Some Mathematical Induction

Here is a problem similar to one was posed in a YouTube video that I watched recently:

Prove that \(a^n-b^n\) is divisible by \(a-b\) for all positive integer values of \(n\) with \(a\) and \(b\) both positive integers and \(a \gt b\).

We can do this using mathematical induction as follows:

It's clearly true when \(n=1\) and so let's assume it's true for some value of \(k \gt 1\). This means that:$$ \frac{a^k-b^k}{a-b}=m$$where \(m\) is an integer. Can we now show that it's true for \(k+1\). Let's begin:$$ \begin{align} \frac{a^{k+1}-b^{k+1}}{a-b} &= \frac{a \, . b^k-b \, . b^k}{a-b} \\ a-b &= c \text{ where c is some positive integer} \\ a & = b+c \\ \frac{a \, . b^{k}-b \, . b^{k}}{a-b} &= \frac{(b+c) \, . b^{k}-b \, . b^{k}}{c} \\ &= \frac{b \, . b^{k} +c \,.  b^{k} - b \, . b^{k}}{c} \\ &= \frac{c \, . b^{k} }{c} \\ &=b^{k} \text{ which is an integer} \end{align} $$Since it is true for \(n=1\) and \(n=k+1\) when it is true for \(n=k\) then it is true for all \(n\).

In the video that I referred to at the start of this post, a specific instance of the more general situation was looked at, namely:$$ \text{prove that } \frac{11^n-4^n}{7} \text{ is always an integer}$$

A Special Class of Semiprimes

A well-known factorisation involves the sum of two cubes:$$ a^3 + b^3=(a+b)(a^2-ab+b^2)$$Assuming that both \(a\) and \(b\) are positive integers, then this sum of two cubes is a semiprime if \(a+b\) and \(a^2-ab+b^2\) are both prime. It's interesting to explore when this happens.

Let's begin with \(b=1\) and restrict ourselves to semiprimes between 4 and 4000. There are only seven that qualify and they are:$$9, 65, 217, 4097, 5833, 10649, 21953 $$These numbers form the initial terms of OEIS A237040. Their factorisations are as follows:$$ \begin{align} 9 &= 3^2 = 2^3 + 1^3\\65 &= 5 \times 13 = 4^3 + 1^3\\217 &= 7 \times 31 = 6^3 + 1^3\\4097 &= 17 \times 241 = 16^3 + 1^3\\5833 &= 19 \times 307 = 18^3 + 1^3\\10649 &= 23 \times 463 = 22^3 + 1^3\\21953 &= 29 \times 757 = 28^3 + 1^3 \end{align}$$The next is \(b=8\) which yields six numbers:$$35, 133, 737, 1339, 3383, 24397$$These numbers factorise as follows:$$ \begin{align} 35 &= 5 \times 7 = 3^3 + 2^3\\133 &= 7 \times 19 = 5^3 + 2^3\\737 &= 11 \times 67 = 9^3 + 2^3\\1339 &= 13 \times 103 = 11 ^3 + 2^3\\3383 &= 17 \times 199 = 15^3 + 2^3\\24397 &= 31 \times 787 = 29^3 + 2^3 \end{align} $$These terms are not listed in the OEIS database. Next we'll let \(b=27\). This yields eight numbers and we see that 35 makes a reappearance as the 2 and 3 swap places. The numbers are:$$35, 91, 1027, 2771, 8027, 17603, 21979, 39331 $$These numbers are not listed in the OEIS database and their factorisations are:$$ \begin{align} 35 &= 5 \times 7 = 2^3 + 3^3\\91 &= 7 \times 13 = 4^3 + 3^3\\1027 &= 13 \times 79 = 10^3 + 3^3\\2771 &= 17 \times 163 = 14^3 + 3^3\\8027 &= 23 \times 349 = 20^3 + 3^3\\17603 &= 29 \times 607 = 26^3 + 3^3\\21979 &= 31 \times 709 = 28^3 + 3^3\\39331 &= 37 \times 1063 = 34^3 + 3^3 \end{align} $$Here is a permalink to an algorithm that allows further exploration using additional values of \(b\). A further avenue for investigation would be to consider semiprimes that are a difference of two cubes since we know that:$$ a^3 - b^3=(a-b)(a^2+ab+b^2)$$Using \(b=-1\), we find that the following semiprimes qualify:$$26, 215, 511, 1727, 2743, 7999, 13823$$These numbers form the initial members of OEIS A242262. Their factorisations are as follows:$$ \begin{align} 26 &= 2 \times 13 = 3^3 -1^3\\215 &= 5 \times 43 = 6^3 - 1^3\\511 &= 7 \times 73 = 8^3 - 1^3\\1727 &= 11 \times 157 = 12^3 - 1^3\\2743 &= 13 \times 211 = 14^3 - 1^3\\7999 &= 19 \times 421 = 20^3-1^3\\13823 &= 23 \times 601 = 24^3-1^3 \end{align} $$

