A Simple BMI Problem

For some reason I thought of this simple algebraic problem while going to sleep last night. Here it is:

A fully grown adult person has a body mass index of 20.2. After gaining 3 kilograms in weight, the same person now has a body mass index of 21.2. How tall is the person to the nearest centimetre?

Let's say the person's initial weight is \(W\) and height is \(h\). This gives us two equations:$$ \begin{align} \frac{W}{h^2} &= 20.2\\ \\ \frac{W+3}{h^2}&=21.2 \end{align} $$ The first equation yields:$$W=20.2 \,h^2$$ and substituting this into the second yields:$$\begin{align} 20.2 \, h^2+3 &= 21.2 \, h^2\\h^2 &= 3\\h &= \pm \sqrt{3}\\ &\approx \pm 1.73 \end{align} $$Thus the person is 1.73 metres tall to the nearest centimetre. I obviously chose easy numbers to work with but the calculation is straightforward for any numbers.

Clearly the problem is of lower secondary high school level of difficulty but for some reason it popped into my head, an echo of my long ago days as a high school teacher of Mathematics. The key to the problem is realising that the height of the person doesn't change because we are dealing with an adult and not a child or adolescent. Units for weight are kilograms and, for height, metres. The problem could be made more difficult by using pounds and inches, forcing conversion between the metric and non-metric systems of measurement.

The issue of accuracy in this problem should be addressed because it must be noted that the BMI is accurate to only one decimal place and yet the height is being reported to two decimal places. BMI is commonly expressed to one decimal place so it's reasonable to suppose that a rounding to the nearest one decimal place has occurred. This means that the first BMI (let's call it BMI_1) and the second BMI (let's call it BMI_2) lie within the following bounds: $$20.15 \leq \text{ BMI}_1 <20.25\\21.15 \leq \text{BMI}_2 <21.25$$Considering the minimum and maximum of these ranges, we get a maximum and minimum difference as follows:$$ \begin{align} \text{minimum}&=21.15-20.25\\ &=0.90\\ \text{maximum} &=21.25-20.15\\ &=1.10 \end{align}$$Going back to our original equation and designating the minimum and maximum heights as \(h_{ \text{min}} \) and  \(h_{ \text{max}} \) respectively, we see that:$$\begin{align} 1.1 \, h_{\text{min}}^2 &= 3 \\ h_{ \text{min}} &=  \pm \sqrt{3 \div 1.1}\\ & \approx \pm 1.65 \\0.9 \, h_{\text{max}}^2 &= 3 \\ h_{ \text{max}} &=\pm \sqrt{3 \div 0.9}\\ & \approx \pm 1.83 \end{align}$$Thus the height could have ranged between 1.65 metres and 1.83 metres if we record the BMI to the nearest one decimal place. The assumption here is that the weight is 3.00 kilograms which is unlikely of course but I didn't want to complicate matters.

Having realised this, let's rephrase the problem using BMI's accurate to two decimal places as well as a weight accurate to two decimal places as well:

A fully grown adult person has a body mass index of 20.23. After gaining 3.11 kilograms in weight, the same person now has a body mass index of 21.19. How tall is the person to the nearest centimetre?

Using these figures, we see that:$$ \begin{align} 0.96 \, h^2 &=3.11\\ h&= \pm \sqrt{3.11 \div 0.96}\\& \approx \pm 1.80 \end{align}$$Thus we have a height of 1.80 metres to the nearest two decimal places. The moral of this problem is that it's meaningless to have one quantity accurate to one decimal place (BMI) and another accurate to the nearest whole number (weight) and to then ask for a quantity that is dependent on these first two quantities to be expressed with an accuracy of two decimal places. This is simply false accuracy and it can creep into calculations unless we are vigilant, which I wasn't when I originally framed the problem. The more we multiply and divide approximations, the more the margin of error grows.

This post marks my 100th of the year and equals my record for 2022. My record of 105 posts is from 2021, a record I seem likely to break given that three months of the year remain.

Pointer Primes and Beyond

I need to acknowledge that the number associated with my diurnal age today, 27211, is the last pointer prime that I will encounter in my current lifetime, the next being 61211. So what is a pointer prime?

