Rational Solutions to a^a = b^b

Here's a problem that I encountered on this site:

Find all pairs of rational numbers \((a,b)\) such that \(a^a=b^b\) for \(0 < a < b\).

One solution that was proposed runs as follows:

Write \(b=ax\) for some \(x>1\). Then \(x=b/a\) so \(x\) is rational.

Then taking logs, we get $$ \begin{align} a\log a&=b\log b\\&=ax\log(ax) \\ \log a&= x \log(ax) \\&=x \log(x)+x \log(a)      \\&=\frac{x\log x}{1-x} \\  a&=x^{\frac{x}{1-x}}\\b&=x^{\frac{1}{1-x}} \end{align} \\ \text{since } x\neq1 \text{ and } a\neq0$$Since \(x\) is rational, it suffices to find the values of \(x>1\) for which \(x^{\frac{1}{1-x}}\) is rational.

We claim that the only such values of \(x\) are \(x=\dfrac{n+1}{n}\) where \(n\) is an integer.

We may write \(x=1+\dfrac1n\) where \(n\in\mathbb{Q}\), \(n>0\).

It is easy to see if \(n\) is a positive integer that \(x^{\frac{1}{1-x}}\) is rational.

If \(n\) is not an integer, write \(n=p/q\) where \(p,q\in\mathbb{Z}^+\) are coprime with \(q>1\).

We have \(x^{\frac{1}{1-x}}=\left(\dfrac{n}{n+1}\right)^n=\left(\dfrac{p}{p+q}\right)^{p/q}\).

Hence \(p\) and \(p+q\) must be \(q \,\)th powers.

No two \(q \,\)th powers can differ by \(q\) since for positive integers \(u,v,q\), we have by the Binomial Theorem, $$(u+v)^q-u^q=qu^{q-1}v+\ldots+v^q>q\,.$$Therefore, if \(n\) is not an integer, \(x^{\frac{1}{1-x}}\) is not rational, so the only rational solutions are given by $$a=\left(\dfrac{n}{n+1}\right)^{n+1} \text{ and } \, b=\left(\dfrac{n}{n+1}\right)^n \text{ with } n\in\mathbb{Z}^+$$The initial values of \(a\) and \(b\) are shown in Figure 1.

Figure 1: permalink

Pseudo-Sphenic Number Sequences

A cubic polynomial with three real roots and rational coefficients can be written in the following form:

(a\(x\) + b) \(\times \)  (c\(x\) + d) \( \times \) (e\(x\) + f) 
where a, b, c, d, e and f are rational numbers

Let's modify the conditions so that a, b, c, d, e and f are integers (positive or negative) and \(x\) can only take integer values greater than 1. Let's change the \(x\) to an \(n\) so that we have:

(a\(n\) + b) \( \times \) (c\(n\) + d) \( \times \) (e\(n\) + f)

Let's take a specific example where a=1, b=0, c=1, d=1, e=2 and f=3. This gives us:$$n \times (n+1) \times (2n+3)$$As we plug in different values for \(n\), starting with \(n=2\), a series of terms arises. In this case, the terms begin:

42, 108, 220, 390, 630, 952, 1368, 1890, 2530, 3300, 4212, 5278, 6510, 7920, 9520, 11322, 13338, 15580, 18060, 20790, 23782, 27048, 30600, 34450, 38610, 43092, 47908, 53070, 58590, 64480, 70752, 77418, 84490, 91980, 99900, 108262, 117078, 126360, 136120

These terms constitute OEIS A163815 (although the terms for \(n=0\) and \(n=1\) are included). These sorts of sequences involve the multiplication of triple linear combinations of \(n\), in this case \(n\), \(n+1\) and \(2n+3\). This permalink leads to a SageMath algorithm that will generate a sequence of terms for varying values of a, b, c, d, e and f. 

If we impose the condition that each of the linear factors must be distinct (and this is the case for the example just shown), then we have a sequence where each member is a sort of pseudo-sphenic number that can be written as a product of three of its divisors but each divisor is a linear combination of an underlying integer. For example, take the number 27048 (my diurnal age yesterday). It can be written as:$$27048 = 23 \times 24 \times 49\\ \text{where } 23=n, 24=n+1 \text{ and } 49=2n+13$$The associated polynomial will cut the \(x\) axis in three locations. Figure 1 shows the situation for \(y=x \times (x+1) \times (2x+3) \) where \(x\)=-1.5, -1 and 0.

Figure 1: Geogebra link

Let's try another example where a, b, c, d, e, f = 1, -1, 1, 1, 1 , 2. The initial terms generated will be of the form \( (n-1) \times (n+1) \times (n+2) \) and are as follows (starting with \(n=3\) because we want to avoid getting a 1 as a factor):

40, 90, 168, 280, 432, 630, 880, 1188, 1560, 2002, 2520, 3120, 3808, 4590, 5472, 6460, 7560, 8778, 10120, 11592, 13200, 14950, 16848, 18900, 21112, 23490, 26040, 28768, 31680, 34782

This sequence does not appear in the OEIS but it is recognised as being generated from a polynomial (I included the terms for \(n=1\) and \(n=2\) when searching in the OEIS). See Figure 2.

