Fibonacci Generating Functions

The Fibonacci Sequence can be generated from the following recurrence relation:$$a(n)=a(n-1)+a(n-2) \text{ with } a(0)=1 \text{ and } a(1)=1$$The generating function for this series is given by:$$ \frac{x^2}{1-x-x^2} $$How is generating function affected if we have a recurrence relation as follows:$$a(n)=a(n-1)+a(n-8)$$I created a post about this titled Fibonacci-like Sequences quite recently on May 13th 2024. The new generating function is now:$$ \frac{x^2}{1-x-x^8} $$What if there are coefficients (let's say \( \alpha \) and \( \beta \) in front of the two terms on the LHS:$$a(n) = \alpha . a(n-1) + \beta . a(n-2)$$In this case, the generating function becomes: $$ \frac{x^2}{1-\alpha . x - \beta . x^2}$$Let's take the following example:$$a(n)=4. \, a(n-1)+10. \, a(n-2)$$The generating function becomes$$ \frac{x^2}{1-4x - 10 x^2}$$The sequence of terms becomes (permalink):

0, 1, 4, 26, 144, 836, 4784, 27496, 157824, 906256, 5203264, 29875616, 171535104, 984896576, 5654937344, 32468715136, 186424233984, 1070384087296, 6145778689024, 35286955629056

This numbers form the initial terms of OEIS A180226:

A180226

a(n) = 4*a(n-1) + 10*a(n-2), with a(1)=0 and a(2)=1.

To check what happens to the generating function when the tribonacci sequence is considered, see my post titled Beyond Fibonacci (March 10th 2019).

A Timely Surprise

When searching for interesting properties associated with the number (27491) representing my diurnal age, I encountered OEIS A335789:

A335789

a(\(n\)) = time to the nearest second at the \(n\)-th instant (n>=0) when the hour and minute hands on a clock face coincide, starting at time 0:00.

The initial members of the sequence are: 0, 3927, 7855, 11782, 15709, 19636, 23564, 27491, 31418, 35345, 39273, 43200, 47127, 51055, 54982, 58909, 62836, 66764, 70691, 74618, 78545, 82473, 86400, 90327, 94255, 98182, 102109, 106036, 109964, 113891, 117818, 121745, 125673, 129600, 133527, 137455, 141382

Clock face after 27491 seconds from midnight

The comment is made that "after 12 hours or 43200 seconds, the hands overlap at 12:00 and the cycle repeats". What surprised me was that I was the author of this sequence, approved on August 14th of 2020, under the pseudonym of Sean Lestrange. I wrote about this sequence and related matters in a blog post titled Sexagesimal Number System on the same date. I'd quite forgotten about until I received this pleasant reminder.

Of course I no longer propose new sequences to the OEIS for reasons I've explained but I continue to make daily use of it. It's not the enormous body of sequences contained in the OEIS that I have a problem with. Instead it's the people who control the approval process.

This number of seconds corresponds to 7 hours 38 minutes 11 seconds. Figure 1 shows the clock's appearance for all those times when the hour and minute hands of the clock coincide (source).

It will be quite a while before the hour and minute hands of the clock overlap again when reckoning one second of clock time equal to one day of my lifetime. Tuesday, April 10th 2035 to be precise.

Hidden Pandigitals

When we think of pandigital numbers, it is a number like 1263480759 that comes to mind. Each of the digits from 0 to 9 occur exactly once. There are 3,265,920 such numbers (leading zeroes not being considered) out of the total of 3,486,784,401 possible ten digit numbers. That's a representation of about 0.094%.

