Beyond Emirp

An emirp is a prime that remains prime when its digits are reversed. Palindromic primes are excluded as the reversed number must be different from the original. For example, 13 and 17 are both emirps because 31 and 71 are prime. Another way to phrase this is say that: 

An emirp is a number with one prime factor (itself) such that its reverse (a different number) also only has one factor. 

This definition allows for a generalisation and a special class of numbers arises, namely:

Numbers with $k$ prime factors, counting multiplicity, such that their reversals are different numbers and also contain $k$ prime factors, counting multiplicity.

It's easy to set up an algorithm to determine all such numbers in a given range for various values of $k$. Let's consider the range up to 40000 and $k=1$. This will generate the emirps. There are 980 of them in the given range so I won't list them here. The sequence members can be found at A006567 or by following this permalink. The first few are 13, 17, 31, 37, 71, 73, 79 and 97.

For $k=2$, we get the semiprimes. There are an impressive 3450 in the range selected so again I won't show them here but the sequence members can be found at A097393 or by following this permalink. The first few are 15, 26, 39, 49, 51, 58, 62, 85, 93 and 94.

When $k=3$, we get 2750 numbers that satisfy starting with 117. Here is a permalink that will generate the numbers. Let's use 117 as an example.$$\begin{array}{rl}117& =3\times 3\times 13\\ 711& =3\times 3\times 79\end{array}$$When $k=4$, we get 1302 numbers beginning with 126. Here is a permalink that will generate these numbers. Let's use 126 as an example.$$\begin{array}{rl}126& =2\times 3\times 3\times 7\\ 621& =3\times 3\times 3\times 23\end{array}$$ For $k=5$, there are 429 numbers starting with 270. Here is a permalink that will generate the numbers. Let's look at 270.$$\begin{array}{rl}270& =2\times 3\times 3\times 3\times 5\\ 72& =2\times 2\times 2\times 3\times 3\end{array}$$For $k=6$, there are 103 numbers in the range and so I'll list them. Here is a permalink that will generate the numbers. The first such number is 2576:$$\begin{array}{rl}2576& =2^4\times 7\times 23\\ 6752& =2^5\times 211\end{array}$$2576, 2970, 4284, 4356, 4410, 4600, 4698, 4824, 5265, 5625, 6534, 6752, 6900, 8250, 8964, 10710, 10890, 13140, 13986, 16236, 16335, 17577, 18504, 19494, 20286, 20574, 21114, 21150, 21160, 21336, 21492, 21576, 21609, 21900, 21996, 22392, 22770, 22788, 22824, 22869, 23058, 23247, 23250, 23496, 23562, 23580, 23598, 24156, 24660, 24975, 25020, 25092, 25104, 25164, 25245, 25300, 25416, 25434, 25608, 25668, 26163, 26334, 26532, 27060, 27108, 27135, 27192, 27240, 27248, 27270, 27405, 27408, 27468, 27588, 27608, 27636, 27816, 28116, 28215, 28314, 28710, 28890, 29052, 29172, 29322, 29340, 29412, 29580, 29750, 29784, 29835, 29900, 29960, 29984, 32967, 34965, 35775, 35937, 36162, 36990, 37026, 38367, 38934

$k=7$ generates 25 and this is a small enough number such that we can factorise them all. Here is the permalink to generate numbers.

  number   factor                 reverse   factor

  8820     2^2 * 3^2 * 5 * 7^2    288       2^5 * 3^2
  21240    2^3 * 3^2 * 5 * 59     4212      2^2 * 3^4 * 13
  21708    2^2 * 3^4 * 67         80712     2^3 * 3^2 * 19 * 59
  21780    2^2 * 3^2 * 5 * 11^2   8712      2^3 * 3^2 * 11^2
  21920    2^5 * 5 * 137          2912      2^5 * 7 * 13
  23280    2^4 * 3 * 5 * 97       8232      2^3 * 3 * 7^3
  23472    2^4 * 3^2 * 163        27432     2^3 * 3^3 * 127
  23625    3^3 * 5^3 * 7          52632     2^3 * 3^2 * 17 * 43
  23800    2^3 * 5^2 * 7 * 17     832       2^6 * 13
  25560    2^3 * 3^2 * 5 * 71     6552      2^3 * 3^2 * 7 * 13
  25584    2^4 * 3 * 13 * 41      48552     2^3 * 3 * 7 * 17^2
  25758    2 * 3^5 * 53           85752     2^3 * 3^3 * 397
  26280    2^3 * 3^2 * 5 * 73     8262      2 * 3^5 * 17
  27432    2^3 * 3^3 * 127        23472     2^4 * 3^2 * 163
  27504    2^4 * 3^2 * 191        40572     2^2 * 3^2 * 7^2 * 23
  27888    2^4 * 3 * 7 * 83       88872     2^3 * 3 * 7 * 23^2
  27900    2^2 * 3^2 * 5^2 * 31   972       2^2 * 3^5
  28836    2^2 * 3^4 * 89         63882     2 * 3^3 * 7 * 13^2
  29250    2 * 3^2 * 5^3 * 13     5292      2^2 * 3^3 * 7^2
  29403    3^5 * 11^2             30492     2^2 * 3^2 * 7 * 11^2
  29736    2^3 * 3^2 * 7 * 59     63792     2^4 * 3^2 * 443
  29970    2 * 3^4 * 5 * 37       7992      2^3 * 3^3 * 37
  30492    2^2 * 3^2 * 7 * 11^2   29403     3^5 * 11^2
  34884    2^2 * 3^3 * 17 * 19    48843     3^6 * 67
  36828    2^2 * 3^3 * 11 * 31    82863     3^5 * 11 * 31