Figure 2

For a given number to be a pseudo-sphenic number, it must have more than three, not necessarily distinct, prime factors. For example, my earlier example of 27048:$$27048=2^3 \times 3 \times 7^2 \times 23=23 \times 24 \times 49$$Sometimes the members of a polynomial sequence will be genuinely sphenic as with the case of:$$n \times (n+6) \times (n+12)$$Figure 3 shows the presence of triplets of so-called "sexy" primes: (5, 11, 17), (11, 17, 23), (17, 23, 29) and (31, 37, 43).

Figure 3: permalink

Any sphenic number will be a member of some polynomial sequence. Let's arbitrarily choose the sphenic number formed by the prime factors 79, 101 and 197. We have:$$1571863=79 \times 101 \times 197$$Let's subtract 77 from each number so that they become smaller. This gives us 2, 24 and 120 so that we consider the polynomial:$$ (n+2) \times (n+24) \times (n+ 120)$$Figure 4 shows the result when we generate the sequence up to 1571863.

Figure 4: permalink

Looking at the results we can see that:

• 106723 = 19 * 41 * 137
• 244807 = 31 * 53 * 149
• 906277 = 61 * 83 * 179
• 1571863 = 79 * 101 * 197

Thus the seemingly unrelated sphenic numbers 106723, 244807, 906277 and 1571863 are in fact very closely related because they all arise when different values of \(n\) are assigned to the polynomial \( (n+2) \times (n+24) \times (n+120) \), specifically \(n=17, 29, 59\) and \(77\). Thus a hidden link between sphenic numbers is revealed.

Concatenating Factors of Sphenic Numbers

The number associated with my diurnal age today, 27047, is a sphenic number and it factorises to 17 x 37 x 43. When these factors are concatenated to form a new number, 173743, then this number is prime. By convention, the concatenation is effected in ascending order but I wondered how the primeness would hold up if all possible orders were considered. The answer is as follows with True meaning is prime and False meaning is not:

173743 True

174337 True

371743 False

374317 True

433717 False

431737 False

So 27047 scores 50% with half the concatenations being prime and half not. The natural follow up was to ask what sphenic numbers hold up 100% of the time. It didn't take too long to construct a SageMath algorithm to determine this in the range up to one million (permalink). There are only 15 such numbers in that range and they are:

3311, 27181, 32153, 41237, 53977, 86507, 110971, 125069, 208579, 256413, 500981, 543337, 853811, 901949, 964481

The factorisations of these numbers are:

3311 = 7 * 11 * 43

27181 = 7 * 11 * 353

32153 = 11 * 37 * 79

41237 = 7 * 43 * 137

53977 = 7 * 11 * 701

86507 = 19 * 29 * 157

110971 = 7 * 83 * 191

125069 = 7 * 17 * 1051

208579 = 7 * 83 * 359

256413 = 3 * 127 * 673

500981 = 13 * 89 * 433

543337 = 17 * 31 * 1031

853811 = 7 * 283 * 431

901949 = 19 * 37 * 1283

964481 = 7 * 211 * 653

Checking with the OEIS, I found that this sequence of numbers is listed as OEIS A180679:

A180679

Numbers with three distinct prime factors which when concatenated in any order form prime numbers.

A longer list of sequence members is provided:

3311, 27181, 32153, 41237, 53977, 86507, 110971, 125069, 208579, 256413, 500981, 543337, 853811, 901949, 964481, 1053787, 1144171, 1197851, 1215731, 1344539, 1385189, 1433659, 1549603, 1674741, 1681547, 1699481, 1973479, 2028181

One could carry out a similar investigation on semiprimes and numbers with four distinct prime factors, although the OEIS comments note that there is no term with four distinct prime factors under \(10^8\). Interestingly, the second number is the sequence, 27181, is coming up very soon in my diurnal age count. It is only 134 days away.

Undulating Numbers

I've only ever made a passing reference to an undulating number in a post titled 26262: A Special Palindrome from February 26th 2021. This post will address that omission. Firstly, let's have Numbers Aplenty define what is meant by an undulating number:

A number is undulating in base \(b\) if it has at least \(3\) digits and it is made of exactly two distinct digits which alternate, like \(252\) or \(373737\) in base \(10\) or \(21=10101_2\).

Undulating numbers can be termed undulants and in base 10 they comprise OEIS A046075:

A046075

Nontrivial undulants; base 10 numbers >100 which are of the form \(aba, abab, ababa, \dots \) where \(a\) and \(b\) are not equal.

 The initial members of this sequence are:

101, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 454, 464, 474, 484, 494, 505, 515, 525, 535, 545, 565, 575, 585, 595, 606, 616, 626, 636, 646, 656

Wikipedia lists these properties of undulating numbers:

There are infinitely many undulating numbers.

For any \(n\) ≥ 3, there are 9 × 9 = 81 non-trivial \(n\)-digit undulating numbers, since the first digit can have 9 values (it cannot be 0), and the second digit can have 9 values when it must be different from the first.

Every undulating number with even number of digits and at least four digits is composite, since: $$ababab \dots ab = 10101 \dots 01 \times ab\\ \text{ e.g. } 171717 = 10101 \times 17$$Undulating numbers with odd number of digits are palindromic. They can be prime, for example 151.