Zeisel Numbers

Digalo con numeros
Say it with Numbers

Investigating the properties of the number associated with my diurnal age yesterday (27559), I noticed that Numbers Aplenty mentions that this number is a Zeisel number with parameters (4, 3). Now this was a type of number that I'd not heard of before. Clicking on the link in previously mentioned resource, I was given this explanation:

Let us define a sequence as:$$\begin{array}{l} p_0 = 1\\ p_n = a\cdot p_{n-1}+b\ \end{array}$$where \(a,b\in\mathbb{Z}\). If the numbers \(p_1,p_2,\dots,p_k\) are all distinct primes and \(k\ge 3\), then their product is a Zeisel number.

So applied to 27599, we start with 1 to form a new number thus \(4 \times 1 + 3 = 7\). This is prime so it is used as our new input to form the next number which is \(4 \times 7 + 3 = 31\). This is also prime so we proceed using 31 as the new input. This generates \(4 \times 31 + 3 = 127\) which is prime and this sequence of prime numbers (\(7, 31,127\)) can thus form a Zeisel number: $$7 \times 31 \times 127 = 27559$$The input 127 does not produce a prime number and so no further Zeisel numbers can be formed using (4, 3) as parameters. Interestingly, the primes 7, 31 and 127 are also consecutive Mersenne primes.

The example is given in the link of \(1419 = 3 \times 11 \times 43\) that is a Zeisel number with parameters of \(a=4\) and \(b=-1\), because \(3 = 4 \times 1-1\), \(11=4 \times 3-1\) and \(43=4 \times 11-1\).

The smallest Zeisel numbers which are the product of 3, 4, 5 and 6 factors are shown in Table 1.

Table 1: link

The first Zeisel numbers are 105, 1419, 1729, 1885, 4505, 5719, 15387, 24211, 25085, 27559, 31929, 54205, 59081, 114985, 207177 (OEIS A051015). Clearly they are few and far between as the previous such number in my diurnal age count was 25085. This was before I began keeping records in my first AirTable database. After 27599, the next is 31929, a long way off.

Compressing a Hollow Cube

Rewatching the movie "Transformers", my attention was caught by the giant (presumably hollow) cube made up out of many tiny "unit" cubes, that reassembled itself into a much smaller solid cube. This got me thinking about what volume a hollow cube could be compressed into to form a solid cube. Let's consider a  hollow cube with a side of \(n\) units. It has a volume of \(n^3\) cubic units with a hollow centre of \( (n-2)^3 \) cubic units . Thus the number of unit cubes are:$$\begin{align} n^3 - (n-2)^3 &= 2 (n^2 + n(n-2) + (n-2)^2) \\ &= 2(n^2+n^2-2n+n^2-4n+4)\\ &=2(3n^2-6n+4) \end{align}$$Using this formula we can determine the size of the solid cube that can be formed from the unit cubes and how many blocks are left over from this process. See Table 1.

Table 1: permalink

From Table 1 we can see that 216 of the 218 unit cubes that comprise a hollow cube of side 7 units can be used to form a solid cube of side 6 units with only two cubes left over. See Table 2.

Table 2

Also we can see that 13824 of the 13826 unit cubes that comprise a hollow cube of side 49 units can be used to form a solid cube of side 24 units with only two cubes left over. See Table 3.

Table 3

Looking beyond the figures in Table 1, we find that hollow cubes with sides of 163, 385, 751 and 1297 units also compress to solid cubes with sides of 54, 96, 150 and 216 units with two cubes left over (permalink).

We can summarise the results for the "2 left over hollow to solid cubes" as follows:

(7, 6) gives 62.97 % ratio of volumes (solid to hollow)
(49, 24) gives 11.75 % ratio of volumes (solid to hollow)
(163, 54) gives 3.636 % ratio of volumes (solid to hollow)
(385, 96) gives 1.550 % ratio of volumes (solid to hollow)
(751, 150) gives 0.7968 % ratio of volumes (solid to hollow)
(1297, 216) gives 0.4619 % ratio of volumes (solid to hollow)

Naturally as the side length of the hollow cube increases, the ratio of the volume of the solid cube to its hollow counterpart decreases rapidly as more and more empty space is formed inside. For example, in the case of the 751 hollow cube, the ratio of the solid to hollow volumes in less than 1%. It can be noted that none of the sides of the "2 left over hollow to solid cubes" have any factors in common.