For $k=8$ there are only three such numbers (permalink):

  number   factor               reverse   factor

  16560    2^4 * 3^2 * 5 * 23   6561      3^8
  25515    3^6 * 5 * 7          51552     2^5 * 3^2 * 179
  27864    2^3 * 3^4 * 43       46872     2^3 * 3^3 * 7 * 31

For $k=9$ there are four suitable numbers in the given range but for $k>9$ there are no suitable numbers in the range.

  number   factor              reverse   factor

  21168    2^4 * 3^3 * 7^2     86112     2^5 * 3^2 * 13 * 23
  23424    2^7 * 3 * 61        42432     2^6 * 3 * 13 * 17
  23616    2^6 * 3^2 * 41      61632     2^6 * 3^2 * 107
  27456    2^6 * 3 * 11 * 13   65472     2^6 * 3 * 11 * 31

Palindromes Within Palindromes

I'm surprised that I've not covered this sequence before but checking through my previous posts it certainly seems as if I haven't. Here is the sequence in question:

A046351  Palindromic composite numbers with only palindromic prime factors.

The initial members of this sequence, up to 40000, are (permalink):

4, 6, 8, 9, 22, 33, 44, 55, 66, 77, 88, 99, 121, 202, 242, 252, 262, 303, 343, 363, 393, 404, 484, 505, 525, 606, 616, 626, 686, 707, 808, 909, 939, 1111, 1331, 1441, 1661, 1991, 2112, 2222, 2662, 2772, 2882, 3333, 3443, 3773, 3883, 3993, 4224, 4444, 5445, 5555, 5775, 6336, 6666, 6776, 6886, 7777, 7997, 8448, 8888, 9999, 10201, 12221, 13231, 14641, 15251, 15851, 18281, 19291, 20402, 20602, 22622, 22822, 23232, 24442, 24842, 25152, 25452, 26462, 26662, 28682, 30603, 30903, 31613, 33933, 34643, 35653, 36663, 37673, 37873, 38683, 39693, 39993

There are 94 terms in all. Let's just look at the numbers with two distinct prime factors but each of which is two digits or longer. It can be noted that these numbers have either 11 or 101 as factors.

  number   factors

  1111     11 * 101
  1441     11 * 131
  1661     11 * 151
  1991     11 * 181
  3443     11 * 313
  3883     11 * 353
  7997     11 * 727
  13231    101 * 131
  15251    101 * 151
  18281    101 * 181
  19291    101 * 191
  31613    101 * 313
  35653    101 * 353
  37673    101 * 373
  38683    101 * 383

From the above it can be noted that multiplying a palindrome by 11 or 101 seems to produce another palindrome. By extension, multiplying a palindrome by 1001, 10001, 100001 etc. will often produce another palindrome. For example:$$10001\times 1340431=13405650431$$Even a series of alternating 1's and 0's may also produce palindromes. For example:$$10101\times 1340431=13539693531$$However, such products of palindromes are NOT always palindromic. For example:$$\begin{array}{rl}11\times 1949999491& =21449994401\\ 101\times 1949999491& =196949948591\\ 1001\times 1940491& =1942431491\\ 10101\times 1940491& =19600899591\end{array}$$If all the digits of the second palindrome are less than 5 then the multiplication will always produce palindromes.

13x+1 Record Breaker

Trajectory length record breakers for the 3$x$ + 1, 5$x$ + 1, 7$x$ + 1, 11$x$ + 1, 13$x$ + 1 and 17$x$ + 1 Collatz mappings are a fairly exclusive set of numbers and its members from 27859 to 40000 are 27859, 28927, 30301, 30771, 32326, 32581, 34239, 35556, 35655, 35803, 37647, 38030, 39053, 39254 and 39281. Why start at 27859? Well this is the number associated with my diurnal age and the next celebration of such an event is about three years away when I reach 28927 days old. 27859 is associated with the 13$x$ + 1 mapping where the numbers that mark the record breaking trajectory lengths are shown in Figure 1.