There are however, other ways in which pandigital numbers can arise and one way is in the decimal approximations of the square roots of whole numbers when approximated to ten digits NOT ten decimal places. Some of the digits will occur in the whole number part of the square root and the rest will occur in the decimal part. An example is the square root of 1362 where we have:$$ \sqrt{1362}=36.90528417 \dots$$Note that the final digit arises from truncation of the infinite decimal and not from rounding. Essentially, the decimal point is ignored and so associated with certain whole numbers are their pandigital square roots expressed as whole numbers themselves with the decimal points ignored. The complete list of such numbers, up to 40000, is as follows:$$ \begin{align} 1362 &\rightarrow 3690528417\\1843 &\rightarrow 4293017586\\2540 &\rightarrow 5039841267\\4280 &\rightarrow 6542170893\\5507 &\rightarrow 7420916385\\6896 &\rightarrow 8304215796\\14601 &\rightarrow 1208345976\\15143 &\rightarrow 1230568974\\17547 &\rightarrow 1324650897\\18393 &\rightarrow 1356207948\\20337 &\rightarrow 1426078539\\22710 &\rightarrow 1506983742\\23560 &\rightarrow 1534926708\\25887 &\rightarrow 1608943752\\27487 &\rightarrow 1657920384\\30728 &\rightarrow 1752940386\\32286 &\rightarrow 1796830542\\32615 &\rightarrow 1805962347\\33144 &\rightarrow 1820549367\\34499 &\rightarrow 1857390642\\37194 &\rightarrow 1928574603 \end{align} $$Of course it's easy enough to see where the decimal point should be. These numbers form OEIS A113507 (permalink). We can extend this idea to cube roots and in so doing the first number to make an appearance is 2017 because:$$ (2017)^{1/3}=12.63480759 \dots$$The numbers up to 40000 are (permalink):$$ \begin{align} 2017 &\rightarrow 1263480759\\3053 &\rightarrow 1450693287\\9950 &\rightarrow 2150837964\\15139 &\rightarrow 2473806519\\15533 &\rightarrow 2495083671\\18357 &\rightarrow 2637954108\\24214 &\rightarrow 2893047156\\24424 &\rightarrow 2901386574\\31457 &\rightarrow 3156742089\\32654 &\rightarrow 3196284750\\39605 &\rightarrow 3408657291 \end{align} $$These numbers form OEIS A119517. This approach can be extended to fourth roots and beyond. We can also consider whole numbers raised to let's say \(1/ \pi \). For example:$$3638^{1/ \pi}=  13.59746028 \dots$$jUpt to 40000, the following whole numbers lead to pandigitals when raised to the power \(1/ \pi\):$$ \begin{align} 3638 &\rightarrow 1359746028\\7109 &\rightarrow 1682940375\\10271 &\rightarrow 1892064735\\11572 &\rightarrow 1965273840\\13818 &\rightarrow 2079431586\\14435 &\rightarrow 2108547963\\20539 &\rightarrow 2359047168\\20981 &\rightarrow 2375089614\\26220 &\rightarrow 2549731608\\27158 &\rightarrow 2578419036\\27313 &\rightarrow 2583094176\\35022 &\rightarrow 2795816403\\35330 &\rightarrow 2803619574\\35901 &\rightarrow 2817964035\\37023 &\rightarrow 2845703691 \end{align} $$Not surprisingly this sequence of numbers is not found in the OEIS.

Rising To The Challenge

From time to time, numbers arise in the count of my diurnal age that seemingly have no properties that interest me. Always however, with a little research, at least one interesting property emerges. Today's number of 27488 is such a number. Now this number factorises as follows:$$27488 = 2^5 \times 859$$The number has two prime factors, 2 and 859, both of which have digit sums that contain only the digit 2 because 859 has a digit sum of 22.

This got me thinking about what other numbers have prime factors where each factor has a digit sum consisting only of the digit 2. I decided to eliminate prime numbers and numbers that are purely powers of 2. This left with 431 numbers in the range up to 40,000 with the majority of these being even numbers. The list can be viewed here. If we exclude the even numbers, there are only 29 odd numbers and these are (permalink):

121, 1111, 1331, 5489, 8459, 8657, 9449, 9647, 10201, 10637, 12221, 14641, 15389, 16379, 17369, 17567, 18359, 19349, 19547, 20537, 21923, 26279, 29249, 29447, 30239, 30437, 31427, 38159, 39149

The first of these numbers is \(121=11^2\). This got me thinking about digits other than 1 so I set about solving this further challenge. Here is what I discovered:

DIGIT ONE

The algorithm I originally used is easily modified so that similar lists can be generated for the other digits. There are 61 numbers that satisfy and the first is \(841 = 29^2\). The last number in the list is \(39527 = 29^2 \times 47\).

