Gaps between Deficient Numbers

An alternative title for this post could have been Runs of Abundant Numbers because the two topics are complementary. Today I turned 26143 days old and one of this number's properties is that it's a member of OEIS A317049.

A317049

Numbers \(k\) such that both \(k\) and \(k\) + 3 are consecutive deficient numbers.

At first, this didn't seem all that significant a property, until one looks at the sequence and realises that this is a relatively rare occurrence. Below are the members less than 100,000:

5774, 5983, 7423, 11023, 21734, 21943, 26143, 27403, 39374, 43063, 49663, 56923, 58694, 61423, 69614, 70783, 76543, 77174, 79694, 81079, 81674, 82003, 84523, 84643, 89774, 91663, 98174, ...

Figure 1

As Figure 1 shows, the usual pattern is a run of deficient numbers punctuated by an abundant number. This is because approximately three out of every four numbers will be deficient. The following SageMath algorithm (permalink) will generate the above sequence of numbers:

L=[]

gap=3

for n in [1..100000]:

    N=[]

    for i in [0..gap]:

        difference=(n+i)-(sigma(n+i)-(n+i))

        N.append(difference)

    if N[0]>0 and N[gap]>0:

        OK=1

        if gap>1:

            for i in [1..(gap-1)]:

                if N[i]>0:

                    OK=0

        if OK==1:

            L.append(n)

print(L)

The obvious question then is where do runs of three abundant numbers occur or where do we find numbers \(k\) such that both \(k\) and \(k\) + 4 are consecutive deficient numbers. Well this occurs between between 171078829 and 171078833, where these two are consecutive deficient numbers. In other words, the consecutive abundant numbers are 171078830, 171078831 and 171078832.

The starting term of the smallest consecutive 4-tuple of abundant numbers is at most:

141363708067871564084949719820472453374

and so 141363708067871564084949719820472453373 to 141363708067871564084949719820472453378 is probably the smallest \(k\) to \(k\)+5 case.

See OEIS A094268 for more information. Thus it will be another 1260 days before there is another run of two abundant numbers. Note that most abundant numbers are even, so if two abundant numbers are to be adjacent then one of them must be odd which is rare. Returning to the deficient 26143 (the number that prompted this post), it can be noted that 26144 is even and abundant while 26145 is odd and abundant. Furthermore, 26145 is an odd primitive abundant number, meaning that none of its proper divisors is abundant. Such numbers form OEIS sequence A006038:

A006038

Odd primitive abundant numbers.

The sequence, up to 26145, runs:

945, 1575, 2205, 3465, 4095, 5355, 5775, 5985, 6435, 6825, 7245, 7425, 8085, 8415, 8925, 9135, 9555, 9765, 11655, 12705, 12915, 13545, 14805, 15015, 16695, 18585, 19215, 19635, 21105, 21945, 22365, 22995, 23205, 24885, 25935, 26145

Phi - Sigma Loops

Yesterday I turned 26140 days old and one of this number's properties is that it is a member of OEIS A095953:

A095953

Initial values for f(x) = phi(sigma(x)) such that iteration of f ends in a cycle of length 3.

In the case of 26140 we get:

• \( \phi(\sigma_1(26140))=15552 \)
• \( \phi(\sigma_1(15552))=18144 \)
• \( \phi(\sigma_1(18144))=15840 \)
• \( \phi(\sigma_1(15840))=15552 \)

Thus there is a cycle of length 3. 

This got me thinking about what other cycles exist. What I found was that cycles of 1, 2, 3, 4 and 6 are possible but not cycles of 5 or cycles greater than 6. Between 1 and 26200, the percentages for the cycles of 6, 4, 3, 2 and 1 are about 5.5%, 3.4%, 6.2%, 15.3% and 69.6% respectively (for a total of 100%). Clearly, a cycle of 1 is by far the most common. See Figure 1.