Figure 1: see blog post

Figure 2 shows the trajectory for 27859 using a logarithmic scale for the vertical axis.

Figure 2: permalink

The trajectory is as follows ending in a 7, 1, 14, 7 loop. The maximum value reached is an impressive $\mathbf{\text{1,004,280,846,804}}$. That's just over a trillion. That's why a logarithmic scale was needed for the vertical axis!

27859, 362168, 181084, 90542, 45271, 588524, 294262, 147131, 1912704, 956352, 478176, 239088, 119544, 59772, 29886, 14943, 4981, 64754, 32377, 420902, 210451, 2735864, 1367932, 683966, 341983, 4445780, 2222890, 1111445, 222289, 2889758, 1444879, 18783428, 9391714, 4695857, 61046142, 30523071, 10174357, 132266642, 66133321, 859733174, 429866587, 5588265632, 2794132816, 1397066408, 698533204, 349266602, 174633301, 2270232914, 1135116457, 14756513942, 7378256971, 95917340624, 47958670312, 23979335156, 11989667578, 5994833789, 856404827, 11133262752, 5566631376, 2783315688, 1391657844, 695828922, 347914461, 115971487, 1507629332, 753814666, 376907333, 34264303, 445435940, 222717970, 111358985, 22271797, 289533362, 144766681, 1881966854, 940983427, 12232784552, 6116392276, 3058196138, 1529098069, 19878274898, 9939137449, 129208786838, 64604393419, 839857114448, 419928557224, 209964278612, 104982139306, 52491069653, 4771915423, 433810493, 5639536410, 2819768205, 939922735, 187984547, 2443799112, 1221899556, 610949778, 305474889, 101824963, 1323724520, 661862260, 330931130, 165465565, 33093113, 430210470, 215105235, 71701745, 14340349, 186424538, 93212269, 1211759498, 605879749, 7876436738, 3938218369, 51196838798, 25598419399, 3656917057, 47539921742, 23769960871, 309009491324, 154504745662, 77252372831, 1004280846804, 502140423402, 251070211701, 83690070567, 27896690189, 362656972458, 181328486229, 60442828743, 20147609581, 1831600871, 23810811324, 11905405662, 5952702831, 1984234277, 25795045602, 12897522801, 4299174267, 1433058089, 18629755158, 9314877579, 3104959193, 443565599, 5766352788, 2883176394, 1441588197, 480529399, 68647057, 892411742, 446205871, 5800676324, 2900338162, 1450169081, 18852198054, 9426099027, 3142033009, 40846429118, 20423214559, 1856655869, 24136526298, 12068263149, 4022754383, 52295806980, 26147903490, 13073951745, 4357983915, 1452661305, 484220435, 96844087, 1258973132, 629486566, 314743283, 4091662680, 2045831340, 1022915670, 511457835, 170485945, 34097189, 4871027, 695861, 9046194, 4523097, 1507699, 19600088, 9800044, 4900022, 2450011, 31850144, 15925072, 7962536, 3981268, 1990634, 995317, 12939122, 6469561, 924223, 12014900, 6007450, 3003725, 600745, 120149, 1561938, 780969, 260323, 37189, 483458, 241729, 3142478, 1571239, 20426108, 10213054, 5106527, 66384852, 33192426, 16596213, 5532071, 71916924, 35958462, 17979231, 5993077, 77910002, 38955001, 506415014, 253207507, 36172501, 470242514, 235121257, 33588751, 4798393, 62379110, 31189555, 6237911, 81092844, 40546422, 20273211, 6757737, 2252579, 321797, 45971, 597624, 298812, 149406, 74703, 24901, 323714, 161857, 2104142, 1052071, 13676924, 6838462, 3419231, 44450004, 22225002, 11112501, 3704167, 48154172, 24077086, 12038543, 1094413, 14227370, 7113685, 1422737, 18495582, 9247791, 3082597, 440371, 5724824, 2862412, 1431206, 715603, 102229, 1328978, 664489, 94927, 13561, 176294, 88147, 1145912, 572956, 286478, 143239, 1862108, 931054, 465527, 6051852, 3025926, 1512963, 504321, 168107, 2185392, 1092696, 546348, 273174, 136587, 45529, 4139, 53808, 26904, 13452, 6726, 3363, 1121, 14574, 7287, 2429, 347, 4512, 2256, 1128, 564, 282, 141, 47, 612, 306, 153, 51, 17, 222, 111, 37, 482, 241, 3134, 1567, 20372, 10186, 5093, 463, 6020, 3010, 1505, 301, 43, 560, 280, 140, 70, 35, 7, 1, 14, 7

A Special Number: 3367

 Figure 1 shows an interesting result via Cliff Pickover.