The undulating number \( abab \dots ab\) with \(n\) repetitions of \(ab\) can be expressed as: $$ ab \times \frac{10^{2n} − 1}{99} \\ \text{ e.g. } 171717 = 17 \times \frac{10^6 − 1}{99}$$The undulating number \(abab \dots aba\) with \( n\) repetitions of \(ab\) followed by one \(a\) can be expressed as$$ ab \times \frac{10^{2n+1} − ba}{99}\\  \text{ e.g. } 989898989 = 98 \times \frac{10^9 − 89}{99}$$Undulating numbers can be generalized to other bases. If a number in base \(b\) with even number of digits is undulating, in base \(b^{2} \) it is a repdigit.

There can be confusion about what constitutes an undulating number. For some, the only requirement is that the numbers alternate between up and down or down and up. For this reason the term smoothly undulating has been introduced as explained below:

Smoothly Undulating Palindromic Primes (or SUPP's for short) are numbers that are primes, palindromic in base 10, and the digits alternate, but why smooth one might ask! The smoothness was added to make a difference with the normal undulating numbers. The description for normal undulating numbers is that the next digits alternately go up and down (or down and up) but the absolute difference values between two adjacent digits may differ e.g. 906343609. In a smoothly undulating number the absolute difference values between two adjacent digits are always equal, therefore only two distinct digits can appear in the number e.g. 74747474747474747. Source.

The smoothly undulating primes begin:

101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 18181, 32323, 35353, 72727, 74747, 78787, 94949, 95959, ...

Apparently there are only  four undulating squares, namely \(121, 484, 676\) and \(69696\) corresponding to \(11^2,22^2,26^2\) and \(264^2\). See https://oeis.org/A016073.

The first numbers which are undulating in at least two bases \(b \leq 16\) are 10, 46, 50, 55, 67, 78, 85, 92, 98, 100, 104, 109, 119, 121, 130, 135, 136, 141, 145, 151, 154, 164, 166, 170, 178, 181, 182, 185, 191, 197, 200, ...

For example, \(10_{10}=1010_2=101_3\) and is thus undulating in bases 2 and 3.

SOD Prime Chains

Over the years, I've looked at many forms of prime chains but, as far as I know, not prime chains formed by successively adding the sum of the digits of the prime. What got me thinking about this type of prime chain was the number associated with my diurnal age today: 27043. This number is prime and if we add its sum of digits, we get a new prime. Thus, where SOD stands for Sum Of Digits, we have:$$ \overbrace{27043}^{\text{prime}} + \overbrace{16}^{\text{SOD}}=27059 \text{ which is also prime}$$Because 27059 is the next prime after 27043, it is known as an a-pointer prime defined by Numbers Aplenty as follows:

A prime number  \(p\) is called a-pointer if the next prime number can be obtained adding  \(p\)  to its sum of digits (here the 'a' stands for additive).

When considering prime chains formed by adding the sum of digits, we are only interested in "prime-ness" and not "a-pointer prime-ness". The earliest example of a prime chain begins with the prime 11. If we add its sum of digits, we get 13 and thus we have a prime chain of length 1: $$\overbrace{11}^{\text{prime}}+\overbrace{2}^{\text{SOD}}=\overbrace{13}^{\text{prime}}$$If we add the sum of digits again we get 17 and thus we have a chain of length 2 namely: $$\overbrace{11}^{\text{prime}}+\overbrace{2}^{\text{SOD}}=\overbrace{13}^{\text{prime}} \text{ and } \overbrace{13}^{\text{prime}}+\overbrace{4}^{\text{SOD}}=\overbrace{17}^{\text{prime}}$$Here, both 11 and 13 are a-pointer primes. We do not get a chain of three primes until 277 where the chain is:$$ 277 \rightarrow 293 \rightarrow 307 \rightarrow 317$$None of these primes are a-pointer primes. The first chain of four occurs with 37783:$$37783 \rightarrow 37811 \rightarrow 37831 \rightarrow 37853 \rightarrow 37879$$The first chain of five occurs with 516493:$$516493 \rightarrow 516521 \rightarrow 516541 \rightarrow 516563 \rightarrow 516589 \rightarrow 516623$$These record chains constitute OEIS A090009:

A090009

Begins the earliest length-\(n\) chain of primes such that any term in the chain equals the previous term increased by the sum of its digits.

The initial members are (permalink - will time out beyond 516493):

2, 11, 11, 277, 37783, 516493, 286330897, 286330897, 56676324799

The progressions for the larger numbers are:

286330897 286330943 286330981 286331021 286331047 286331081 286331113 286331141 

56676324799 56676324863 56676324919 56676324977 56676325039 56676325091 56676325141 56676325187 56676325243 

In conclusion, we must say that 27043 has the unusual property that its successor, 27044, also produces a prime (27061) when its sum of digits is added. Thus 27059 and 27061 form a pair of twin primes. Given this property of 27043, a new sequence could be formulated as follows:

Numbers \(n\) such that \(n\) plus digit sum of \(n\) and \(n+1\) plus digit sum of \(n+1\) are both prime.

