and graph the result. It also contains other interesting information relating to the Collatz conjecture. My spreadsheet will graph the trajectory but one has to manually alter the upper bound to get the best looking graph. I haven't figured out a way to adjust it automatically but I'll keep working on it.

Remember that the rule is to divide by 2 if the number is even and multiply by 3 and add 1 if the number is odd (hence the "3x+1 problem" as an alternative moniker). However, the site mentioned also allows one to customise the algorithm, so that for example instead of multiplying by 3, one can multiply by 2.

Interestingly, the trajectory still reaches 1 but it takes 669 steps and it's graph is quite different to that followed using the standard algorithm. Using larger multipliers like 4 doesn't seem to lead to convergence. For example after 10000 iterations using 4 as the multiplier, one gets 6,922,158,704,601,770. I'm not sure what happens with more iterations. The site also has a page for testing Lychrel numbers. I've looked at these sorts of numbers before but hadn't realised that they were called Lychrel numbers. I'd been referring to the algorithm to find them, namely reverse and add. See this post and this post to view.

See also: https://voodooguru23.blogspot.com/2018/03/the-px1-map.html

Read about Terence Tao's latest discovery: https://t.co/h8cMC9QKes

The Missing Dollar Riddle

Desy showed me this problem that she had encountered on an Internet post and asked for my thoughts:

This is a variation on the famous Missing Dollar Riddle described in Wikipedia as having a long history. It's not immediately apparent why the amounts in the two columns should differ. However, first the translation of the Indonesian into English. 

A boy has 50,000 rupiah. He buys some bread for 20,000 rupiah and thus has 30,000 left. He buys some juice for 15,000 and has 15,000 left. He buys a drink for 9,000 and has 6,000 left. Finally, he spends his remaining 6,000 rupiah on cake and has no money left. The question asked is why is there a difference of 1,000 rupiah. What is the answer to this question?

In attempting to answer the question, I thought I'd look at what happens when the amounts spent are different to those shown above. Firstly though, let's get rid of the 1000's and work with 20, 15, 9 and 6. Suppose the boy spent 20.5, 15.5, 8.5 and 5.5 instead. The progressive amounts left at each stage are 29.5, 14.0, 5.5 and 0, totalling 49. In this case, the progressive totals underestimate by 1 instead of overestimating by 1. I then considered the case where the boy spends 20.25, 15.25, 8.75 and 5.75. The corresponding amounts left are 29.75, 14.5, 5.75 and 0. These total 50 and so there is no difference.

The conclusion is that the progressive totals for the amount of money left are sometimes in excess of the amount of money spent, sometimes less than it and sometimes equal to it. Clearly the progressive totals are measuring something that's close to the amount of money spent but generally not the same as it. At this point, I turned to algebra to explore the generality of what is really going on.

To do this consider the initial amount of money are being M and the amounts spent are a, b, c and d. Because all the money is spent, we can say that d=M-a-b-c and so the amounts spent are a, b, c and M-a-b-c. The progressive amounts of money left are then M-a, M-a-b, M-a-b-c and 0. These amounts total 3M-3a-2b-c and clearly do not necessarily equal M.

However, what happens when they do? What happens if 3M-3a-2b-c = M? In this case, we get 2M = 3a+2b+c. For the particular case referred to above, M=50 and so 2M=100. For the case where a=29.75, b=14.5 and c=5.75, it can be seen that 3a+2b+c = 60.75+30.50+8.75 = 100 and so the equation is satisfied. There are many values of a, b and c that satisfy the equation. For example, a=20, b=15 and c=10 but a=20, b=15 and c=9 (the problem as it appeared in the Instagram post above) does not.

The Gaussian Correlation Inequality (GCI)

As he was brushing his teeth on the morning of July 17, 2014, Thomas Royen, a little-known retired German statistician, suddenly lit upon the proof of a famous conjecture at the intersection of geometry, probability theory and statistics that had eluded top experts for decades.

 Thus begins a most interesting article I came across in Quanta Magazine. The article continues:

Known as the Gaussian correlation inequality (GCI), the conjecture originated in the 1950s, was posed in its most elegant form in 1972 and has held mathematicians in its thrall ever since. “I know of people who worked on it for 40 years,” said Donald Richards, a statistician at Pennsylvania State University. “I myself worked on it for 30 years.” 

Royen hadn’t given the Gaussian correlation inequality much thought before the “raw idea” for how to prove it came to him over the bathroom sink. Formerly an employee of a pharmaceutical company, he had moved on to a small technical university in Bingen, Germany, in 1985 in order to have more time to improve the statistical formulas that he and other industry statisticians used to make sense of drug-trial data. In July 2014, still at work on his formulas as a 67-year-old retiree, Royen found that the GCI could be extended into a statement about statistical distributions he had long specialised in. On the morning of the 17th, he saw how to calculate a key derivative for this extended GCI that unlocked the proof. “The evening of this day, my first draft of the proof was written,” he said. 

Not knowing LaTeX, the word processor of choice in mathematics, he typed up his calculations in Microsoft Word, and the following month he posted his paper to the academic preprint site arxiv.org. He also sent it to Richards, who had briefly circulated his own failed attempt at a proof of the GCI a year and a half earlier. “I got this article by email from him,” Richards said. “And when I looked at it I knew instantly that it was solved.” 

Upon seeing the proof, “I really kicked myself,” Richards said. Over the decades, he and other experts had been attacking the GCI with increasingly sophisticated mathematical methods, certain that bold new ideas in convex geometry, probability theory or analysis would be needed to prove it. Some mathematicians, after years of toiling in vain, had come to suspect the inequality was actually false. In the end, though, Royen’s proof was short and simple, filling just a few pages and using only classic techniques. Richards was shocked that he and everyone else had missed it. “But on the other hand I have to also tell you that when I saw it, it was with relief,” he said. “I remember thinking to myself that I was glad to have seen it before I died.” He laughed. “Really, I was so glad I saw it.”

Here is the diagram that accompanies the article:

Interesting story. I won't include the full text of the article here but I'll quote from near its end:

Royen represented the amount of correlation between variables in his generalised GCI by a factor we might call C, and he defined a new function whose value depends on C. When C = 0 (corresponding to independent variables like weight and eye colour), the function equals the product of the separate probabilities. When you crank up the correlation to the maximum, C = 1, the function equals the joint probability. To prove that the latter is bigger than the former and the GCI is true, Royen needed to show that his function always increases as C increases. And it does so if its derivative, or rate of change, with respect to C is always positive. 

His familiarity with gamma distributions sparked his bathroom-sink epiphany. He knew he could apply a classic trick to transform his function into a simpler function. Suddenly, he recognised that the derivative of this transformed function was equivalent to the transform of the derivative of the original function. He could easily show that the latter derivative was always positive, proving the GCI. “He had formulas that enabled him to pull off his magic,” Pitt said. “And I didn’t have the formulas. 

Any graduate student in statistics could follow the arguments, experts say. Royen said he hopes the “surprisingly simple proof … might encourage young students to use their own creativity to find new mathematical theorems,” since “a very high theoretical level is not always required.”

I could have helped him with his LaTeX if he'd asked. Anyway, it's good to see an old timer like Thomas Royen making a significant contribution to Mathematics.