More about Golden Semiprimes

In June of 2016, I posted about Golden Semiprimes. Today I was reminded of this class of numbers once again because today's number, 24709, is a prime that forms the greater prime in a series of semiprimes that give increasingly better approximations to the golden ratio. This OEIS series is A165570: successively better golden semiprimes and begins:

6, 15, 77, 589, 851, 1363, 15229, 201563, 512893, 644251, 1366553, 3416003, 7881197, 377331139, 400711231, 2963563859, 4035221017, 28862500577, 52027213697, 133793658289, 418298061641, 1363588753103, 1970239102459

The OEIS series that gives the greater primes in these semiprimes is A165572: greater prime factor of Successively Better Golden Semiprimes. The series begins:

3, 5, 11, 31, 37, 47, 157, 571, 911, 1021, 1487, 2351, 3571, 24709, 25463, 69247, 80803, 216103, 290141, 465277, 822691, 1485373, 1785473

For today's number 24709, the associated semiprime is 377331139 and the factorisation is 15271×24709. In a little over two years time, the next prime in series A165572 will pop up, namely 25463.

A Curious Coincidence

My first wife, and mother of my daughter, was born on the 23rd November 1953 and will be turning 63 shortly. My daughter was born on the same date in 1980 and will be turning 36. I've written about the mathematical aspects of this digit reversal in an earlier post. 

Getting back to my first wife however, she will be turning 23011 days old. A curious coincidence indeed. Furthermore, the number is prime and marks the start of five consecutive primes: 23011, 23017, 23021, 23027 and 23029. It is also the first member of a prime quadruple in a 2p-1 progression: 23011, 46021, 92041 and 184081. This is a Cunningham chain of the second kind. I've written about such chains in an earlier post.

However, I'm interested in the general question of what are the conditions for such coincidences to occur? It can be noted that 23011/365.242199 is about 63.00203 so such coincidences can only occur when the day count is almost exactly divisible by 365.242199. Now a person must be born between the 1st of January and the 31st December, thus the possible numbers generated (using leading zeroes) range from 01001 to 31012. 

On a spreadsheet, I tested division over every number in this range by 365.242199, looking for remainders that were less 0.01. A divergence of just two days either side produces a difference of around 0.005 so 0.01 is quite generous. A few pockets of numbers satisfied the condition but only 23011 held up in the end. 23012, corresponding to 23rd December, came close but birthdays fell on 23rd November or 23rd October so it didn't satisfy.

Even having this birthday does not guarantee the coincidence. For example, my daughter will turn 63 in 2043 but on that day she will be 23010 days old. While on the topic of coincidences, I must note that on the 23rd November 2016, I will be 24706 days old. This number factorises to 2 x 11 x 1123. Amazing.

The \(3x+1\) Problem

Today I was struggling to find something of interest to say about the daily number 24684 (see comment at end of this post) and so I considered not just its prime factorisation: \(2^2 \times 3 \times 11^2 \times17\), that consisted only of small primes, but also the factorisation \(4 \times 6171\). I wanted to find out if there was something of interest about the number 6171. It turns out that there is.

OEIS A006877 stated that in the \(3x+1\) problem, these values for the starting value set new records for number of steps to reach 1 and it listed these initial values:

1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, 129, 171, 231, 313, 327, 649, 703, 871, 1161, 2223, 2463, 2919, 3711, 6171, 10971, 13255, 17647, 23529, 26623, 34239, 35655

But what is the \(3x+1\) problem that's referred to. I'd heard of it but checked on Wikipedia to clarify my understanding. It's referred to there as the Collatz conjecture. Here is the introduction to that Wikipedia article:

The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz. The conjecture is also known as the \(3n + 1\) conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.

The conjecture can be summarized as follows. Take any positive integer \(n\). If \(n\) is even, divide it by 2 to get \(n/2\). If n is odd, multiply it by 3 and add 1 to obtain \(3n + 1\). Repeat the process (which has been called "Half Or Triple Plus One", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. 

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems." He also offered $500 for its solution. Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, "this is an extraordinarily difficult problem, completely out of reach of present day mathematics."

For 6171, 261 steps are required to reach 1. At this stage, nobody has proved the Collatz conjecture. Here are some more records:

The longest progression for any initial starting number

less than 10 is 9, which has 19 steps,
less than 100 is 97, which has 118 steps,
less than 1,000 is 871, which has 178 steps,
less than 10,000 is 6,171, which has 261 steps,
less than 100,000 is 77,031, which has 350 steps,
less than 1 million is 837,799, which has 524 steps,
less than 10 million is 8,400,511, which has 685 steps,
less than 100 million is 63,728,127, which has 949 steps,
less than 1 billion is 670,617,279, which has 986 steps,
less than 10 billion is 9,780,657,631, which has 1132 steps,
less than 100 billion it is 75,128,138,247, which has 1228 steps.

See also:
https://voodooguru23.blogspot.com/2018/03/the-collatz-conjecture-revisited.html

https://voodooguru23.blogspot.com/2018/03/the-px1-map.html

https://t.co/h8cMC9QKes this link relates to Terence Tao's discovery of late 2019.

Comment regarding 24684: 

In December of 2019, I discovered that by entering a number like 24684 into the Google search box along with the acronym OEIS, a greater range of results will appear compared to just typing the number into the OEIS search bar. The reason for this is that Google will index the text files associated with the sequence. These files typically involve much larger numbers than are displayed for each sequence (possibly the first 1,000 or 10,000 rather than just the first 50).

Using 24684, as an example we find that it is a member of OEIS A228844: smallest sets of 3 consecutive abundant numbers in arithmetic progression. The initial abundant number is listed. However, only the following numbers are listed:

24, 42, 80, 100, 104, 114, 120, 126, 144, 162, 180, 196, 200, 220, 228, 234, 240, 246, 272, 282, 288, 304, 324, 348, 350, 364, 392, 402, 420, 426, 440, 460, 504, 572, 582, 588, 594, 608, 616, 624, 640, 654, 660, 666, 684, 700, 708, 714, 728, 736, 740, 786

The example given in the comments is:

24, 30, 36 is the smallest set of 3 consecutive abundant numbers in arithmetic progression so 24 is in the list.

24684 is nowhere to be found in the previous list but it does turn up in the Google search results as the 1729th member of the sequence. The arithmetic progression is 24684, 24690 and 24696 because all three are abundant with a common difference of 6 between terms.