These numbers constitute about 1.335% of numbers in the range up to 40000. This is to be expected since the probability of any number having this property is about 0.1, so two in succession would have a probability of about 0.01. The primes resulting from this process are generally twin primes, although perhaps not exclusively. Numbers ending in 9 such as 299 (with sod = 20) will change to 300 (with sod = 3). However, looking at the output below, there are no numbers ending in 9. Interesting. Triplets are not possible as this would mean three successive primes separated by only a single number. The 534 members up to 40000 are:

10, 13, 34, 52, 58, 91, 94, 100, 103, 127, 142, 166, 181, 184, 217, 232, 256, 271, 295, 304, 340, 412, 418, 451, 508, 583, 610, 631, 787, 811, 814, 838, 1024, 1042, 1048, 1081, 1138, 1222, 1264, 1285, 1312, 1420, 1441, 1465, 1468, 1591, 1597, 1600, 1606, 1648, 1681, 1711, 1771, 1861, 1915, 1933, 1975, 2017, 2071, 2074, 2095, 2104, 2122, 2128, 2230, 2254, 2293, 2302, 2326, 2365, 2638, 2671, 2692, 2701, 2767, 2782, 2947, 2980, 3112, 3154, 3241, 3244, 3283, 3316, 3355, 3373, 3445, 3448, 3514, 3538, 3751, 3796, 3805, 3913, 3976, 3994, 4012, 4036, 4075, 4120, 4144, 4210, 4216, 4231, 4255, 4324, 4411, 4495, 4504, 4528, 4618, 4633, 4696, 4705, 4765, 4780, 4945, 4984, 5002, 5008, 5083, 5221, 5263, 5395, 5404, 5425, 5461, 5482, 5623, 5641, 5827, 5845, 5860, 6073, 6121, 6181, 6253, 6277, 6343, 6433, 6547, 6637, 6670, 6676, 6742, 6760, 6766, 6811, 6850, 6925, 6940, 7114, 7192, 7201, 7285, 7315, 7333, 7441, 7465, 7531, 7537, 7570, 7735, 7930, 7978, 8071, 8215, 8272, 8365, 8413, 8521, 8611, 8788, 8815, 8836, 8944, 8968, 8983, 9001, 9025, 9223, 9262, 9394, 9403, 9421, 9442, 9598, 9607, 9688, 9799, 9910, 9976, 10003, 10027, 10060, 10081, 10132, 10261, 10285, 10318, 10420, 10444, 10483, 10516, 10687, 10843, 10867, 10918, 11050, 11056, 11098, 11107, 11146, 11161, 11341, 11473, 11677, 11695, 11704, 11761, 11815, 11923, 11947, 12028, 12061, 12088, 12151, 12226, 12241, 12358, 12592, 12601, 12796, 12805, 12976, 13210, 13324, 13381, 13657, 13672, 13690, 13696, 13705, 13741, 13813, 13855, 13876, 13984, 14002, 14065, 14311, 14371, 14428, 14533, 14572, 14608, 14845, 15124, 15253, 15271, 15343, 15499, 15562, 15631, 15721, 15946, 16045, 16048, 16171, 16399, 16627, 16666, 16798, 16807, 16954, 16996, 17014, 17167, 17185, 17272, 17365, 17470, 17560, 17635, 17656, 17725, 17764, 17815, 17878, 17893, 17902, 17962, 18025, 18028, 18043, 18115, 18265, 18286, 18517, 18883, 18886, 19057, 19123, 19162, 19186, 19360, 19411, 19450, 19522, 19672, 19726, 19816, 19858, 19963, 19987, 20014, 20140, 20218, 20344, 20431, 20458, 20491, 20500, 20695, 20704, 20728, 20785, 20875, 20962, 20986, 21004, 21007, 21046, 21175, 21310, 21358, 21511, 21538, 21571, 21577, 21592, 21601, 21628, 21718, 21823, 22030, 22075, 22093, 22102, 22144, 22255, 22258, 22348, 22525, 22618, 22675, 22723, 22837, 22942, 23020, 23026, 23044, 23353, 23518, 23608, 23647, 23665, 23884, 24091, 24100, 24163, 24892, 24901, 25021, 25153, 25282, 25285, 25390, 25447, 25555, 25576, 25774, 25825, 25912, 25972, 26092, 26101, 26233, 26656, 26674, 26698, 26707, 26836, 26854, 26926, 27043, 27085, 27223, 27262, 27460, 27511, 27517, 27556, 27664, 27715, 27730, 27886, 27916, 28075, 28090, 28162, 28255, 28327, 28384, 28525, 28546, 28588, 28636, 28726, 28987, 29005, 29113, 29374, 29644, 29734, 29848, 29977, 30004, 30130, 30373, 30448, 30538, 30823, 30847, 31108, 31141, 31165, 31234, 31297, 31306, 31492, 31501, 31525, 31696, 31705, 31708, 31747, 31831, 32020, 32044, 32110, 32131, 32173, 32281, 32311, 32353, 32392, 32401, 32425, 32515, 32590, 32776, 32884, 32917, 32950, 33055, 33163, 33271, 33328, 33565, 33580, 33727, 33745, 33784, 33811, 34021, 34114, 34138, 34192, 34201, 34243, 34282, 34351, 34447, 34480, 34486, 34570, 34627, 34735, 34822, 34825, 34936, 35035, 35257, 35299, 35431, 35566, 35698, 35707, 35815, 35872, 35983, 36001, 36085, 36445, 36511, 36754, 36868, 36910, 36991, 37180, 37321, 37342, 37525, 37546, 37564, 37783, 37963, 38221, 38311, 38425, 38440, 38578, 38626, 38644, 38683, 38887, 39142, 39211, 39217, 39322, 39346, 39478, 39814 