Let's see if there's a pattern in the sides of the hollow and solid cubes:

• 7 = 7 and 6 = 2 * 3
• 49 = 7^2 and 24 = 2^3 * 3
• 163 = 163 and 54 = 2 * 3^3
• 385 = 5 * 7 * 11 and 96 = 2^5 * 3
• 751 = 751 and 150 = 2 * 3 * 5^2
• 1297 = 1297 and 216 = 2^3 * 3^3

It can be seen that the solid cubes all contain "6" as a factor. Let's investigate further. Is it possible that a hollow cube with integer sides can be collapsed into a solid cube of integer side \(x\) with no unit cubes left over? If it were possible then we would have:$$ \begin{align} 2(3n^2-6n+4) &= x^3\\6n^2-12n+8 &= x^3\\x &= (6n^2-12n+8)^{1/3}  \end{align}$$Now testing up to \(n=10000\), the expression on the LHS of the equation above never produces a whole number. I'll test for larger values of \(n\) later.

What about when there are two unit cubes left over. In that case we have:$$ \begin{align} 2(3n^2-6n+4) &= x^3+2\\6n^2-12n+6 &= x^3 \\ x &= (6n^2-12n+6)^{1/3} \end{align} $$Now in this case, up to \(n=10000\), we get all the solutions shown above plus some more. These are shown in Table 4.

Table 4: permalink

So the key equation to work with is \(6n^2-12n+(8-a) \) where \(a\) is the number of unit cubes left over. Up to 10000, there are no solutions for \(a=0\) and \(a=1\) but with \(a=2\) we get the solutions shown in Table 4. Going back to Table 1 and using the values of \(a\) shown there in the column "blocks wasted" will produce other tables of solutions as shown in Table 4. There's more to be discovered here but this post is at least a start.

New Telephone Number

It's always exciting to get a new telephone number because of the properties of the number that may turn up. Having arrived in Australia for a temporary stay, I needed a telephone number and the number that I was given was:$$0451591949 \rightarrow 451591949$$Now this number is prime but has additional "primeness" embedded in it because:$$ \text{Sum of digits is }47 \text{ and prime}\\47 \rightarrow 4 + 7 = 11 \\ 11 \rightarrow 1+1=2 \\ \text{ 2 (the digital root) is prime}$$Now 1949 and 1951 form a pair of twin primes but it turns out that:$$ 451591949 \text{ and } 4511591951 \\ \text{ also form a pair of twin primes}$$Furthermore:$$451591949 \text{ is a Sophie Germain prime} \\ 451591949 \times 2 + 1 = 903183899 \text{ a prime}$$\(451591949\) is also a Chen prime defined as a prime number  \(p\)  such that  \(p+2\)  is either a prime or a semiprime. Jing Run Chen, after which they are named, proved in 1966 that there are infinitely many such primes. Binbin Zhou has proved in 2009 that the Chen primes contain arbitrarily long arithmetic progressions.

So overall I'm very happy with my new prime telephone number even though it will lapse and be discarded once I leave the country. For now though it's mine and prime!

Composites as Linear Combinations of Prime Numbers

Goldbach's Conjecture states every even number can be written as a sum of two primes. This has not yet been proven but it appears highly likely. In fact, it seems that every number greater than 12 can be written as a sum of two primes in at least two different ways. For example:$$\begin{align} 14 &=3 + 11 \\ &= 7+7\\16 &= 3+13 \\ &=5 + 11 \end{align}$$What about expressing primes as linear combinations of pairs of successive primes? Let's say that \( (p, q) \) and \((r, s) \) are pairs each made up of two successive primes. For a given number \(n\) can we find constants \(a, b, c, d\) such that:$$ \begin{align} n &=a \cdot p + b \cdot q \\ &=c \cdot  r + d \cdot s \end{align}$$Generally speaking we can and particular examples can be found in OEIS  A283392 where \(a=3\), \(b=5\), \(r=5\) and \(s=7\):

A283392

Integers m of the form \(m = 3 \cdot p + 5 \cdot q = 5 \cdot r + 7 \cdot s \) where \( (p, q) \) and \( (r, s) \) are pairs of consecutive primes.