Figure 1

This is not to say that other cycles are not possible but the percentages shown in Figure 1 hold true for the first 26200 numbers. The following SageMath algorithm (permalink) will determine those numbers with a cycle of 3 within a given range. The range and cycle size are easily modified:

L,cycle=[],3

for n in [26100..26200]:

    count,number,M = 0,n,[n]

    for i in [1..7]:

        number=euler_phi(sigma(number))

        if number in M:

            break

        M.append(number)

    for i in range(len(M)):

        if M[i]==number:

            count=len(M)-i

    if count==cycle:

        L.append(n)

print(L)

[26118, 26120, 26140, 26166]

Sums of Cubes

 I've written about cubic numbers before, specifically in posts titled:

• Cubic Numbers from the 11th January 2016 in which I discuss turning \(29^3\) days old
• Cuban Primes from the 11th July 2016 in which I discuss primes that are the difference of two consecutive cubes
• Sums of Cubes and Squares of Sums from the 31st July 2018 in which I discuss with the property that the sum of the cubes of their digits is equal to the square of the sums of their digits
• Platonic Numbers from the 20th November 2018 in which I discuss cubes or cubic numbers as being one type of platonic number along with tetrahedral, octahedral, icosahedral and dodecahedral numbers.
• 42 is the new 33 from the 1st April 2019 in which I report on the discovery that 33 is expressible as a sum of three signed cubes and that only 42 remains to be cracked.
• 42 from the 9th September 2019 in which I report on the discovery is expressible as a sum of three signed cubes and that a new way of expressing 3 as a sum of three signed cubes was discovered
• The Original Taxi Cab Number in a New Light from the 21st December 2019 in which I discuss 1729 and centred cube numbers

In this post, I'll examine 26132 that has the interesting property that it can be expressed as a sum of six cubes in three possible ways. It is a member of OEIS A048931

A048931

Numbers that are the sum of 6 positive cubes in exactly 3 ways.

What's remarkable about 26132 is that it's the last such number with this property. Beginning with 221, the sequence has 1141 members and 26132 is the last member. It is expressible as the sum of the following cubic numbers: [8, 2744, 2744, 3375, 8000, 9261], [64, 1331, 2744, 4913, 4913, 12167] and [512, 512, 2197, 2744, 8000, 12167].

From WolframMathWorld we learn that:

• 23 and 239 are the only integers requiring nine positive cubes
• only 15 integers require eight cubes: 15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, and 454 

The same source also provides the following table that gives the first few numbers which require at least N = 1, 2, 3, ..., 9 (i.e., N or more) positive cubes to represent them as a sum.

	OEIS	numbers
1	A000578	1, 8, 27, 64, 125, 216, 343, 512, ...
2	A003325	2, 9, 16, 28, 35, 54, 65, 72, 91, ...
3	A047702	3, 10, 17, 24, 29, 36, 43, 55, 62, ...
4	A047703	4, 11, 18, 25, 30, 32, 37, 44, 51, ...
5	A047704	5, 12, 19, 26, 31, 33, 38, 40, 45, ...
6	A046040	6, 13, 20, 34, 39, 41, 46, 48, 53, ...
7	A018890	7, 14, 21, 42, 47, 49, 61, 77, ...
8	A018889	15, 22, 50, 114, 167, 175, 186, ...
9	A018888	23, 239

Again, the same source provides the following table that gives the numbers which can be represented in exactly \(W\) different ways as a sum of \(N\) positive cubes.

		OEIS	numbers
1	0	A007412	2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, ...
1	1	A000578	1, 8, 27, 64, 125, 216, 343, 512, ...
2	0	A057903	1, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, ...
2	1		2, 9, 16, 28, 35, 54, 65, 72, 91, ...
2	2	A018850	1729, 4104, 13832, 20683, 32832, ...
2	3	A003825	87539319, 119824488, 143604279, ...
2	4	A003826	6963472309248, 12625136269928, ...
2	5		48988659276962496, ...
2	6		8230545258248091551205888, ...
3	0	A057904	1, 2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, ...
3	1	A025395	3, 10, 17, 24, 29, 36, 43, 55, 62, ...
3	2		251, ...
4	0	A057905	1, 2, 3, 5, 6, 7, 8, 9, 10, 12, 13, 14, ...
4	1	A025403	4, 11, 18, 25, 30, 32, 37, 44, 51, ...
4	2	A025404	219, 252, 259, 278, 315, 376, 467, ...
5	0	A057906	1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...
5	1	A048926	5, 12, 19, 26, 31, 33, 38, 40, 45, ...
5	2	A048927	157, 220, 227, 246, 253, 260, 267, ...
6	0	A057907	1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 14, 15, ...
6	1	A048929	6, 13, 20, 27, 32, 34, 39, 41, 46, ...
6	2	A048930	158, 165, 184, 221, 228, 235, 247, ...
6	3	A048931	221, 254, 369, 411, 443, 469, 495, ...