To see why 3367 can be written as the various fractions shown, one approach is to consider a number of the form $0a0a0a$ where $1\ge a\ge 99$. Let's manipulate this number in the following way:$$\begin{array}{rl}\frac{0a0a0a}{a+a+a}& =\frac{a\times 010101}{3a}\\ \\ & =\frac{010101}{3}\\ \\ & =3367\end{array}$$Thus when $a=11$ we have:$$\frac{0a0a0a}{a+a+a}=\frac{111111}{11+11+11}$$If $a>99$, then the pattern breaks down.

A Peculiarity

Sometimes, when looking for properties of the number associated with my diurnal age, I struggle to find a property that interests me. Today was one such occasion and the number was 27857. However, I did notice a "peculiarity" as I call it and it involved both the sum of the number's digits cubed and the number's proper divisors. Firstly, the digits cubed:$$\begin{array}{rl}27857& \to 2^3+7^3+8^3+5^3+7^3\\ & =1331\end{array}$$Next, let's consider the number's proper divisors:$$\begin{array}{rl}27857& \to 1,89,313\text{ as proper divisors}\\ \text{sum of proper divisors }& =403\\ & =13\times 31\\ & \to 1331\text{ when concatenated}\end{array}$$

This peculiarity is quite rare and in fact up to $\mathbf{\text{one hundred million}}$, there is only one other number with this same property and that is 150809. The details are:$$\begin{array}{rl}150809& \to 1^3+5^3+0^3+8^3+0^3+9^3\\ & =1367\\ \\ 150809& \to 1,239,631\text{ as proper divisors}\\ \text{sum of proper divisors }& =871\\ & =13\times 67\\ & \to 1367\text{ when concatenated}\end{array}$$So 27857 has a very interesting property after all, even if it is rather obscure. I still find it rather amazing that I was able to check up to one hundred million using SageMath on my M1 Macbook Air.

Partitions and Generating Functions

I asked Gemini to explain the statement:

“It’s long been known that you can associate

a polynomial equation to any partition.”

Here's what it came up with.

The statement refers to the relationship between an integer partition and a generating function, which is a way of encoding an infinite sequence of numbers as the coefficients of a formal power series. In this context, the "polynomial equation" is more accurately a generating function, which can be represented as a polynomial-like expression.

A partition of a positive integer $n$ is a way of writing $n$ as a sum of positive integers. The order of the addends (the parts of the partition) does not matter. For example, the partitions of the number 4 are:

• 4

• 3 + 1

• 2 + 2

• 2 + 1 + 1

• 1 + 1 + 1 + 1

The statement means that for any given integer, we can construct a polynomial-like expression (a generating function) that "encodes" all of its possible partitions.

How the Association Works

The connection is established through the concept of generating functions. The generating function for the number of partitions of an integer $n$, denoted as $p(n)$, can be expressed as an infinite product:$$P(x)=\sum _{n=0}^{\mathrm{\infty }}p(n)x^n=\prod _{k=1}^{\mathrm{\infty }}\frac{1}{1-x^k}$$Let's break down this formula:

• $p(n)$ This represents the number of partitions for an integer $n$

• ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}p(n)x^n}$ This is the power series where the coefficient of $x^n$ is the number of partitions of $n$.

• ${\displaystyle \prod _{k=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{1-x^k}}}$ This is the infinite product that generates these coefficients.

Each term in the product ${\displaystyle \frac{1}{1-x^k}}$ can be expanded as a geometric series:

• ${\displaystyle \frac{1}{1-x}}=1+x+x^2+x^3+\dots$ This represents using the number 1 as a part in the partition

• ${\displaystyle \frac{1}{1-x^2}}=1+x^2+x^4+x^6+\dots$ This represents using the number 2 as a part

• ${\displaystyle \frac{1}{1-x^3}}=1+x^3+x^6+x^9+\dots$ This represents using the number 3 as a part

• And so on for all integers $k$.

When you multiply these series together, the coefficient of $x^n$ in the resulting series will be the total number of ways to form $n$ by adding these parts.

Examples

Let's find the partitions for the number 3. The partitions are:

• 3

• 2 + 1

• 1 + 1 + 1

There are 3 partitions of 3.

Now, let's see how the generating function gives us this result. We look at the expansion of the product:$$(1+x+x^2+x^3+\dots )(1+x^2+x^4+\dots )(1+x^3+x^6+\dots )\dots$$To get the term $x^3$, we can combine terms from these expansions in the following ways:

• From the first factor, take $x^3$. This corresponds to the partition 1 + 1 + 1.

• From the first factor, take $x^1$, and from the second factor, take $x^2$. This corresponds to the partition 1 + 2.

• From the third factor, take $x^3$. This corresponds to the partition 3.

The coefficient of $x^3$ is therefore 3, which matches the number of partitions.

Here is another example for the partitions of 4, which we know is 5.