The product of the digits of the prime number 23 is 2 x 3 = 6. Note that 23 + 6 = 29, the next prime number. The prime 23 can be said to "point" to the next prime number 29 in the sense that when 23 is added to the product of its digits, 29 is obtained. We define the prime number \(p\) to be a pointer prime if the next prime after \(p\) can be obtained from \(p\) by adding to \(p\) the product of the digits of \(p\). Source.

The pointer primes up to one million are as follows(permalink):

23, 61, 1123, 1231, 1321, 2111, 2131, 11261, 11621, 12113, 13121, 15121, 19121, 21911, 22511, 27211, 61211, 116113, 131231, 312161, 611113

The conjecture is that 23 is the only pointer prime that does not contain a 1 and this seems likely to be the case. In the case of my diurnal age number we have 27211 + 28 = 27239.

It's easy enough to extend the idea of pointer primes to pointer semiprimes. In this case we can say that:

We define the semiprime number \(n\) to be a pointer semiprime if the next semiprime after \(n\) can be obtained from \(n\) by adding to \(n\) the product of the digits of \(n\).

In the range up to one million, the only such pointer semiprimes are 22, 123, 1411, 11521, and 111161 as shown below (permalink with the algorithm only considering semiprimes with distinct prime factors):

22 = 2 * 11 --> 26 = 2 * 13

123 = 3 * 41  --> 129 = 3 * 43

1411 = 17 * 83  --> 1415 = 5 * 283

11521 = 41 * 281  --> 11531 = 13 * 887

111161 = 89 * 1249  --> 111167 = 7 * 15881

It can be seen that all except 22 contain the digit 1 as was the case with 23 and the other pointer primes.

So let's go on to pointer sphenics in which case we can say that:

We define the sphenic number \(n\) to be a pointer sphenic if the next sphenic after \(n\) can be obtained from \(n\) by adding to \(n\) the product of the digits of \(n\).

In the range up to one million, the only such pointer sphenics are 222, 21122, 31114, 111122, 111711, 125111 and 912111 as shown below (permalink):

222 = 2 * 3 * 37 --> 230 = 2 * 5 * 23

21122 = 2 * 59 * 179 --> 21130 = 2 * 5 * 2113

31114 = 2 * 47 * 331 --> 31126 = 2 * 79 * 197

111122 = 2 * 11 * 5051 --> 111126 = 2 * 3 * 18521

111711 = 3 * 23 * 1619 --> 111718 = 2 * 83 * 673

125111 = 7 * 61 * 293 --> 125121 = 3 * 179 * 233

912111 = 3 * 23 * 13219 --> 912129 = 3 * 47 * 6469

Once again it is only 222 that does not contain the digit 1 while all others do. Of course, we could change the definition of a pointer prime so that the prime needs only to point to any higher prime not just the next higher prime. We need to exclude primes containing zero in this definition because the product of digits is then zero and adding zero to the original prime leaves us with the original prime! By this definition, the pointer primes would be far more frequent. For example, there would be 899 of them in the range up to 40000 (permalink). 

Periodic Unitary Aliquot Sequences.

At first glance, the phrase "periodic unitary aliquot sequences" can sound intimidating so it needs to be broken down into its individual components. Let's start with a definition of aliquot taken from study.com:

An aliquot is a portion or part of a larger whole. An aliquot, or the aliquot part as it is referred to in mathematics, is defined as a positive proper divisor of a number. A divisor refers to a whole number that can be divided evenly into a number.

Using the number associated with my diurnal age today, 27208, the aliquot parts of this number are 1, 2, 4, 8, 19, 38, 76, 152, 179, 358, 716, 1432, 3401, 6802 and 13604. 

The next term to deal with is unitary. The unitary divisors of a number are defined by Wikipedia as follows:

\(a\) is a unitary divisor (or Hall divisor) of a number \(b\) if \(a\) is a divisor of \(b\) and if \(a\) and \(b/a\) are coprime, having no common factor other than 1. Thus, 5 is a unitary divisor of 60, because 5 and 60/5 =12 have only 1 as a common factor, while 6 is a divisor but not a unitary divisor of 60, as 6 and 60/6 = 10 have a common factor other than 1, namely 2. 1 is a unitary divisor of every natural number. 

In the case of 27208, the unitary divisors are 1, 8, 19, 152, 179, 1432, 3401, 27208 but the unitary aliquot divisors are 1, 8, 19, 152, 179, 1432, 3401. See my blog post Unitary Divisors.