In the final row of the above table, we see the sequence to which 26132 belongs. There's a lot more of course that could be said about numbers formed from the sum of cubes but what we do know is that, after 26132, there are no more numbers that can be formed from six cubes in only three ways. How we know this, I don't know!

Heronian Triangles and Tetrahedra

I have mentioned Heronian triangles tangentially in a post of Saturday, 2nd December 2017 when dealing with the number 25080. This number is the area of a so-called almost equilateral triangle with sides of 240, 241 and 241. In that post, I stated that:

A close-to-equilateral integer triangle is defined to be a triangle with integer sides and integer area such that the largest and smallest sides differ in length by unity. The first five close-to-equilateral integer triangles have sides (5, 5, 6), (17, 17, 16), (65, 65, 66), (241, 241, 240) and (901, 901, 902).

Figure 1

This triangle is also an example of an Heronian triangle which is simply any triangle with integer sides that also has an integer area. The simplest example is such as a triangle is right-angled with sides of 3, 4 and 5 (see Figure 1). The area of the triangle is 6 square units. From any right-angled Heronian triangle, two isosceles Heronian triangle can be constructed. For example, the 3, 4, 5 right-angled triangle leads to the 5, 5, 6 and the 5, 5, 8 isosceles triangles.

I was reminded of Heronian triangles by my diurnal age today which is 26129. This number has the following property:

A306626

Numbers that set a record for occurrences as longest side of a primitive Heronian triangle.

A primitive Heronian triangle is one in which the greatest common divisor of the three sides is 1. The 3, 4, 5 triangle is primitive but the 6, 8, 10 triangle, though still Heronian, is not primitive because the greatest common divisor is 2. Figure 2 shows these record numbers up to 15725:

Figure 2

The list of numbers up to 26129 is as follows:

1, 5, 13, 17, 37, 52, 65, 85, 119, 125, 145, 221, 325, 481, 697, 725, 1025, 1105, 1625, 1885, 2465, 2665, 3145, 5525, 6409, 15457, 15725, 26129

As can be seen from the table in Figure 2, the number of possible triangles with a longest side of 15725 is 463 but no figure is available for the number of Heronian triangles with a longest side of 26129. The algorithm I developed to calculate the number works fine for smaller sides but times out for larger numbers. Here is a permalink to the code.

Of course the Heron whose name is being used to identify these types of triangles is the same person responsible for Heron's formula that states the area \(A\) of a triangle whose sides have lengths \(a, b, c\) is:$$A=\sqrt{s(s-a)(s-b)(s-c)} \text{ where } s=\frac{a+b+c}{2}$$It was interesting to learn that a shape is called equable if its perimeter equals its area. There are only five equable Heronian triangles and these have sides (5,12,13), (6,8,10), (6,25,29), (7,15,20), and (9,10,17). As another former Mathematics teacher pointed out in a blog post on Heronian triangles:

Students, or teachers, who liked "The Hitchhiker's Guide to the Galaxy" may enjoy the 7-15-20 triangle which has both perimeter and area of 42. Author Douglas Adams set 42 as the answer to "life, the universe, and everything" and Tony Crilly and Colin Fletcher have dubbed this the "hitchhiker triangle."

There is another category of Heronian triangles that is of interest. Since the area of an equilateral triangle with rational sides is an irrational number, no equilateral triangle is Heronian. However, there is a unique sequence of Heronian triangles that are "almost equilateral" because the three sides are of the form \(n − 1, n, n + 1\). Some of these are listed in Figure 3. These are very similar to the "close-to-equilateral" triangles, mentioned at the start of this blog, that are of the form \(n,n,n+1\) or \(n-1,n-1,n\). 

Figure 3: source

The numbers in column 2, the \(n\) column, form a Lucas sequence defined by$$a_n = 4 \times a_{n-1} - a_{n-2} \text{ with } a_0 = 2 \text{ and } a_1 = 4$$Alternatively, the formula \( (2+\sqrt{3})^n+(2-\sqrt{3})^n\) will generate all the terms. 

There's a lot more to said about Heronian triangles. For example, let's venture into three dimensions and an Heronian tetrahedron defined as a not necessarily regular tetrahedron whose sides, face areas, and volume are all rational numbers. It therefore is a tetrahedron all of whose faces are Heronian triangles and additionally that has rational volume. Figure 4 shows the integer Heronian tetrahedron having smallest maximum side length. It has edge lengths 51, 52, 53, 80, 84, 117; faces (117, 80, 53), (117, 84, 51), (80, 84, 52), (53, 51, 52); face areas 1170, 1800, 1890, 2016; and volume 18144. 