841, 1363, 2209, 2407, 3901, 3973, 5017, 5539, 6439, 6583, 6889, 7627, 8131, 8149, 8977, 9193, 10237, 10669, 11371, 12361, 12847, 13207, 13369, 14359, 14899, 15853, 16591, 18589, 18769, 18841, 20821, 21667, 21829, 23323, 23701, 23809, 24389, 26167, 26311, 26419, 29299, 29551, 29929, 30127, 31099, 31639, 32161, 33043, 33727, 34249, 35293, 36031, 36481, 36769, 37903, 38263, 38497, 38587, 39271, 39469, 39527

DIGIT FOUR

No numbers satisfy when the digit is 3 and 22 satisfy when the digit is 4. An example is \(169 = 13^2\) and \(39013 = 13 \times 3001\). Numbers must composite and these numbers up to 40000 are:

169, 403, 961, 1339, 2197, 2743, 3193, 5239, 6541, 10609, 12493, 13273, 15613, 17407, 21733, 26143, 28561, 29791, 31651, 35659, 37231, 39013

DIGIT FIVE

63 numbers satisfy when the digit is 5 and examples are \(115 =  5 \times 23\) and \(38875 = 5^3 \times 311\). Again no numbers that are purely powers of 5 are allowed and all numbers are composite. These numbers up to 40000 are:

115, 205, 529, 565, 575, 655, 943, 1025, 1555, 1681, 2005, 2599, 2645, 2825, 2875, 3013, 3275, 4633, 4715, 5065, 5125, 5155, 5371, 5515, 6505, 7153, 7775, 8405, 9223, 10015, 10025, 10555, 12167, 12751, 12769, 12995, 13225, 14125, 14375, 14803, 15055, 15065, 16375, 16441, 17161, 20005, 21689, 23165, 23299, 23575, 23713, 25325, 25369, 25625, 25775, 26855, 27575, 29923, 32525, 35143, 35765, 38663, 38875

DIGIT SEVEN

None satisfy when the digit is 6 and 63 satisfy when the digit is 7 and examples are \(301 = 7 \times 43\) and \(36661 =  61 \times 601\).  Again no numbers that are purely powers of 7 are allowed and all numbers are composite. Up to 40000, these numbers are:

301, 427, 1057, 1561, 1687, 1849, 2107, 2191, 2317, 2623, 2947, 2989, 3721, 4207, 6493, 7231, 7357, 7399, 7861, 8491, 8617, 9121, 9211, 9247, 9589, 10363, 10927, 11809, 12943, 13459, 13603, 14233, 14701, 14749, 14791, 14917, 15337, 15421, 15547, 16177, 16219, 18103, 18361, 19093, 20191, 20629, 20923, 21847, 22801, 23107, 25681, 25843, 26047, 28021, 28147, 28777, 29407, 29449, 33673, 35077, 35707, 36391, 36661

DIGIT EIGHT

There are 40 numbers that satisfy when the digit is 8 and examples are \(289 = 17^2\) and \(37621 = 17 \times 2213\). These composite numbers up to 40000 are:

289, 901, 1207, 1819, 2809, 3763, 3961, 4267, 4913, 5041, 5671, 7327, 7597, 8551, 8857, 11449, 11917, 12349, 13303, 15317, 16543, 17821, 18037, 19567, 20519, 20791, 22843, 24931, 25687, 26659, 26857, 27217, 27613, 30601, 30923, 35713, 36397, 36991, 37153, 37621

No numbers arise when the digit is 9. So why are there no numbers arising when the digits 3, 6 and 9 are considered. When a digit sum consists only of 3’s or multiples of 3, then it must be divisible by 3 in the first place. This causes it to break down into smaller prime factors.

The Primes Within

The following little problem caught my attention yesterday. I found it in New Scientist’s Brain Twister 16: Order, Order!

Arrange the digits 1–9 in a line so that each pair of adjacent digits sums to a prime number.

It's not a difficult problem. Here is a permalink to some SageMathCell code that does the job of finding all possible such arrangements of the digits. There are 140 out of a total of 362,880 permutations. Every number of course has its reversal within this list as well. Thus our first number (123476589) has its reversal (985674321) as the last member of the list.