Next we need to tackle aliquot sequences. Here is a definition:

Now in the case of 27208, the aliquot sequence terminates in zero. Here is the trajectory (permalink):

27208, 26792, 26668, 21212, 15916, 13316, 9994, 5846, 3274, 1640, 2140, 2396, 1804, 1724, 1300, 1738, 1142, 574, 434, 334, 170, 154, 134, 70, 74, 40, 50, 43, 1, 0

This sequence is terminating and not periodic. However, a unitary aliquot sequence uses the unitary divisors and can behave quite differently. In the case of 27208, the sequence becomes periodic with the following trajectory (permalink):

27208, 5192, 1288, 440, 208, 30, 42, 54, 30

27208 is a member of OEIS A003062:

A003062

Beginnings of periodic unitary aliquot sequences.

The initial members are:

6, 30, 42, 54, 60, 66, 78, 90, 100, 102, 114, 126, 140, 148, 194, 196, 208, 220, 238, 244, 252, 274, 288, 292, 300, 336, 348, 350, 364, 374, 380, 382, 386, 388, 400, 420, 436, 440, 476, 482, 484, 492, 516, 528, 540, 542, 550, 570, 578, 592, 600, 612, 648, 660, 680, 688, 694, 708, 720, 722, 740, 756, 758, 764, 766, 770, 780, 784, 792, 794, 812

Thus periodicity is fairly common for these types of sequences, quite unlike the aliquot sequences which nearly all end in 0. See my post Aliquot Sequences.

Graham-Pollak Sequences

I  was reminded of my previous post titled Pisot Sequences when I came across the topic of this current post relating to Graham-Pollak sequences. An example of this type of sequence is the following:$$a(n) = \text{ floor} \sqrt{2 \times a(n-1) \times (a(n-1)+1)}\\ \text{ where } a(0)=1$$The initial members of this sequence (OEIS A001521) are (permalink):

1, 2, 3, 4, 6, 9, 13, 19, 27, 38, 54, 77, 109, 154, 218, 309, 437, 618, 874, 1236, 1748, 2472, 3496, 4944, 6992, 9888, 13984, 19777, 27969, 39554, 55938, 79108, 111876, 158217, 223753, 316435, 447507, 632871, 895015, 1265743, 1790031, 2531486, 3580062, 5062972

This sequence has apparently some amazing properties that can be read about here in Wolfram Mathworld but I have to confess to not understanding their significance. Note that from 1 to 9, the numbers 5 and 8 are missing and so two others sequences, listed in the OEIS, can be generated using these numbers as starting points instead of 1. Starting with 5 produces OEIS A091522 (permalink):

A091522

Graham-Pollak sequence with initial term 5.

The initial members are:

5, 7, 10, 14, 20, 28, 40, 57, 81, 115, 163, 231, 327, 463, 655, 927, 1311, 1854, 2622, 3708, 5244, 7416, 10488, 14832, 20976, 29665, 41953, 59331, 83907, 118663, 167815, 237326, 335630, 474653, 671261, 949307, 1342523, 1898614, 2685046 

Starting with 8 produces OEIS A091523 (permalink):

A091523

Graham-Pollak sequence with initial term 8.

The initial members are:

8, 12, 17, 24, 34, 48, 68, 96, 136, 193, 273, 386, 546, 772, 1092, 1545, 2185, 3090, 4370, 6180, 8740, 12360, 17480, 24721, 34961, 49443, 69923, 98886, 139846, 197772, 279692, 395544, 559384, 791089, 1118769, 1582179, 2237539, 3164358

I came across the Graham-Pollak sequences through a reference in the OEIS to the number associated my diurnal age of 27207. The number arises from the following formula:$$a(n) = \text{ floor} \sqrt{3 \times a(n-1) \times (a(n-1)+1)}\\ \text{ where } a(0)=1$$Notice that the multiplier is not 2 but 3. The numbers generated by this formula produce OEIS A100671 (permalink):

A100671

A Graham-Pollak-like sequence with multiplier 3 instead of 2.

1, 2, 4, 7, 12, 21, 37, 64, 111, 193, 335, 581, 1007, 1745, 3023, 5236, 9069, 15708, 27207, 47124, 81622, 141374, 244867, 424122, 734601, 1272367, 2203805, 3817103, 6611417, 11451311, 19834253, 34353934, 59502759, 103061802, 178508278, 309185407, 535524834 

Apparently it's not known whether this sequence has properties similar to the original.