If we impose the restriction that \(n\) must be a prime number, then only 66 numbers qualify. Permalink. These numbers are:

13, 103, 127, 181, 271, 631, 787, 811, 1597, 1861, 1933, 2017, 2293, 2671, 2767, 3373, 4231, 5623, 5641, 5827, 6073, 6121, 6277, 6343, 6547, 6637, 7333, 7537, 8521, 9001, 9403, 9421, 10687, 10867, 11161, 11677, 11923, 12241, 12601, 13381, 14533, 15271, 17167, 18043, 18517, 19963, 20431, 21577, 21601, 22093, 24091, 25153, 25447, 27043, 32173, 32353, 32401, 32917, 33811, 34351, 35257, 35983, 37321, 37783, 37963, 39217

If we impose the restriction that \(n+1\) must be a prime number, then 73 numbers qualify. Permalink. These numbers are:

10, 52, 58, 100, 166, 232, 256, 418, 508, 838, 1048, 1222, 1600, 1606, 2128, 2692, 3448, 3538, 3796, 4012, 4210, 4216, 5002, 5008, 5482, 5860, 6760, 7192, 8272, 8836, 8968, 9688, 10060, 10132, 11056, 12226, 13690, 13696, 13876, 17470, 17656, 17902, 18286, 19162, 19726, 20218, 20962, 22030, 22258, 22348, 22618, 22942, 23020, 23026, 23608, 25390, 25576, 25912, 26698, 26926, 27916, 28162, 28546, 30448, 30538, 31306, 33328, 33580, 34282, 34486, 37180, 37546, 39322

Up to 10 million, no two consecutive prime numbers (that is a pair of twin primes) can produce another pair of twin primes.

Striking a Balance

Composite numbers have four or more divisors. For example, the number 6 has divisors of 1, 2, 3 and 6. The first three are deficient and the final divisor, the number itself, is perfect. Figure 1 shows the situation:

Figure 1: divisors of 6
permalink

The number 48 has divisors of 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48. The balance of deficient, perfect and abundant divisors is shown in Figure 2.

Figure 2: divisors of 48
permalink

The question can be asked as to what numbers have an equal balance of deficient and abundant divisors. It turns out that 144 is the first number to satisfy this criterion. The divisors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144. The balance of deficient, perfect and abundant divisors is shown in Figure 3.

Figure 3: divisors of 144
permalink

The numbers with an equal balance of deficient and abundant divisors constitute OEIS A335543 (permalink):

A335543

Numbers with an equal number of deficient and abundant divisors.

The initial members are 144, 324, 336, 756, 900, 1176, 1848, 2100, 2184, 2940, 3200, 3520, 4000, 4160, 4400, 5200, 5952, 10880, 11440, 12160, 12348, 12544, 13600, 14720, 15200, 16368, 17360, 18304, 18400, 18560, 19344, 19360, 19404, 22932, 23200, 27040, 28600, 29988, 33516, 40572, 47124.

It was only today that I became acquainted with this sequence because my diurnal age, 27040, happens to be a member. It's clear to see why I haven't come across the sequence before. The previous member is 23200 corresponding to a time when I wasn't keeping track of the numbers associated with my diurnal age. Figure 4 shows the breakdown for 27040 with divisors of 1, 2, 4, 5, 8, 10, 13, 16, 20, 26, 32, 40, 52, 65, 80, 104, 130, 160, 169, 208, 260, 338, 416, 520, 676, 845, 1040, 1352, 1690, 2080, 2704, 3380, 5408, 6760, 13520 and 27040.

Figure 4: divisors of 27040
permalink

As can be seen, these numbers are rare. There are only 39 of them in the range up to 40000. The OEIS entry has some interesting comments:

• This sequence is infinite. For example, \(3200 \times p \) is a term for all primes \(p \geq 257\). 

• The least odd term of this sequence is a(1273824) = 3010132125.

Checking out 3200 x 257 = 822400 we find that it does indeed have the required balance. See Figure 5:

Figure 5: divisors of 822400
permalink

Zeckendorf Representation Revisited

I've dealt with the topic of the Zeckendorf representation of numbers in earlier posts: namely The Fibonacci Number Base on July 30th 2019 and Goldbach's Conjecture and Zeckendorf's Theorem on November 15th 2015. As can be seen, the topic doesn't come up all that much and so, when it does, it's time to make a note of it because the opportunity may not come again for a long time. There are so many interesting topics in number theory that it's easy to forget even some of the really interesting ones like the Zeckendorf representation of a number.

The reason I was reminded was that yesterday I turned 27038 days old and one the properties of this number is that it's a member of OEIS A179250:

A179250

Numbers that have 10 terms in their Zeckendorf representation.

Such numbers are uncommon as the following list of the initial members attests:

10945, 15126, 16723, 17333, 17566, 17655, 17689, 17702, 17707, 17709, 17710, 21891, 23488, 24098, 24331, 24420, 24454, 24467, 24472, 24474, 24475, 26072, 26682, 26915, 27004, 27038, 27051, 27056, 27058, 27059, 27669, 27902, 27991

Notice how the numbers are clumped together. For example: 27004, 27038, 27051, 27056, 27058, 27059. This is seen more clearly in a plot. See Figure 1.