To find the coefficient of $x^4$, we would consider combinations like:

• $x^4$ from the first factor: 1 + 1 + 1 + 1

• $x^2$ from the first factor and $x^2$ from the second: 1 + 1 + 2

• $x^1$ from the first factor and $x^3$ from the third: 1 + 3

• $x^4$ from the second factor: 2 + 2

• $x^4$ from the fourth factor: 4

The total number of ways to get $x^4$ is 5, so the coefficient is 5.

In essence, the "polynomial" or generating function provides a systematic way to count the number of partitions for any integer by looking at the coefficients of the series expansion.

Prime Factor Sequences

I'm surprised I've not come across this type of sequence before. It has two variants and they are generated iteratively as follows:

• number --> sum of prime factors without multiplicity
  For example, 24 with factors of 2 and 3 gives 5 and terminates after just one step
  
  
• number --> sum of prime factors with multiplicity
  For example, 24 with factors of 2, 2, 2 and 3 gives 11 and terminates after just one step

Larger numbers of course take more than one step to terminate and it's of interest to consider those numbers that set records in term of trajectory lengths. In this context, let's consider OEIS A047830.

A047830  least number which becomes prime after exactly $n$ iterations of f($x$) = sum of prime factors of $x$.

The members of this sequence are 4, 14, 26, 62, 134, 393, 1774, 13682, 41037 up to 100,000. The trajectories are shown in Figure 1 with length indicating the number of steps or iterations:

Figure 1: permalink

If the sum of prime factors with multiplicity is considered then we get OEIS A121360 with sequence members 1, 8, 14, 26, 62, 134, 393, 1257, 4659, 9314, 27933 up to 100,000. The trajectories are shown in Figure 2 with length indicating the number of steps or iterations:

Figure 2: permalink

The algorithms used to find the trajectories of record lengths can be easily modified to find numbers with trajectories of a specified length (number of steps or iterations). For example, how many numbers in the range up to 40000 require eight steps to reach a prime under the sum of prime factors without multiplicity algorithm. Here are the numbers (with 13682 being the first as we know already from Figure 1):

13682, 18002, 19137, 22934, 24014, 24787, 27364, 27849, 30062, 30993, 32577, 33477, 35410, 35798, 36004, 36398, 36706, 39206

These numbers are the initial members of OEIS A047827. Figure 3 shows the details of their trajectories:

Figure 3: permalink

Building Sequences from a Seed Pair

 FIRST EXAMPLE

The numbers 1 and 4 have the interesting properties that:

• their sum is prime: 1 + 4 = 5
• their difference is prime: 4 - 1 = 3
• their product (4) is the average of a pair of twin primes (3 and 5)

Let's make this the starting point of a sequence and let the third member of the sequence be $x$. This gives us: $1,4,x$. We want 4 and $x$ to share the properties that 1 and 4 enjoyed. Namely:

• 4 + $x$ is prime
• $x$ - 4 is prime
• 4 $\times x$ is the average of a pair of twin primes

A little trial and error shows that the smallest value of $x$ we are looking for is 15 because:

• 4 + 15 = 19 is prime
• 15 - 4 = 11 is prime
• 4 $\times$ 15 = 60 is average of a pair of twin pairs (59 and 61)

By using not trial and error but a simple algorithm we can find further terms. The result is OEIS A154493 and the initial terms are:

1, 4, 15, 28, 39, 50, 81, 350, 459, 512, 675, 944, 987, 1040, 1917, 1936, 2325, 2378, 2421, 2588, 2745, 2812, 3459, 3488, 3495, 3506, 5667, 5804, 6027, 6074, 24765, 24832, 25479, 25552, 27621, 27848, 27951, 27980, 34101, 34720, 34773, 35344

SECOND EXAMPLE

Let's take another seed pair with the simple property that the two numbers must add to a cubic number. We'll use 1 and 7 as our seed pair because: $1+7=8=2^3$. Here we see that the next number must be 20 because $7+20=27=3^3$. Using an algorithm, the sequence generated begins thus (permalink):

1, 7, 20, 44, 81, 135, 208, 304, 425, 575, 756, 972, 1225, 1519, 1856, 2240, 2673, 3159, 3700, 4300, 4961, 5687, 6480, 7344, 8281, 9295, 10388, 11564, 12825, 14175, 15616, 17152, 18785, 20519, 22356, 24300, 26353, 28519, 30800, 33200, 35721, 38367

THIRD EXAMPLE

Let's start with seed numbers 1 and 2 this time with the property that:

• the sum of the two numbers has a digit sum that is prime
• the product of the two numbers plus 1 has a digit product that is prime

The seed pair 1 and 2 satisfy since:

• the sum of 1 and 2 is 3 and 3 is prime
• the product of 1 and 2 plus 1 is 3 and 3 is prime

This leads to the following sequence: 1, 2, 9, 12, 13, 16, 18, 23, 24, 25, 27, 29, 32, 33, 34, 40, 45, 47, 51, 60, 62, 66, 100, ... (permalink). There are 2211 terms in the range up to 40000.