123476589, 123498567, 123856749, 123894765, 125674389, 125674983, 129438567, 129476583, 129834765, 129856743, 143298567, 143892567, 147652389, 147652983, 147658329, 147658923, 149238567, 149832567, 165238947, 165298347, 165832947, 165892347, 167432589, 167432985, 167438529, 167438925, 167492385, 167492583, 167498325, 167498523, 321476589, 321498567, 321658947, 321674985, 325894167, 325894761, 329856147, 329856741, 341298567, 347612589, 347612985, 347658921, 349852167, 385216749, 385294167, 385294761, 385612947, 385674129, 385674921, 389214765, 389256147, 389256741, 389412567, 389476125, 389476521, 521674389, 521674983, 523894167, 523894761, 529834167, 529834761, 561238947, 561298347, 567412389, 567412983, 567438921, 567498321, 583216749, 583294167, 583294761, 583476129, 583492167, 589216743, 589234167, 589234761, 589432167, 589476123, 741652389, 741652983, 741658329, 741658923, 743216589, 743298561, 743856129, 743892165, 743892561, 749216583, 749238561, 749832165, 749832561, 749856123, 761234985, 761258349, 761258943, 761294385, 761432589, 761432985, 761438529, 761438925, 761492385, 761492583, 761498325, 761498523, 765214389, 765214983, 765238941, 765298341, 765832149, 765832941, 765834129, 765834921, 765892143, 765892341, 765894123, 765894321, 921438567, 921476583, 921658347, 921674385, 923856147, 923856741, 925834167, 925834761, 941238567, 943852167, 947612385, 947612583, 947658321, 983214765, 983256147, 983256741, 983412567, 983476125, 983476521,    985216743,    985234167,    985234761,    985612347,    985674123,    985674321

This got me thinking about five digit numbers with this property. The reason for focusing on five digit numbers is that my diurnal age, since I turned 10000 days old, always consists of five digits. So what are the five digit numbers with this property given that zero is not allowed and no digits can repeat. The earlier code can be easily modified (permalink) to give us the numbers we want. It turns out that there are 222 numbers out of a total of 15120. Here they are:

12347, 12349, 12385, 12389, 12567, 12583, 12589, 12943, 12947, 12983, 12985, 14325, 14329, 14385, 14389, 14765, 14923, 14925, 14983, 14985, 16523, 16529, 16583, 16589, 16743, 16749, 21438, 21476, 21498, 21658, 21674, 23416, 23476, 23498, 23856, 23894, 25614, 25674, 25834, 25894, 29416, 29438, 29476, 29834, 29856, 32147, 32149, 32165, 32167, 32561, 32567, 32589, 32941, 32947, 32985, 34125, 34129, 34165, 34167, 34761, 34765, 34921, 34925, 34985, 38521, 38529, 38561, 38567, 38921, 38925, 38941, 38947, 41238, 41256, 41258, 41298, 41652, 41658, 43216, 43256, 43258, 43298, 43852, 43856, 43892, 47612, 47652, 47658, 49216, 49238, 49256, 49258, 49832, 49852, 49856, 52143, 52147, 52149, 52167, 52341, 52347, 52349, 52389, 52941, 52943, 52947, 52983, 56123, 56129, 56143, 56147, 56149, 56741, 56743, 56749, 58321, 58329, 58341, 58347, 58349, 58921, 58923, 58941, 58943, 58947, 61234, 61238, 61258, 61294, 61298, 61432, 61438, 61492, 61498, 65214, 65234, 65238, 65294, 65298, 65832, 65834, 65892, 65894, 67412, 67432, 67438, 67492, 67498, 74123, 74125, 74129, 74165, 74321, 74325, 74329, 74385, 74389, 74921, 74923, 74925, 74983, 74985, 76123, 76125, 76129, 76143, 76149, 76521, 76523, 76529, 76583, 76589, 83214, 83216, 83256, 83294, 83412, 83416, 83476, 83492, 85214, 85216, 85234, 85294, 85612, 85614, 85674, 89214, 89216, 89234, 89256, 89412, 89416, 89432, 89476, 92143, 92147, 92165, 92167, 92341, 92347, 92385, 92561, 92567, 92583, 94123, 94125, 94165, 94167, 94321, 94325, 94385, 94761, 94765, 98321, 98325, 98341, 98347, 98521, 98523, 98561, 98567

Again it can be noted that every number in this list has its reversal in this list because both a number and its reversal will satisfy the criterion. Thus for 12347, we will also find 74321.

Of these 222 numbers, 30 are prime:

12347, 12583, 12589, 12983, 14389, 14923, 14983, 16529, 32561, 32941, 34129, 38561, 38567, 38921, 52147, 56123, 56149, 58321, 58921, 58943, 74923, 76123, 76129, 92143, 92347, 92567, 94321, 98321, 98347, 98561

Of these 30, six are emirps:

• 14923 and 32941
• 34129 and 92143
• 12983 and 38921

So these six five digit numbers are indeed rather special. They are prime. Their reversals are prime and both have the property that the sums of their adjacent digits are prime (permalink).