Figure 4: source

The integer Heronian tetrahedron with smallest possible surface area and volume has edges 25, 39, 56, 120, 153, and 160; areas 420, 1404, 1872, and 2688 (for a total surface area of 6384) and volume 8064. The smallest examples of integer Heronian tetrahedra composed of four identical copies of a single acute triangle (i.e., disphenoids) have pairs of opposite sides (148, 195, 203), (533, 875, 888), (1183, 1479, 1804), (2175, 2296, 2431), (1825, 2748, 2873), (2180, 2639, 3111), (1887, 5215, 5512), (6409, 6625, 8484), and (8619, 10136, 11275). An Heronian tetrahedron is sometimes called a perfect tetrahedron.

Subfactorials, Semifactorials and Others

I've written about subfactorials before in a post titled Derangements on Sunday, 13th January 2019 where I explained that:

There are formulae and examples of subfactorials (derangements) aplenty there, so in this post I'm going to focus on what rekindled my interest. It was a YouTube video that I'll link to later and the problem posed was to connect the 3's shown in Figure 1 by mathematical operations so the LHS = RHS.

Figure 1

It wasn't made clear what constituted mathematical operations. Certainly the basic arithmetical operations of addition, subtraction, multiplication, division and exponentiation are sufficient, together with brackets, to balance most of the equations in Figure 1. For example:

• \( (3-3) \times 3 =0\)
• \(3^{3-3} = 1 \)
• \(3+3-3=3 \)
• \( \frac{3}{3}+3=4\)
• \(3 \times 3 - 3=6 \)
• \( 3 + 3 + 3 =9 \)

However, for 5, 7 and 8, the factorial is required as shown in Figure 2 where some alternative solutions for the numbers 0, 1, 2, 3, 4, 6 and 9 are shown:

Figure 2

The final problem of balancing the three 3's with 10 does not yield even to the factorial. It's here that the subfactorial is required. Firstly, let's list the subfactorials, up to ten using SageMathCell (permalink) as shown in Figure 3.

Figure 3

It can be seen that subfactorial 3, which we'll write as !3, is equal to 2 and that's what we need to balance the equation. Specifically:$$3! + !3 +!3=6+2+2=10$$If other mathematical operations such as log, square root and differentiation are allowed, then some other solutions are possible, as shown in Figure 4.

Figure 4

Clearly, the subfactorial offers the simplest solution. Here is the video from which the above screenshots were taken:

So much for subfactorials. What are semifactorials? Well, the semifactorial or double factorial of a number \(n\), denoted by \(n!! \) is the product of all the integers from \(1\) up to \(n\) that have the same parity (odd or even) as \(n\). Thus \(5!! = 5 \times 3 \times 1 \) and \( 4!! = 4 \times 2 \). The factorial function iterated twice should be written as \((5!)! = 6.689503 \times 10^{198}\), a very large number. However, the Google Calculator will return this same result when 5!! is entered and thus does not recognise the semifactorial.

The nomenclature associated with factorials can be quite confusing. For example, when the double factorial or semifactorial is odd, it's sometimes called the odd factorial. Thus the odd factorial of 5 is 5 x 3. To confuse matters even more, there's the falling factorial and the rising factorial and this pair goes by a variety of names as well. The former can be referred to as the descending factorial, the falling sequential product, or the lower factorial while the latter can be referred to as the Pochhammer function, the Pochhammer polynomial, the ascending factorial, the rising sequential product, or the upper factorial. 

Again, there are different ways of representing the falling and rising factorials. The respective different representations and definitions are shown below:$$(x)_n = x^{\underline{n}} = x(x-1)(x-2)\cdots(x-n+1) = \prod_{k=1}^n(x-k+1) = \prod_{k=0}^{n-1}(x-k)$$ $$x^{(n)} = x^{\overline{n}} = x(x+1)(x+2)\cdots(x+n-1) = \prod_{k=1}^n(x+k-1) = \prod_{k=0}^{n-1}(x+k)$$Clearly, there's some need for standardisation in the world of factorials. Anyway, let's apply these different sorts of factorials to the number 5, just to highlight the differences:

• factorial \(5\) is represented as \(5! =120\)
• factorial \(5\) iterated twice is represented as \((5!)! = 6.689503 \times 10^{198}\)
• double factorial \(5\) is represented as \(5!! = 5 \times 3 \times 1 = 15 \)
• \( 5^{\underline{3}}\) would be a falling factorial, equal to \(5 \times 4 \times 3 = 60\)
• \( 5^{\overline{3}}\) would be a rising factorial, equal to \(5 \times 6 \times 7 = 210\)