An Usual Application of Continued Fractions

Yesterday I turned 27487 days old. This is a prime number of days and this number together with the next three prime numbers form the following sequence:$$27847, 27851, 27883, 27893$$Now let's combine these numbers together to from a continued fraction:

                    1            
27847 + -------------------------
                        1        
         27851 + ----------------
                             1   
                  27883 + -------
                           27893 

This continued fraction is equivalent to the improper fraction:$${\displaystyle \frac{603189725547991331}{21660851250413}}$$with both numerator and denominator being prime. The first of the primes, 27847, qualifies it for membership in OEIS A270884:

A270884: smallest of FOUR consecutive prime numbers that when represented as a simple continued fraction, generates prime numbers in the numerator and denominator, when reduced.

The initial members are (permalink):

41, 367, 619, 659, 701, 2267, 2789, 3253, 3463, 6917, 8969, 9221, 11959, 13499, 14431, 17359, 17851, 20143, 22283, 23669, 26107, 27847, 28547, 28879, 29537, 32503, 32717, 32987, 37549

Let's look at the first of these numbers, 41, for another example. The four primes are thus 41, 43, 47 and 53.

            1        
41 + ----------------
               1     
      43 + ----------
                  1  
            47 + ----
                  53 

Here the continued fraction is equal to:$$\frac{4398061}{107209}$$with both numerator and denominator being prime. Up to one million, the continued fractions of five and six runs of consecutive primes do not produce improper fractions with numerators and denominators that are prime. However, some runs of seven consecutive primes do. Up to 40,000 these are 223, 1579, 5881, 8293, 9013, 12347, 15121, 17783, 25523 and 38903 (permalink). Let's look at the first of these: 223.

                         1                    
223 + ----------------------------------------
                             1                
       227 + ---------------------------------
                                1             
              229 + --------------------------
                                    1         
                     233 + -------------------
                                       1      
                            239 + ------------
                                           1  
                                   241 + -----
                                          251 

This produces the fraction:$$\frac{39053357680436977}{175123705045789}$$with both numerator and denominator prime. No numbers satisfy for runs of eight and nine consecutive primes but for runs of ten we have 5519, 13037, 18743 and 39857 (permalink). Let's look at 5519. The continued fraction (this time in compact format) is:$$\begin{array}{rl}& 5519;5521,5527,5531,5557,5563,5569,5573,5581,5591]\\ \\ & =\frac{27886154132102326260514832393879709899}{5052754705003449540186402260387567}\end{array}$$with the numerator and denominator prime.

Harmonic Numbers

On January 1st 2024, I made a post titled Unitary Harmonic Numbers which are defined as numbers whose unitary divisors have a harmonic mean that is an integer. Oddly, I have never made a post simply about harmonic numbers defined as numbers whose divisors have a harmonic mean that is an integer. Like unitary harmonic numbers, they are quite rare. The number associated with my diurnal age today (27846) is one such harmonic number. These numbers make up OEIS A001599 and the initial members up to one million are (perfect numbers are shown in red):

1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190, 18600, 18620, 27846, 30240, 32760, 55860, 105664, 117800, 167400, 173600, 237510, 242060, 332640, 360360, 539400, 695520, 726180, 753480, 950976

Here is what Numbers Aplenty had to say about them:

A number $n$  is called harmonic divisor number if the harmonic mean of its divisors is an integer. This is equivalent to saying that the average of the divisors of $n$ divides $n$: $$\frac{n}{\sigma (n)/\tau (n)}=\frac{n\times \tau (n)}{\sigma (n)}\text{ is an integer}$$Harmonic divisor numbers are also called harmonic numbers, for brevity, or Ore numbers, after O.Ore who studied them. He proved that all the perfect numbers are also harmonic and conjectured that 1 is the only odd harmonic number. This conjecture has been verified by G.L.Cohen et al. for $n<{10}^{24}$ and if true, it will imply that no odd perfect numbers exist. Jaycob Coleman has observed that all the Ore numbers up to ${10}^{14}$ are also practical numbers and conjectured this holds in general. T. Goto and K. Okeya have computed a list of the 937 harmonic numbers up to ${10}^{14}$.

In the case of the number (27846) associated with my diurnal age, we have:$$\begin{array}{rl}\tau (27846)& =48\\ \sigma (27846)& =78624\\ \frac{27846\times 48}{78624}& =17\end{array}$$

A Novel Number Run

Every fourth number is a multiple of 4 and so there cannot be a run of four consecutive numbers that are the product of two distinct primes. But how often do we observe runs of FIVE consecutive numbers that have only two prime divisors ignoring multiplicity.  Today, when I turned 27845 days old, marked the end of one such run. See Figure 1.