Pisot Sequences

Thanks to the number associated with my diurnal age today, 27204, I was introduced to so-called Pisot sequences that I'm still coming to terms with. Let's start with one of the properties of this number, namely its membership in OEIS A048589:

A048589

Pisot sequence L(7, 9).

The members of the sequence are derived using the following formula:

$$a(n)= \Biggl \lceil \frac{a(n - 1)^2}{a(n - 2)} \Biggr \rceil \\ \text{ where } n \geq 2 \text{ with }a(0)=7 \text{ and } a(1)=9$$The initial members of the sequence are:

7, 9, 12, 16, 22, 31, 44, 63, 91, 132, 192, 280, 409, 598, 875, 1281, 1876, 2748, 4026, 5899, 8644, 12667, 18563, 27204, 39868, 58428, 85629, 125494, 183919, 269545, 395036, 578952, 848494, 1243527, 1822476, 2670967, 3914491, 5736964, 8407928, 12322416, 18059377

However, exactly the same sequence of numbers can be derived from the recurrence relation:$$a(n)=2a(n-1)-  a(n-2) + a(n-3) - a(n-4)\\ \text{ with } a(0)=7, a(1)=9, a(2)=12 \text{ and } a(3)=16$$However, the equivalence may last only up to \(n=1000\) according to the OEIS comments. After that, there is uncertainty. Permalink.

Let's take another example using OEIS A018910:

A018910

Pisot sequence L(4, 5).

Once again, the previous formula can be used with different starting values:$$a(n)= \Biggl \lceil \frac{a(n - 1)^2}{a(n - 2)} \Biggr \rceil \\ \text{ where } n \geq 2 \text{ with }a(0)=4 \text{ and } a(1)=5 $$The initial members of the sequence are:

4, 5, 7, 10, 15, 23, 36, 57, 91, 146, 235, 379, 612, 989, 1599, 2586, 4183, 6767, 10948, 17713, 28659, 46370, 75027, 121395, 196420, 317813, 514231, 832042, 1346271, 2178311, 3524580, 5702889, 9227467, 14930354, 24157819, 39088171, 63245988, 102334157, 165580143

Again, the same sequence can be generated using the recurrence relation (permalink):$$a(n)=2a(n-1)- a(n-3)\\ \text{ with } a(0)=4, a(1)=5 \text{ and } a(2)=7 $$Just to confuse matters we can also generate the sequence using:$$a(n) = \text{ Fibonacci}(n+3)+2$$What the letter L signifies I don't know. Other letters are used with Pisot sequences, including E, P and T. Not all Pisot sequences use the ceiling function. Some use the floor function and others add or subtract a constant. There's not a great deal of introductory information out there about these sorts of sequences. You find yourself in deep mathematical matters very quickly once you start to explore the topic. 

Here is what Google's Bard had to say about them:

Pisot sequences are a type of infinite sequence of real numbers that have a number of interesting and useful properties. They are named after the French mathematician Charles Pisot, who first studied them in the 1930s.

Simple examples of Pisot sequences include:

• The Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
• The Lucas sequence: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, ...
• The Parry sequence: 0.1101001000101101...

Pisot sequences are important because they have a number of unique properties. For example:

• They are all irrational numbers.
• They are all aperiodic (meaning that they never repeat themselves exactly).
• They are all dense in the real numbers (meaning that between any two real numbers, there is always a Pisot number).

Pisot sequences are connected to Pisot numbers and Salem numbers in the following way:

• A Pisot number is a real number that is greater than 1 and has the property that all of its conjugates are less than 1.
• A Salem number is a real number that is greater than 1 and has the property that all of its conjugates are either less than 1 or equal to 1.

Every Pisot number is the limit of a Pisot sequence, and every Pisot sequence converges to a Pisot number. Salem numbers are a special type of Pisot number.

Here are some of the applications of Pisot sequences:

• They can be used to generate random numbers.
• They can be used to model complex systems, such as the growth of populations and the spread of diseases.
• They can be used to design efficient algorithms for tasks such as sorting and searching.
• Pisot sequences are a fascinating and important area of mathematics, and they have a wide range of applications in other fields.