As one might expect, there are variations on this theme. What if we require instead that the absolute values of the differences between adjacent digits be prime? If we revert the case of the digits 1 to 9, there is only one number that satisfies and that is 864297531.

However, restricting ourselves to only five digit numbers (again with no zeroes and no repeating digits), we find surprisingly that there are 792 such numbers that satisfy. I won't list them all but here is a permalink. Of this total, 92 are prime:

13649, 13697, 14683, 14753, 16427, 16927, 24169, 24631, 24683, 24697, 25741, 27581, 27583, 27941, 27961, 29641, 29683, 29741, 29753, 31469, 31649, 35279, 35729, 35869, 36479, 36497, 36857, 36947, 38149, 41357, 41863, 42961, 46183, 46381, 46831, 46853, 47581, 47963, 49253, 49613, 49681, 52963, 53147, 53149, 53681, 53861, 57241, 57413, 58147, 58169, 58369, 58613, 58631, 61357, 63149, 63527, 63857, 64279, 64927, 68147, 68531, 69247, 69257, 69427, 72461, 72469, 75869, 79241, 79613, 79631, 81463, 81647, 81649, 83579, 83641, 85247, 85297, 85361, 85369, 86357, 86413, 86927, 92413, 92461, 92581, 92753, 94253, 94613, 96857, 97241, 97463, 97583

Of these 92, the following 12 are emirps:

• 13697 and 79631
• 16427 and 72461
• 31649 and 94613
• 35729 and 92753
• 36479 and 97463
• 75869 and 96857

Many other variations are possible but that's enough for this post in which I've focused on two variations involving prime numbers.

The Number Plate Game

Figure 1

Most number plates in Indonesia display a four digit number as shown in Figure 1 where the number is 1234. A game I like to play as I drive around or am driven around is to try to make a digit equation from these four digits. My rules are that only the four operations of addition, subtraction, multiplication and division are allowed together with exponentiation and brackets. Concatenation of digits is not allowed. Division can be of two types:

• a digit divided into the next digit e.g. 2 | 4 = 2 which reads 2 divided into 4
• a digit divided by the next digit e.g. 4 / 2 = 2 which reads 4 divided by 2

For the 1234 in Figure 1, the digit equation is very straightforward:$$ \begin{align}  -1 + 2 &= -3 + 4 \\ 1 &=1 \end{align}$$Theoretically there are \(10^4 = 10000 \) possible numbers ranging from 0000 to 9999. Any numbers with at least two zeroes will form a digit equation. Take 1009 as an example:$$ \begin{align} 1 \times 0 &= 0 \times 9 \\ 0 &=0 \end{align} $$If the digit 1 appears at least twice in the first three digits, then a digit equation is always possible. Take 1814 as an example:$$\begin{align} 1^8 &= 1^4 \\ 1 &=1 \end{align}$$Two consecutive digits are often helpful because they can be collapsed to a 1. Take 6599 as an example:$$\begin{align} 6 - 5 &= 9 \, /  \,9 \\ 1 &=1 \end{align}$$One or more zeroes are also helpful because raising any digit to the zero power produces a 1. Take 9032 as an example:$$\begin{align} 9^0 &= 3-2 \\ 1 &=1 \end{align}$$Brackets also prove very useful as shown in the example of 4894 where we bracket the 8 and 9 and make the 8 negative:$$\begin{align} 4^{-8+9} &= 4 \\ 4^1 &=4 \\ 4&=4  \end{align}$$Of course, forming a digit equation can require a little trickery. Take 3649 as an example:$$\begin{align} 3^{ 6-4}&= 9 \\ 3^2 &=9 \\ 9 &= 9 \end{align}$$Sometimes a digit equation is not possible according to the rules imposed. Take 3637 as an example. I can't see how to make a digit equation out of those four digits.

On the other hand, it's often possible to form digit equations in several different ways. Take our original 1234 as an example where another representations is possible:$$\begin{align} 1^2 &= -3+4 \\ 1 &=1 \end{align}$$It would be an interesting exercise to work out digit equations for all the numbers between 0000 and 9999 and to make a note of those numbers where an equation is not possible. In fact I've began this process in an Airtable database (link) and it would appear that 0124 is the first four digit number to thwart the creation of a digit equation. I'll keep working on the database and make additions to this post when I discover numbers of interest. It will of course take some time to complete.