Up to 40,000, the members of this sequence are 50, 146, 866, 2162, 4178, 8362, 14372, 17138, 19094, 22504, 25346, 26764, 27544 and 35074 (permalink). 

For example:$$ \begin{align} 27544 &=3 \cdot 3433 + 5 \cdot 3449 \\ &= 5 \cdot 2293 + 7 \cdot 2297 \end{align}$$This is equivalent to finding the solutions to the two separate equations with \(x\) and \(y\) necessarily being consecutive primes:$$ \begin{align} 3x_1+5y_1 &= 27544 \text{ with } x_1 = 3433 \text{ and } y_1=3449\\5x_2+7y_2 &=27544 \text{ with }x_2=2293 \text{ and } y_2=2297 \end{align}$$The algorithm that generated the earlier list is easily modified to accommodate different values of  \(a, b, c, d\). Let's take the case of \(a=3\), \(b=5\), \(r=7\) and \(s=9\) where the following numbers (not listed in the OEIS) satisfy (permalink):

98, 482, 2314, 8966, 9766, 10562, 12148, 12962, 13066, 14372, 14548, 25588, 31202, 31414, 31846, 34082, 37174, 37588, 39526

An example is 98 where:$$ \begin{align} 98 &=3 \cdot 11 + 5 \cdot 13 \\ &= 7 \cdot 5 + 9\cdot 7 \end{align}$$Of course 98 itself cannot be written as sum of two consecutive primes because:$$ \begin{align} 98 &= 19 + 79 \\ &= 31 + 67 \\ &= 37+61 \end{align} $$Clearly (19, 79), (31, 67) and (37, 61) are not consecutive primes. However, by considering linear combinations of the two consecutive primes, we see (11, 13) and (5, 7) satisfy.

Here is another variation \(a=2\), \(b=3\), \(r=4\) and \(s=5\) where 38 numbers are generated in the range up to 40000. The numbers are (permalink):

407, 541, 1351, 1757, 1783, 2161, 4973, 6077, 6349, 6511, 6889, 7427, 8083, 8723, 10427, 11747, 12751, 13069, 13339, 14833, 18203, 20657, 20791, 21283, 24953, 25111, 29171, 29297, 29323, 31703, 33511, 34157, 35623, 36047, 37397, 37423, 38507, 39181

An example is 407 where:$$ \begin{align} 407 &=2 \cdot 79 + 3 \cdot 83 \\ &= 4 \cdot 43 + 5\cdot 47 \end{align}$$Yet another variations is \(a=3\), \(b=4\), \(r=5\) and \(s=6\) where 23 numbers are generated in the range up to 40000. The numbers are (permalink):

1123, 1157, 4327, 4621, 5671, 5987, 11867, 15877, 19111, 21451, 22157, 23477, 23731, 24257, 26887, 27097, 30199, 30367, 33707, 35773, 36961, 37031, 39607

An example is 1123 where:$$ \begin{align} 1123 &=3 \cdot 157 + 4\cdot 163 \\ &= 5 \cdot 101 + 6\cdot 103 \end{align}$$I'll leave it there but the permalink provides the opportunity to play around with other combinations of \(a, b, c, d\). The number properties described in this post are not base dependent and so are possibly of serious mathematical interests. It seems to be that there should be some practical applications of this technique as a useful way of decomposing or building composite numbers.

Here is an alternative algorithm that generates the same output but uses a different method. It can be used to discover numbers that are linear combinations of two consecutive primes without being equal to a linear combination of two other consecutive primes. For example, it will show that 844 is a linear combination of 3 times 103 and 5 times 107. In other word we have solved the linear equation:$$ 3x+5y = 844$$where \(x\) and \(y\) are prime and \(y\) is the next prime after \(x\). 

Obviously, by choosing prime values of \(a, b\) and \(c,d\) we end up with a composite number that is the sum of the products of two pairs of primes. Thus for 844 we have:$$844 = 3 \cdot 103 + 5 \cdot 107$$ which defines the number in the similar way to its prime factorisation:$$844 = 2 \cdot \ 2 \cdot 211$$Unlike the product of a number's prime factors however, this representation is not necessarily unique as we've seen that some numbers have more than one representation as the sum of the products of two pairs of primes e.g. 2162 (permalink):$$ \begin{align} 2162 &=3 \cdot 269 + 5\cdot 271 \\ &= 5 \cdot 179 + 7\cdot 181 \end{align}$$