Figure 1: horizontal red lines added for emphasis

Before going on however, we should remind ourselves what the Zeckendorp representation is all about. Let's restate Zeckendorf's Theorem:

Every positive integer can be represented uniquely as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers.

The theorem has two parts:

Existence: every positive integer \(n\) has a Zeckendorf representation.

Uniqueness: no positive integer \(n\) has two different Zeckendorf representations.

Source 

So 27038 has ten terms in its Zeckendorf representation. What are these terms? We can use this site to quickly identify them. The terms are:

1, 3, 8, 21, 89, 233, 610, 1597, 6765, 17711 

and thus

27038 = 1 +3 + 8 + 21 + 89 + 233 + 610 + 1597 + 6765 + 17711 

The first few Fibonacci numbers are 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368 and so the Zeckendorf representation, using the same site as listed earlier, becomes:

\(27038 = 101001010101001010101_{Zeck} \text{ with 10 ones} \)

There are more such numbers coming up in the near future (27051, 27056, 27058 and 27059) but after that there is a big gap of 610 days to 27669. It can be noted that 27058 and 27059 are a consecutive pair. The thing about ten terms is that this number is relatively uncommon. Generally, fewer terms are required as can be seen from the following list of numbers from 27038 to 27045 (link) and this is in an area where such numbers are clumped together:

27038 = 101001010101001010101Zeck with 10 ones

27039 = 101001010101010000000Zeck with 7 ones

27035 = 101001010101001010001Zeck with 9 ones

27036 = 101001010101001010010Zeck with 9 ones

27037 = 101001010101001010100Zeck with 9 ones

27038 = 101001010101001010101Zeck with 10 ones

27039 = 101001010101010000000Zeck with 7 ones

27040 = 101001010101010000001Zeck with 8 ones

27041 = 101001010101010000010Zeck with 8 ones

27042 = 101001010101010000100Zeck with 8 ones

27043 = 101001010101010000101Zeck with 9 ones

27044 = 101001010101010001000Zeck with 8 ones

27045 = 101001010101010001001Zeck with 9 ones

Finding Fibonacci

The Fibonacci numbers are few and far between. Up to a little over two million, the Fibonacci numbers are:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309

However, we can find Fibonacci numbers in all sorts of places. For example, I recently turned 27034 days old and this number is a member of OEIS A272412:

A272412

Numbers \(n\) such that \( \sigma_1(n)\) is a Fibonacci number.

It so happens that \( \sigma_1(27034) = 46368 \) which is a Fibonacci number. There are only 41 such numbers in the range up to one million. They are (permalink):

1, 2, 7, 9, 66, 70, 94, 115, 119, 2479, 18084, 19180, 19290, 22060, 23156, 23178, 24934, 24956, 25756, 26715, 27034, 28678, 28965, 29578, 30094, 32253, 32793, 34113, 35365, 38635, 39319, 40963, 42493, 44413, 45223, 45653, 322032, 429424, 503175, 624027, 670975

The sum of the aliquot parts of a number is the sum of its proper divisors and so Fibonacci numbers will show up here as well. We have to exclude prime numbers in our search because their only proper divisor is 1 and so they would need to be included. It turns out that there are 175 composite numbers up to one million whose sum of proper divisors are a Fibonacci number. They are:

1, 4, 10, 18, 27, 35, 36, 49, 51, 62, 90, 91, 171, 329, 415, 473, 533, 629, 687, 713, 902, 1119, 1135, 1207, 1214, 1605, 1711, 1927, 2936, 2949, 3436, 6083, 6103, 6845, 7831, 8119, 9487, 10063, 10207, 12367, 12531, 13231, 17069, 18373, 18703, 20283, 20579, 24319, 26843, 28783, 29719, 32743, 33823, 35263, 45443, 53121, 57683, 61573, 66779, 71653, 72803, 80785, 81779, 90949, 95593, 95611, 99937, 109093, 111179, 130153, 134149, 145403, 153779, 156613, 159323, 162083, 167579, 169699, 173353, 194251, 196393, 199883, 200543, 208723, 210649, 215603, 218731, 225923, 227173, 228649, 230053, 233579, 235993, 238643, 240133, 242149, 242495, 243013, 246179, 275603, 287617, 306179, 313043, 325726, 346415, 356963, 364099, 365363, 372359, 378646, 381779, 395723, 401579, 405443, 408883, 411979, 424283, 433403, 435811, 444083, 451043, 456179, 459179, 461243, 464579, 485483, 488443, 503579, 510779, 512749, 525119, 527243, 530419, 535043, 540083, 547403, 549779, 553283, 558815, 573803, 578723, 581579, 587963, 592283, 597203, 602579, 604763, 612779, 617483, 619459, 622163, 628883, 630563, 632579, 633323, 633779, 635123, 635963, 636179, 636683, 646840, 649869, 670171, 686083, 693211, 716179, 724429, 761899, 825143, 830183, 842899, 919651, 935821, 975143, 986179

Take 51 as an example. It's proper divisors are 1, 3 and 17. These add to 21 which is a Fibonacci number. There is no associated OEIS sequence for these numbers.