Returning to the 3 3 3 = 10 problem, the rising factorial can be used to provide another solution, namely:$$\frac{3^{\overline{3}}}{3!}=\frac{3 \times 4 \times 5}{3 \times 2 \times 1}=\frac{60}{6}=10$$

Nude Numbers

It was only in my previous post that I mentioned Friedman numbers, named after the former Associate Professor of Mathematics at Stetson University in DeLand, Florida. His name popped up again this morning when I was investigating the number associated with my diurnal age: 26124. Before discussing the mathematical property of this number, namely its "nudity", I'll include some biographical information about Erich Friedman that I found on his website:

My name is Erich Friedman. For 26 years, I was a Professor of Mathematics at Stetson University, located in DeLand, Florida. I retired in 2018 to spend more time on my other interests, including recreational mathematics, puzzles, trivia, and my girlfriend of 30 years. I was born in 1965 in West Lafayette, Indiana. I grew up in Indianapolis and went to North Central High School. I got my bachelor's degree from Rose-Hulman in 1987, and my Ph.D. from Cornell University in 1991, and have been at Stetson ever since.

There's more information on his website but that's enough for this post. Suffice to say that his website looks interesting with many links to mathematical topics. Eric Friedman is the author of OEIS A034838: numbers \(n\) that are divisible by every digit of \(n\). 26214 is a member of this sequence because 1, 2, 4 and 6 do indeed divide into it without remainder.

I was lead to this sequence by a link in Numbers Aplenty concerning what are colorfully called nude numbers, so called because such numbers expose some of their factors. The explanation on the Numbers Aplenty website runs like this:

Y.Katagiri calls a number nude if it is divisible by all of its digits (which should be nonzero) like \(672=6\cdot112=7\cdot96=2\cdot 336\). The number is called "nude" because it exposes some of its factors. There are only \(9039\) such numbers below one million, however there are infinite nude numbers since all repdigits are nude. The smallest nude number which contains all the odd digits is \(1117935\). Note that if a nude number contains a \(5\), then all the other digits must be odd. The smallest nude \(n\) which contains the maximal (8) number of distinct digits is \(1123449768\). The smallest triple of consecutive nontrivial nude numbers is \((1111, 1112, 1113)\). It is easy to see that there cannot be four consecutive nude numbers greater than 10.

The entry goes on to depict the smallest 3 × 3 magic square whose entries are nontrivial consecutive nude numbers (that is, not the numbers from 1 to 9). See Figure 1.

Figure 1

It's easy enough to generate all the nude numbers from 1 to 26124 using the SageMath code shown below (permalink to SageMathCell):

L=[]

for n in [1..26124]:

    N=n.digits()

    OK=1

    for i in range(len(N)):

        if N[i]==0:

            OK=0

            break

        else:

            if n%N[i]!=0:

                OK=0

                break

    if OK==1:

        L.append(n)

print(L)

print("The percentage of nude numbers up to",n,"is",numerical_approx(len(L)/n*100,digits=2))