Figure 1: a run of five numbers

So 27841 marked the start of the sequence of five consecutive numbers, each of has only two prime divisors ignoring multiplicity:

• 27841 has prime divisors of 11 and 2531
• 27842 has prime divisors of 2 and 13921
• 27843 has prime divisors of 3 and 9281
• 27844 has prime divisors of 2 and 6961
• 27845 has prime divisors of 5 and 5569

 How many other numbers in the range up to 40000 mark the start of a similar sequence. It turns out that there are 67 such numbers (permalink).

54, 91, 115, 141, 158, 205, 212, 295, 301, 323, 391, 535, 685, 721, 799, 1135, 1345, 1465, 1535, 1711, 1941, 1981, 2101, 2215, 2302, 2425, 2641, 3865, 4411, 5461, 6505, 6625, 6925, 7165, 7231, 7261, 7441, 7855, 7891, 8575, 9121, 9355, 9571, 10081, 10375, 11755, 11851, 13435, 13951, 14371, 15115, 16291, 17635, 18805, 20575, 20665, 24575, 24865, 25291, 27841, 30373, 31285, 32125, 35821, 36961, 38941, 39391

These numbers make up OEIS A383400. 

What about runs of six such numbers? There are only five numbers and they are 91, 141, 212, 323 and 2302. See Figure 2 where 91 is used as an example:

Figure 2: a run of six numbers

Let's investigate runs of seven such numbers. Here we find 91 drops out because 97 is prime but 141, 212, 323 and 2302 remain. Figure 3 shows 141 as the example.

Figure 3: a run of seven numbers

For runs of eight such numbers only 141 and 212 remain. Figure 4 shows the results using 141 again.

Figure 4: a run of eight numbers

149 is prime so that's the end of 141 but what about 212? Unfortunately, it doesn't make it either as Figure 5 shows.

Figure 5: a run of eight but not nine

Even extending the range to TEN million, no suitable numbers are found. Just the fact that I can use SageMath on my M1 Macbook Air to factorise ten million numbers is pretty amazing. Are there runs of nine consecutive numbers? I didn't know so I put the following question to Google's Gemini:

The numbers 212 to 219 inclusive all have only two distinct prime divisors. This represents a run of eight consecutive numbers with this property. According to my analysis there are only two such runs in the range from zero up to ten million. The other starts with 141 and ends with 148. Here is my (extremely simple) code:

L=[]
k=0
for n in [1..10000000]:
    if len(prime_factors(n))==2:
        k+=1
    else:
        k=0
    if k==8:
        L.append(n-k+1)
print(L)

The output is simply [141, 212], numbers which mark the starts of the runs of eight numbers with two prime divisors, ignoring multiplicity of factors.

Are there runs of nine or more such numbers? Has anyone discovered such runs or proven that they are not possible. I'd be interested to find out.

This was Gemini's response:

Runs of Consecutive Integers with Two Distinct Prime Divisors: A Mathematical Exploration 

Your discovery of two runs of eight consecutive integers, each having exactly two distinct prime divisors, is a sharp observation that touches upon a fascinating area of number theory. Your analysis, which identified the starting points of these runs as 141 and 212 within the first ten million integers, is indeed correct.

The question of whether runs of nine or more such numbers exist has been explored by mathematicians. The definitive answer is that no, there are no runs of nine or more consecutive integers with only two distinct prime divisors.

This conclusion stems from a 2018 paper by Roger B. Nelsen titled "Consecutive Integers with Equally Many Principal Divisors." In his work, Nelsen proves that the maximum possible length of a run of consecutive integers, each with exactly two distinct prime factors (ignoring multiplicity), is eight.

Your independent discovery of the runs beginning at 141 and 212 is particularly noteworthy as Nelsen's paper also highlights these specific sequences:

• 141 to 148:

  • $141=3\times 47$
  • $142=2\times 71$
  • $143=11\times 13$
  • $144=2^4\times 3^2$ (Distinct prime factors are 2 and 3)
  • $145=5\times 29$
  • $146=2\times 73$
  • $147=3\times 7^2$ (Distinct prime factors are 3 and 7)
  • $148=2^2\times 37$ (Distinct prime factors are 2 and 37)

• 212 to 219:

  • $212=2^2\times 53$ (Distinct prime factors are 2 and 53)
  • $213=3\times 71$
  • $214=2\times 107$
  • $215=5\times 43$
  • $216=2^3\times 3^3$ (Distinct prime factors are 2 and 3)
  • $217=7\times 31$
  • $218=2\times 109$
  • $219=3\times 73$

Nelsen's paper further conjectures that the two runs you identified are the only runs of length eight in existence. Your computational search up to ten million provides empirical support for this conjecture.