Let's look at the totients of numbers. The totient of a number \(n\) is a count of how many numbers \(1 \leq k \leq n \) have the property that \( \text{gcd}(n,k)=1\) where gcd stands for greatest common divisor. The totient of 6 is 2 because 1 and 5 have this property. These numbers form OEIS A280592: 

A280592

Numbers \(n\)  such that \( \phi(n)\) is a Fibonacci number.

 Here is the list of the 134 sequence members up to one million.

1, 2, 3, 4, 6, 15, 16, 20, 24, 30, 185, 219, 273, 285, 292, 296, 304, 315, 364, 370, 380, 432, 438, 444, 456, 468, 504, 540, 546, 570, 630, 3235, 5176, 6470, 7764, 46843, 47423, 47693, 48053, 50431, 52403, 56231, 57965, 59555, 62855, 67655, 67865, 70735, 72123, 72297, 73473, 75387, 77691, 78819, 81207, 84651, 85869, 86985, 89535, 89655, 89817, 90945, 92744, 93686, 94846, 95288, 95386, 95504, 95632, 96106, 96164, 97964, 100516, 100568, 100862, 101535, 102165, 103588, 103635, 104806, 105092, 108248, 108304, 108584, 108976, 112462, 112868, 113176, 115930, 119110, 119380, 119540, 119756, 125710, 135310, 135380, 135730, 136220, 139116, 139176, 139212, 139248, 141470, 142932, 143256, 143448, 144246, 144594, 145116, 145512, 146946, 147204, 150774, 150852, 155382, 157638, 162372, 162414, 162456, 162876, 163464, 165816, 169302, 169764, 171738, 173970, 174060, 179070, 179310, 179634, 181890, 203070, 204330, 207270

Why are there no numbers from 207271 up to one million that are members of the sequence? If we extend the range to two million, there are some additional members, namely 1040075, 1304859, 1372899, 1739812 and 1830532.

Of course, we don't need to confine ourselves to the Fibonacci numbers. We could consider the Lucas numbers instead which begin with 2, 1 rather than 0, 1 like the Fibonacci. The initial Lucas numbers are:

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196

Up to one million, there are only nine numbers that qualify and they are 1, 2, 3, 4, 10, 17, 688, 1075 and 103681.

If we consider the sum of the aliquot parts however, we get 151 in the range up to one million. These are:

4, 8, 9, 21, 48, 72, 92, 115, 129, 146, 165, 187, 205, 289, 493, 965, 999, 1143, 1337, 1417, 1495, 1749, 1957, 2517, 2527, 2722, 3077, 3397, 3401, 5177, 5599, 6437, 6609, 7097, 8201, 8357, 8551, 8777, 9017, 9485, 9701, 9797, 10777, 14239, 15637, 17549, 19639, 24751, 25141, 27199, 31879, 37499, 38359, 38825, 39149, 42319, 46241, 46715, 48946, 50959, 52471, 53627, 53851, 55505, 56137, 56693, 58951, 60031, 65387, 66511, 67159, 67519, 67591, 75605, 76117, 79897, 81581, 102689, 102707, 104341, 109709, 109869, 114109, 119641, 127957, 130721, 133141, 138037, 140377, 144689, 144841, 151661, 154741, 175477, 177097, 186401, 207149, 224593, 248429, 270251, 275789, 283453, 287033, 340513, 344507, 350369, 357101, 362833, 370541, 377249, 390641, 449233, 459709, 470321, 486476, 500893, 535841, 555341, 560509, 577197, 598813, 606989, 613409, 621827, 648569, 658667, 663073, 664481, 670829, 698209, 704737, 708749, 746381, 748753, 758861, 796109, 798869, 802289, 826489, 833069, 851441, 863773, 869041, 869741, 887969, 894029, 931453, 950353, 956509, 962593, 988787

As for totients, there are 21 numbers in the range up to one million whose totient is a member of the Lucas sequence. These numbers are:

1, 2, 3, 4, 5, 6, 8, 10, 12, 19, 27, 38, 54, 2049, 2732, 4098, 5779, 11558, 36717, 48956, 73434

Another approach is to look at the determinant formed by the circulant matrix of a number. For example, 27255 has a circulant matrix as shown in Figure 1.

Figure 1

This matrix has a determinant of 21 which is a Fibonacci number. It turns out that there are 59500 numbers in the range up to one million that have this property (permalink).

Instead of the determinant, the permanent of the matrix could be considered. For example, 19140 has the circulant matrix shown in Figure 2.