The output tells us that approximately 2.9% of the numbers between 1 and 26124 are nude. Here is output:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99, 111, 112, 115, 122, 124, 126, 128, 132, 135, 144, 155, 162, 168, 175, 184, 212, 216, 222, 224, 244, 248, 264, 288, 312, 315, 324, 333, 336, 366, 384, 396, 412, 424, 432, 444, 448, 488, 515, 555, 612, 624, 636, 648, 666, 672, 728, 735, 777, 784, 816, 824, 848, 864, 888, 936, 999, 1111, 1112, 1113, 1115, 1116, 1122, 1124, 1128, 1131, 1144, 1155, 1164, 1176, 1184, 1197, 1212, 1222, 1224, 1236, 1244, 1248, 1266, 1288, 1296, 1311, 1326, 1332, 1335, 1344, 1362, 1368, 1395, 1412, 1416, 1424, 1444, 1448, 1464, 1488, 1515, 1555, 1575, 1626, 1632, 1644, 1662, 1692, 1715, 1722, 1764, 1771, 1824, 1848, 1888, 1926, 1935, 1944, 1962, 2112, 2122, 2124, 2128, 2136, 2144, 2166, 2184, 2196, 2212, 2222, 2224, 2226, 2232, 2244, 2248, 2262, 2288, 2316, 2322, 2328, 2364, 2412, 2424, 2436, 2444, 2448, 2488, 2616, 2622, 2664, 2688, 2744, 2772, 2824, 2832, 2848, 2888, 2916, 3111, 3126, 3132, 3135, 3144, 3162, 3168, 3171, 3195, 3216, 3222, 3264, 3276, 3288, 3312, 3315, 3324, 3333, 3336, 3339, 3366, 3384, 3393, 3432, 3444, 3492, 3555, 3612, 3624, 3636, 3648, 3666, 3717, 3816, 3864, 3888, 3915, 3924, 3933, 3996, 4112, 4116, 4124, 4128, 4144, 4164, 4172, 4184, 4212, 4224, 4236, 4244, 4248, 4288, 4332, 4344, 4368, 4392, 4412, 4416, 4424, 4444, 4448, 4464, 4488, 4632, 4644, 4824, 4848, 4872, 4888, 4896, 4932, 4968, 5115, 5155, 5355, 5515, 5535, 5555, 5775, 6126, 6132, 6144, 6162, 6168, 6192, 6216, 6222, 6264, 6288, 6312, 6324, 6336, 6366, 6384, 6432, 6444, 6612, 6624, 6636, 6648, 6666, 6696, 6762, 6816, 6864, 6888, 6912, 6966, 6984, 7112, 7119, 7175, 7224, 7266, 7371, 7448, 7476, 7644, 7728, 7777, 7784, 8112, 8128, 8136, 8144, 8184, 8224, 8232, 8248, 8288, 8328, 8424, 8448, 8488, 8496, 8616, 8664, 8688, 8736, 8824, 8832, 8848, 8888, 8928, 9126, 9135, 9144, 9162, 9216, 9288, 9315, 9324, 9333, 9396, 9432, 9612, 9648, 9666, 9864, 9936, 9999, 11111, 11112, 11115, 11122, 11124, 11128, 11133, 11136, 11144, 11155, 11166, 11172, 11184, 11196, 11212, 11222, 11224, 11226, 11232, 11244, 11248, 11262, 11288, 11313, 11316, 11322, 11328, 11331, 11355, 11364, 11412, 11424, 11436, 11444, 11448, 11488, 11515, 11535, 11555, 11616, 11622, 11664, 11676, 11688, 11711, 11824, 11832, 11848, 11872, 11888, 11916, 12112, 12122, 12124, 12126, 12128, 12132, 12144, 12162, 12168, 12184, 12212, 12216, 12222, 12224, 12244, 12248, 12264, 12288, 12312, 12324, 12336, 12366, 12384, 12412, 12424, 12432, 12444, 12448, 12488, 12492, 12612, 12624, 12636, 12648, 12666, 12712, 12726, 12768, 12816, 12824, 12848, 12864, 12888, 12924, 12996, 13113, 13116, 13122, 13128, 13131, 13155, 13164, 13212, 13224, 13236, 13248, 13266, 13272, 13311, 13326, 13332, 13335, 13344, 13362, 13368, 13377, 13392, 13416, 13464, 13488, 13515, 13626, 13632, 13644, 13662, 13713, 13755, 13776, 13797, 13824, 13848, 13896, 13932, 13968, 13995, 14112, 14124, 14128, 14136, 14144, 14184, 14212, 14224, 14232, 14244, 14248, 14288, 14292, 14316, 14328, 14364, 14412, 14424, 14436, 14444, 14448, 14488, 14616, 14664, 14688, 14728, 14784, 14824, 14832, 14848, 14888, 15115, 15135, 15155, 15315, 15515, 15555, 15575, 15715, 16116, 16122, 16128, 16164, 16212, 16224, 16236, 16248, 16266, 16326, 16332, 16344, 16362, 16368, 16416, 16464, 16488, 16626, 16632, 16644, 16662, 16716, 16824, 16848, 16992, 17115, 17122, 17136, 17171, 17199, 17248, 17262, 17444, 17472, 17535, 17717, 17724, 17766, 17955, 18112, 18128, 18144, 18168, 18184, 18216, 18224, 18248, 18264, 18288, 18312, 18336, 18384, 18424, 18432, 18448, 18488, 18624, 18648, 18816, 18824, 18848, 18864, 18872, 18888, 18936, 19116, 19224, 19296, 19332, 19368, 19395, 19692, 19719, 19926, 19935, 19944, 19962, 19971, 21112, 21122, 21124, 21126, 21128, 21132, 21144, 21162, 21168, 21184, 21212, 21216, 21222, 21224, 21244, 21248, 21264, 21288, 21312, 21324, 21336, 21366, 21384, 21412, 21424, 21432, 21444, 21448, 21488, 21492, 21612, 21624, 21636, 21648, 21666, 21672, 21728, 21784, 21816, 21824, 21848, 21864, 21888, 21924, 21996, 22112, 22116, 22122, 22124, 22128, 22144, 22164, 22176, 22184, 22212, 22222, 22224, 22236, 22244, 22248, 22266, 22288, 22326, 22332, 22344, 22362, 22368, 22392, 22412, 22416, 22424, 22444, 22448, 22464, 22488, 22626, 22632, 22644, 22662, 22722, 22764, 22824, 22848, 22888, 22896, 22932, 22968, 23112, 23124, 23136, 23166, 23184, 23226, 23232, 23244, 23262, 23292, 23316, 23322, 23328, 23364, 23412, 23424, 23436, 23448, 23616, 23622, 23664, 23688, 23772, 23832, 23922, 24112, 24124, 24128, 24132, 24144, 24168, 24184, 24192, 24212, 24216, 24224, 24244, 24248, 24264, 24276, 24288, 24312, 24324, 24336, 24384, 24412, 24424, 24432, 24444, 24448, 24472, 24488, 24612, 24624, 24636, 24648, 24696, 24724, 24816, 24824, 24848, 24864, 24888, 24912, 24984, 26112, 26124] 