See post titled Forming Equations from the Digits of a Number from the 14th of March 2024.

ADDENDUM: July 10th 2024

A variation of this game would be to see if one can get the numbers to total zero. This is very similar to the game already described and every solution to that game becomes a solution to this game because \(a=b \implies a-b=0\). This new, let's call it "zero game", allows for additional numbers plates containing the digit zero to qualify. For example 2470:$$2470 \rightarrow 2 \times 4 \times 7 \times 0 = 0$$Under the old rules this doesn't qualify but here the product of the digits is zero so it does. Another example would be 6772:$$6772 \rightarrow 6 \times  (7 - 7) \times 2 = 0$$Under the old rules, a digit equation could not be formed from these digits but here it can.

Unity as a Sum of Egyptian Fractions

Continuing with Achmad Damar's "104 Number Theory", I found this little problem interesting. The author asks: let \(k\) be an even number. Is it possible to write 1 as the sum of the reciprocals of \(k\) odd integers? He approaches the problem by assuming that:$$ 1=\frac{1}{n_1}+ \cdots + \frac{1}{n_k}$$for odd integers \(n_1, \dots, n_k \). Removing denominators produces:$$n_1 \cdots n_k=s_1+ \cdots + s_k $$where all \(s_i\) are odd because numbers that are the products of odd numbers are themselves odd. This is impossible because the LHS is odd and the RHS is even because there are an even number (\(k\)) of integers and the sum of each pair of odd  integers is even. Thus it is not possible to represent 1 as the sum of the reciprocals of an even number of odd integers. However, if \(k\) is odd, then it is possible. The example is given of unity expressed as the sums of reciprocals of nine odd integers. See Figure 1.

Figure 1

This got me thinking about how this result was obtained. There are 114 odd integers between 3 and 231 inclusive and 7,032,112,662,630 ways to sample 9 reciprocals at a time (that's over seven trillion ways). Nonetheless, when running this program in SageMathCell, the above combination of fractions was quickly spat out and after that the program timed out. Running the same program in my jupyter notebook (to avert the program timing out), no further combinations were generated. Is this combination of reciprocals unique? I'm not sure.

Certainly we can state that there are infinitely many disjoint sets of positive integers in which the sum of those reciprocals is equal to unity (source). I have a program that generates Egyptian fractions for rational numbers \(p/q\) where \(p<q\). Using this program, I can determine that: $$ \begin{align} \frac{5}{7} &= \frac{1}{2}+\frac{1}{5}+\frac{1}{70}\\ \frac{2}{7} &= \frac{1}{4}+\frac{1}{48}\\1 &= \frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{48}+\frac{1}{70} \end{align}$$Alternatively, I could take another pair of fractions that add to 1 such as 7/11 and 4/11. This gives:$$ \begin{align} \frac{7}{11} &= \frac{1}{2} + \frac{1}{8} + \frac{1}{88} \\ \frac{4}{11} &= \frac{1}{3} + \frac{1}{33} \\ 1 &= \frac{1}{2}+ \frac{1}{3}+ \frac{1}{8} + \frac{1}{33} + \frac{1}{88} \end{align} $$Obviously we could continue this process indefinitely. This Mathematics Stack Exchange source states that:

R. L. Graham showed that \( a(n)>0 \) for \( n>77 \), where \( a(n) \) is the number of ways to express 1 as the sum of distinct unit fractions such that the sum of the denominators is \( n \). This implies that for values of \( n \leq 77 \) such representations may not be possible. In the two examples shown earlier the sum of denominators is greater than 77. However, representations where \(n \leq 77 \) are certainly possible. For example from the same Stack Exchange source we see that:$$ 2+3+11+22+33 = 2+5+8+12+20+24 \text{ and both total }71\\2+4+9+12+18 =2+5+6+12+20 \text{ and both total }45$$and the sum of reciprocals of LHS's and RHS's all total 1.

The simplest representation of 1 using Egyptian fractions is:$$1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$$but once we impose special conditions such as all denominators must be odd or prime or whatever, then things get more complicated. This site offers some useful insights into a problem that can well be explored in far more depth than I've attempted here.