The limitation on the length of such runs is fundamentally tied to the distribution of prime numbers and how they combine to form composite numbers. As numbers get larger, they are more likely to be divisible by a greater variety of small primes. For a long sequence of consecutive integers, it becomes increasingly difficult to avoid a number with one, three, or more distinct prime factors. For instance, any sequence of 30 consecutive integers must contain a multiple of 30, which has three distinct prime factors (2, 3, and 5). This simple argument illustrates why arbitrarily long runs are not possible.

In summary, your computational findings are in line with established mathematical results. The quest for longer runs of this specific type has been settled, with the maximum length proven to be eight. Your work serves as an excellent example of how computational exploration can lead to and verify deep and sometimes surprising results in number theory.

As the response mentions, every 30th number must have three distinct prime factors (2, 3 and 5) and so 29 is clearly an upper limit for runs of two prime factors. However, thanks to the research paper we now know that the actual upper bound is 8. Similarly for runs of three, four, five etc. prime factors, the primorials will dictate maximum possible upper bounds.

I then went on to ask Gemini the following question:

Is there information about the maximum lengths of runs of numbers with three prime divisors and beyond? Clearly the primorials will dictate the maximum possible lengths. So in the case of three prime divisors, 209 will be the maximum possible but does Eggleton's paper shed any light on the actual limit which I imagine would be considerably less that 210.

Here was the response:

An excellent and insightful question that delves deeper into the fascinating patterns of prime factorizations in consecutive integers. You are absolutely correct in your reasoning about primorials providing a theoretical upper bound on the length of such runs. Let's explore what is known about runs of integers with three or more distinct prime divisors.

Runs of Integers with Three Distinct Prime Divisors 

For the case of three distinct prime divisors, your intuition is spot on. While the primorial $P_4=2\times 3\times 5\times 7=210$ guarantees that in any sequence of 210 consecutive integers, one will be divisible by 2, 3, 5, and 7 (and thus have at least four distinct prime factors), this provides a very loose upper bound. The actual longest known run is significantly shorter.

Based on computational searches, most notably from communities like Project Euler, the longest known run of consecutive integers with exactly three distinct prime factors is 16. This sequence begins at the number 127,890,362.

The first such run of three, for comparison, is:

• $644=2^2\times 7\times 23$
• $645=3\times 5\times 43$
• $646=2\times 17\times 19$

It's clear that finding these runs requires extensive computation, and they are quite rare.

Runs of Integers with Four or More Distinct Prime Divisors 

The investigation into runs for four or more distinct prime divisors follows the same pattern of computational discovery. Here are some of the longest known runs for k distinct prime factors:

• Four distinct prime factors: A run of 11 is known, starting at 163,459,742.
  • The first run of four begins at 130842.
• Five distinct prime factors: A run of 5 has been found, starting at 129,963,314.
• Six distinct prime factors: The longest known run is shorter, at 3, starting at 323,567,034.

A crucial observation here is that as the number of required distinct prime factors (k) increases, the integers involved become much larger, and the maximum known run lengths do not necessarily increase.

The Role of Primorials 

Your point about primorials is the key to understanding why these runs must be finite. Let ω(n) be the number of distinct prime factors of n. You are looking for the maximum length of a sequence $m+1,m+2,\dots ,m+L$ where $\omega (n)=k$ for all n in the sequence.

Let Pk​# denote the k-th primorial (the product of the first k primes). The smallest integer with $k+1$ distinct prime factors is the $(k+1)$-th primorial, Pk+1​#. Any sequence of Pk+1​# consecutive integers must contain a multiple of Pk+1​#. This multiple will have at least $k+1$ distinct prime factors. Therefore, a run of integers with exactly k distinct prime factors cannot have a length of Pk+1​# or longer.

This gives a theoretical, albeit very large, upper bound on the possible length of such a run. For $k=3$, the bound is $P_4\#-1=209$, as you correctly deduced. However, as the computational evidence shows, the actual maximum lengths are far smaller.

Does Eggleton's Paper Shed Light on This? 

The 2008 paper by Eggleton and Macdougall, "Consecutive Integers with Equally Many Principal Divisors," which proved the maximum run length for $k=2$ is 8, appears to be quite specific to the case of two distinct prime divisors. The methods used in that proof do not seem to readily extend to the case of three or more prime factors.

While Eggleton and his collaborators have published other papers on related topics, such as "consecutive integers with no large prime factors," these address a different question—not the number of distinct prime factors, but their size.

As of now, for $k\ge 3$, there are no known published proofs that establish the maximum possible length of such runs. The current state of knowledge is primarily based on computational searches and the ever-growing records they produce. It is an active area of interest in computational number theory, and it's entirely possible that longer runs than those currently known exist, though they would almost certainly start at very large numbers.