The percentage of nude numbers up to 26124 is 2.9

Wolfram MathWorld has some auxiliary information:

Numbers in base-10 which are divisible by their digits are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99, 111, 112, 115, 122, ... (OEIS A034838). Numbers which are divisible by the sum of their digits are called Harshad numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, ... (OEIS A005349). Numbers which are divisible by both their digits and the sum of their digits are 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 24, 36, 48, 111, 112, 126, 132, 135, 144, ... (OEIS A050104). Numbers which are equal to (i.e., not just divisible by) the product of their divisors and the sum of their divisors are called sum-product numbers and are given by 1, 135, 144, ... (OEIS A038369).

Forming Equations from Integers

Read on!

Once upon a time, before I started this exclusively mathematical blog in 2015, I used to post occasional mathematical content to my Pedagogical Posturing blog at https://voodoo-guru.blogspot.com. I was looking back at some of these posts and noticed this one, from Saturday, 24 August 2013, titled Forming Equations from Integer Sequences. It's only short so I'll quote it in full:

Recently I've been using Twitter to create a daily tweet that records my "day count" (number of days I've been alive) plus its factors (if not prime) and some interesting facts about the number itself or one of its factors. Sometimes there's little to say about the number and in such cases I've found that I can usually form an equation by inserting mathematical operators between one or more of the digits. 

For example, yesterday the count was \(23518\) and \(23 - 5 = 18.\) Today the count is \(23519\) and \(2 + 3 + 5 - 1 = 9\). I was wondering if it's always possible to create an equation from five digits using the standard mathematical operators (addition, subtraction, multiplication, division and exponentiation in combination with brackets). Obviously with just two digits, it's only possible when the digits are repeated e.g. \(99\) becomes \(9 = 9\). With three digits, it's sometimes possible e.g. \(819\) becomes \(8 + 1 = 9\) but generally it isn't e.g. \(219\). With four digits, it's more possible e.g. \(2119\) becomes \(-2 + 11 = 9\) but I'm doubtful whether this is always so. There must come a point however, where the number of digits is sufficient to ensure that it's always so. Maybe five digits is that point.

From now on, I'll try each day to form an equation to test out this theory. For example, tomorrow the count is \(23520\) which becomes \(2+3-5=2 \times 0 \) and it works for tomorrow but beyond that let's see.