Misuse of Mathematical Symbols

I often see arithmetical challenges like:$$ \begin{align} 2+5&=14\\3+6&=27\\5+1&=30\\7+2&=\text{ ?} \end{align}$$What's annoying about this is the abuse of the addition and equal signs. Clearly the numbers shown on the left hand side of the equation do NOT add up the number shown on the right hand side and so the equal sign should NOT be used.

Instead some operation, not addition, is being performed on the two numbers on the left in order to generate the outcome shown. In other words, we have a function of two variables. Let's call the variables \(x\) and \(y\) and the function \(f\). This can be represented then as \(f(x,y)\) and we knows that:$$ \begin{align} f(2,5)&=14\\f(3,6)&=27\\f(5,1)&=30\\f(7,2)&=\text{ ?} \end{align}$$This is the correct way to present the problem and then by a little bit of experimentation, it can be seen that:$$ \begin{align} f(x,y)&=x \times (x+y)\\&=x^2+xy \end{align}$$The solution to the original problem is then:$$ \begin{align} f(7,2)&=49+14\\&=63 \end{align}$$Using this method, it's easy to generate new challenges. For example let's suppose that:$$ \begin{align} f(x,y)&=x \times (x-y)\\&=x^2-xy \end{align} $$ We could then formulate the challenge: $$ \begin{align} f(3,1)&=6\\f(5,3)&=10\\f(7,6)&=7\\f(8,5)&=\text{ ?} \end{align}$$The correct answer is then:$$ \begin{align} f(8,5)&= 64-40\\&=24 \end{align} $$Of course, the function could be of three variables. For example:$$f(x,y,z)=(x+y-z)^2$$A challenge could then be formulated as follows:$$ \begin{align} f(2,3,1)&=16\\f(1,4,2)&=9\\f(3,5,1)&=49\\f(5,2,3)&= \text{ ?} \end{align}$$The answer is then:$$ \begin{align} f(5,2,3)&=(5+2-3)^2\\&=4^2\\&=16 \end{align} $$And so it goes.

Quadruple Lucky Numbers

It's known that the frequency of lucky numbers is similar to that of prime numbers. The frequency of twin lucky numbers is also similar to that of twin primes. The similarities do not end there. Quadruple prime numbers occur in a \(p, p+2, p+6 \text{ and } p+8\) pattern. For example: 101, 103, 107 and 109 or 191, 193, 197 and 199. 

Quadruple lucky numbers follow a similar pattern: \(n, n+2, n+6 \text{ and } n+8 \). The first members of these quadruplets populate OEIS A139783:

A139783

Quadruple lucky numbers (lower terms). Numbers \(n\) such that \(n, n+2, n+6, n+8\) are all Lucky numbers.

The initial members of the sequence are (permalink):

1, 7, 67, 127, 613, 925, 1495, 1765, 2209, 2815, 3403, 5965, 6661, 8827, 9115, 15229, 16387, 18145, 19153, 21925, 23563, 24637, 27031, 27199, 28987, 31381, 32635, 34717, 35701, 36673, 40447, 43225, 43975, 47419, 50317, 51157, 56263, 64495

For example, my diurnal age today (27199) belongs to the sequence and thus 27199, 27201, 27205 and 27207 are all lucky numbers. All four numbers are composite but I got to thinking whether there are quadruple lucky numbers that are also quadruple prime numbers?

To begin with the initial primes would need to end in the digit 1 and up to one million the only candidates fail as can be seen below:

6661

6663 = 3 * 2221

6667 = 59 * 113

6669 = 3^3 * 13 * 19

27031

27033 = 3 * 9011

27037 = 19 * 1423

27039 = 3 * 9013

70351

70353 = 3^2 * 7817

70357 = 7 * 19 * 23^2

70359 = 3 * 47 * 499

125311

125313 = 3 * 41771

125317 = 113 * 1109

125319 = 3 * 37 * 1129

321721

321723 = 3^2 * 35747

321727 = 7 * 19 * 41 * 59

321729 = 3 * 107243

602221

602223 = 3 * 191 * 1051

602227 = 602227

602229 = 3 * 197 * 1019

728911

728913 = 3 * 242971

728917 = 7 * 101 * 1031

728919 = 3^4 * 8999

786661

786663 = 3^2 * 87407

786667 = 7 * 41 * 2741

786669 = 3 * 13 * 23 * 877

This is a limited range on which to make a judgement as to whether such a thing is possible and there may well be a theoretical proof that shows that it's impossible. However, that will have to do for now.