Figure 2

In the range up to one million, there are only 68 such numbers (as opposed to the 59500 for the determinant). Here are the numbers:

[1, 2, 3, 5, 8, 10, 11, 12, 21, 22, 23, 32, 35, 53, 58, 85, 100, 101, 110, 200, 1000, 1001, 1021, 1100, 1102, 1120, 1201, 1222, 2011, 2022, 2110, 2122, 2202, 2212, 2220, 2221, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10419, 10941, 11000, 11001, 11010, 11094, 11100, 11490, 14019, 14901, 19104, 19140, 40191, 41109, 41910, 49011, 90114, 91041, 91401, 94110, 100000, 100001, 100100, 110000, 1000000]

Here are the Fibonacci numbers associated with each of these numbers:

1 --> 1

2 --> 2

3 --> 3

5 --> 5

8 --> 8

10 --> 1

11 --> 2

12 --> 5

21 --> 5

22 --> 8

23 --> 13

32 --> 13

35 --> 34

53 --> 34

58 --> 89

85 --> 89

100 --> 1

101 --> 2

110 --> 2

200 --> 8

1000 --> 1

1001 --> 2

1021 --> 34

1100 --> 2

1102 --> 34

1120 --> 34

1201 --> 34

1222 --> 233

2011 --> 34

2022 --> 144

2110 --> 34

2122 --> 233

2202 --> 144

2212 --> 233

2220 --> 144

2221 --> 233

10000 --> 1

10001 --> 2

10010 --> 2

10011 --> 13

10100 --> 2

10101 --> 13

10110 --> 13

10419 --> 75025

10941 --> 75025

11000 --> 2

11001 --> 13

11010 --> 13

11094 --> 75025

11100 --> 13

11490 --> 75025

14019 --> 75025

14901 --> 75025

19104 --> 75025

19140 --> 75025

40191 --> 75025

41109 --> 75025

41910 --> 75025

49011 --> 75025

90114 --> 75025

91041 --> 75025

91401 --> 75025

94110 --> 75025

100000 --> 1

100001 --> 2

100100 --> 8

110000 --> 2

Mathematical Rebuses

A definition of a rebus is a riddle or puzzle made up of letters, pictures, or symbols whose names sound like the parts or syllables of a word or phrase. I discussed rebus puzzles in my most recent post on my Pedagogical Posturing blog. However, in my mathematical blog I wanted to try to create some rebus puzzles that involved only numbers and mathematical symbols. 

The following are what I came up with. There are five questions with a blank space in each and five clues are provided that will help in inserting the correct phrase into the blank space.

1.     The number of items in _________ does not total twelve.

2.     John and his brother are always _________  with each other.

3.     You should learn your _________  at school.

4.     The bank had a _________  rating before it went bankrupt.

5.     It's _________  as to what method you use.

Here are the clues: 

1 .     \(6 + 3 \times 3\)

2.     @ \(6, 6, 6, \dots + 7, 7, 7, \dots \)

3.     \(13 \)

4.     \(2730_{16} \)

5.    \(2748_{16} \)

OK, they may be a bit lame but scroll down to find the answers:

*

*

*

*

*

*

*

*

*

*

1.     The number of items in a baker's dozen does not total twelve (clue 3).

2.     John and his brother are always at sixes and sevens with each other (clue 1).

3.     You should learn your ABC at school (clue 5).

4.     The bank had a AAA rating before it went bankrupt (clue 4).

5.     It's six and two threes as to what method you use (clue 1).

So that's my first attempt at creating some mathematical rebuses. My next effort can only be an improvement.

Highly Composite Deficient Numbers

My diurnal age today is 27027 and this factorises as follows:$$27027=3^3 \times 7 \times 11 \times 13$$Although this number, with its many factors and 32 divisors, looks as though it should be abundant, it's not. It just misses the mark because the ratio of the sum of its proper divisors to the number itself just falls short of unity:$$ \begin{align} \frac{ \sigma(27027, 1)-27027}{27027}&=\frac{53760-27027}{27027}\\ &=\frac{26733}{27027} \\ & \approx 0.989121989121989 \end{align}$$On March 24th 2023, I wrote about Balanced Numbers and 27027 is such a number because:$$27027=\overbrace{27}^{2+7=9} \cdot 0 \cdot \overbrace{27}^{2+7=9}$$However, 27027 has a greater claim to fame because it's a member of OEIS A302934:

A302934

Highly composite deficient numbers: deficient numbers \(k\) whose number of divisors \(d(k) \gt d(m) \) for all deficient numbers \(m \lt k\).

The table below shows a list of deficient numbers up to one million that have a record number of divisors. The ratio of the sum of proper divisors to the number is also shown (permalink).

 number   divisors   ratio

  1        1          0.000000000000000
  2        2          0.500000000000000
  4        3          0.750000000000000
  8        4          0.875000000000000
  16       5          0.937500000000000
  32       6          0.968750000000000
  64       7          0.984375000000000
  105      8          0.828571428571429
  225      9          0.791111111111111
  315      12         0.980952380952381
  1155     16         0.994805194805195
  2475     18         0.953939393939394
  4455     20         0.955555555555556
  8775     24         0.978347578347578
  26325    30         0.994833808167142
  27027    32         0.989121989121989
  63063    36         0.974025974025974
  106029   40         0.971988795518207
  247401   48         0.990614427589217
  693693   54         0.988980716253444
  829521   60         0.995464852607710
  969969   64         0.995280261534132

Looking at the table, the status of 27027 as a record breaker can be seen. Deficient numbers can be ranked by their number of divisors or by how close they approach unity (or how close they approach 2 if we prefer to deal with abundancy). I've investigated the latter in a post titled Odd Deficient Numbers from April 30th 2021. Another post on deficient numbers is Gaps Between Deficient Numbers from October 30th 2020. The post Multiperfect, Hyperfect and Superperfect Numbers from July 24th 2019 is also relevant.