Well, I didn't keep my promise of trying to form an equation each day from the digits making up my diurnal age. I have however, written about selfie numbers in a post of Friday, 27th March 2020 to my Mathematical Meanderings blog site. These are somewhat similar in spirit. In that post, I also mention Friedman numbers that can be described as follows:

Consider \(28547 =(8+5)^4−(7 \times 2)\) expressed in base 10, both sides use the same digits. An integer is a Friedman number if it can be put into an equation such that both sides use the same digits but the right hand side has one or more basic arithmetic operators (addition, subtraction, multiplication, division, exponentiation) interspersed. Brackets, as usual, are essential to clarify the order of operations. These numbers are named after Erich Friedman, Assoc. Professor of Mathematics at Stetson University. With the help of his students he has researched Friedman numbers in bases 2 through 10 and even with Roman numerals. When both sides use the digits in the same order, the number is called a ”nice” or ”strong” Friedman number. For example, \( 3125=(3+[1\times2])^5.\)

My approach is similar to this except I'm trying to create an equation from the digits (using them in the same order as they appear in the number). As another example, today I'm \(26121\) days old and this is an easy one because \( (2 \times 6)/12=1 \) if we allow concatenation of digits. If only individual digits are allowed, then \(-(2-6)=1+2+1 \) satisfies. It's probably better to use the individual digits as is done with the Friedman numbers. I'll try to include this as part of my daily number analysis. 

Going back a few days, we have:

• \(26120 \text{ --> } 2 = \frac{6}{1 + 2 +0} \)
• \(26119 \text{ --> } 2 + 6 \times 1 = -1 + 9 \)
• \(26118 \text{ --> } 2 + 6 + 1 = 1 + 8 \)
• \(26117 \text{ --> } 2 + 6 = 1 \times 1 + 7 \)
• \(26116 \text{ --> } 2 + 6 = 1 + 1 + 6 \)
• \(26115 \text{ --> } -2 + 6 + 1 \times 1 = 5 \)
• \(26114 \text{ --> } -2 + 6 = 1 - 1 + 4 \)
• \(26113 \text{ --> } -2 + 6 = 1 \times 1 + 3 \)
• \(26112 \text{ --> } -2 + 6 = 1 + 1 + 2 \)
• \(26111 \text{ --> } 2 = \frac{6}{1+1+1}\)
• \(26110 \text{ --> } 2 \times 6 \times (1-1) = 0 \)

It seems that it's always possible to form an equation using only brackets and the basic arithmetic operators of addition, subtraction, multiplication, division and exponentiation. Perhaps even the exponentiation is not needed, as the examples above show. Let's see if this assumption holds true for future numbers. This is hardly high level mathematics but it's a simple yet oddly satisfying activity.

Honaker: Primes and Problem

Figure 1

G. L. Honaker Jr is a rather shadowy figure who, as we learn from his LinkedIn profile, is a Maths/Science graduate who has been teaching for 28 years. He lives in Bristol, Virginia, and must be around 50 years old. There's not much other biographical data about him, except that he is the co-author of a book that I'll talk about later.

The reason I mention him is that his name came up today as part of my diurnal age number investigation. Today I'm 26119 days old and this turns out to be a Honaker prime. It's pretty cool to have a type of prime number named after you. So what is a Honaker prime? It's defined as a prime \(p_n\) whose index \(n\) and \(p_n\) itself have the same sum of digits. For example, \(p_{32}=131\) is a Honaker prime because \(3+2=1+3+1\).

Unlike primeness, the property that defines a Honaker prime is base specific. For example in base 2, 32 becomes 100000 and 131 becomes 10000011, but the sums of their digits are different. The smallest prime which is Honaker in all the bases from 2 to 10 is  \(p_{277308991}= 5949670231\). I'm thankful to NumbersAplenty for this information.

Not only does Honaker have a type of prime named after him, he also has an eponymous problem. Wolfram MathWorld explains that Honaker's problem asks for all consecutive prime number triples \((p,q,r)\) with \(p<q<r\) such that \(p|(qr+1)\). Caldwell and Cheng (2005) showed that the only Honaker triplets for \(p<=2×10^{17} \text{ are }(2, 3, 5), (3, 5, 7) \text{ and } (61, 67, 71)\). It is conjectured that these three triplets may be the only such triplets with this property.

Interestingly, I'm currently 71 years old and 71 forms part of a Honaker triplet in addition to the fact that my diurnal age (26119) is a Honaker prime. Now getting back to Honaker's book that is available on Amazon. There is an accompanying website that boasts that "there are currently 25464 curios corresponding to 18548 different numbers in our database, that leaves an infinite number for you to discover!"

Naturally I typed in 26119 and discovered that "The number of Honaker primes less than or equal to 26119 is the smallest Honaker prime." This indeed true as the smallest Honaker prime is 131 and there are exactly 131 Honaker primes less than or equal to 26119. I was impressed. Clearly, this is a book to buy and a website to visit for anyone seriously interested in prime numbers.