Write down the divisors of any number larger than 1. Include the number itself and 1 among the divisors. Now sum only the largest digit from each of the divisors. Continue this process until a number repeats itself. You will eventually reach the number 14 and never escape. Here is an example:
1. The divisors of 27 are 1, 3, 9, 27. Summing the largest digit from each divisor gives 1 + 3 + 9 + 7 = 20.
2. The divisors of 20 are 1, 2, 4, 5, 10, 20, and summing gives 1 + 2 + 4 + 5 + 1 + 2 = 15.
3. The divisors of 15 are 1, 3, 5, 15, and 1 + 3 + 5 + 5 = 14.
4. The divisors of 14 are 1, 2, 7, 14, and 1 + 2 + 7 + 4 = 14 once again - a repeat.
Hence, the number 27 requires 3 iterations to reach the black hole 14. Try the procedure out on a few more numbers, then write a computer program to speed up the process.
Will every integer eventually reach 14 under this procedure?
Yes.
Below I have provided a rigorous proof, but before I list it we need some terminology. Let sl(n) denote the sum of the largest digit in each divisor of n. We will call the full process of repeatedly summing the largest digit of divisors, which was outlined above, the "sl-mapping." And we will refer to the number 14 as a "fixed point," since we will always get 14 after summing the largest digit from each of its divisors.
Theorem: For n >= 2, iteration of the sl-mapping leads to the fixed point 14.
(a) For all numbers n from 2 to 324, iteration of the sl-mapping leads to the fixed point 14. Proved by finite computation.
(b) The largest number of divisors n can have is 2*sqrt(n). (This fact is given in the "Number Theory Functions" section of the Wolfram Research web site.)
(c) For n > 324, sl(n) is at most 2*sqrt(n)*9. This follows from (b) and the definition of sl(n), (the largest digit from each divisor will obviously never be greater than 9).
(d) Hence, for n > 324 we have sl(n) 324, to a number < 324.
(e) The theorem then follows from (d) combined with (a).
Now that we know every number will reduce to 14 under our iteration process, we can ask another question: How many steps will be required to reduce every number to the black hole 14? Is there some kind of upper bound to the sl-mapping? This is where our computers come in and we can have some fun performing various experiments. (It should be noted that for the majority of my numbery theory work, I use the freely available program Pari/GP, which can be downloaded from their web site.)
Let's compute the exact sequence we are curious about. Let sli(n) denote the number of iterations required for any n > 1 to reach 14 under the sl-mapping. The following sequence gives sli(n) for n = 2 to 100.
6, 5, 4, 12, 11, 3, 2, 6, 7, 7, 10, 5, 0, 1, 7, 3, 9, 8, 2, 6, 12, 5, 13, 8, 11, 3, 12, 8, 12, 5, 14, 3, 1, 10, 14, 3, 3, 8, 4, 13, 9, 13, 1, 13, 11, 3, 2, 4, 9, 8, 7, 12, 14, 11, 5, 3, 3, 8, 5, 4, 11, 13, 13, 2, 14, 3, 12, 8, 15, 3, 6, 3, 4, 12, 6, 8, 4, 8, 14, 13, 2, 7, 9, 7, 2, 7, 13, 8, 4, 3, 9, 8, 9, 14, 6, 8, 15, 12, 12, ...
It seems to be rather chaotic and does not tell us much other than all numbers less than 100 require not more than 15 iterations to reduce to 14.
If we compute the least number k that requires n iterations, we don't see much of a pattern either:
n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
k: 14, 8, 7, 4, 3, 2,10, 19, 18, 12, 6, 5, 24, 32, 70, 288, 15624, ?,
What the above means is that, taking the case of n = 8 and k = 19 for example, the sl-trajectory of 19 is 19 => 9 => 13 => 4 => 7 => 8 => 15 => 14, which has 8 total terms in its trajectory. Hence, 19 is the least number that requires 8 steps to reach the blackhole 14.
What is the least number that requires 18 iterations? I don't know. No solution was ever found for all n up to 10^6. And after my systematic search yielded nothing, I employed a number of other polynomials and functions to try and "cheat" my way into finding an 18. I tried various perfect powers, the factorial function, repunits r(n) = (10^n - 1)/9, and other repdigit numbers. I placed them in different combinations but an 18 never appeared. Below is one example of an experiment I tried in which repeated 4s, powers of two, and factorials were used.
sli(4 * r(n) * 2^n * n!) for n = 1 to 19:
2, 9, 16, 11, 6, 4, 13, 11, 11, 16, 7, 6, 13, 14, 16, 6, 6, 7, 10,
The largest term, 4 * r(19) * 2^19 * 19!, has 42 decimal digits and 131,328 divisors, yet it still only requires 10 iterations to reach 14! Amazing!
Although I never found an 18, I mentioned this iteration problem to Klaus Brockhaus and he managed to find the following number with a 19 sl-step trajectory: 1690399999998309600 => 72072 => 585 => 70 => 32 => 24 => 30 => 26 => 12 => 18 => 29 => 10 => 9 => 13 => 4 => 7 => 8 => 15 => 14.
But this number is most likely not the smallest integer requiring 19 steps to reach the 14 black hole..
I conjecture that there is no upper bound to sli(n). Or in other words, as soon as it becomes computationally feasible, I believe someone will be able to find large numbers that require 100 steps or more to reach 14 under the sl-mapping.
Now I will close this article with another related sequence. The numbers that require exactly one step to reach 14. It begins
15, 34, 44, 106, 115, 134, 142, 155, 213, 214, 314, 321, 334, 404, 453, 515, 713, 1006, 1046, 1114, 1115, 1142, 1214, 1234, 1263, 1402, 1403, 1555, 1703, 1849, 2021, 2031, 2103, 2123, ...
And the number of divisors for every term is either 3, 4, or 6. A computer search up to 10^6 revealed that all further solutions had the same set of number of divisors as well. Are there infinitely many n such that sli(n) = 1, and will they always have either 3, 4, or 6 divisors?
References
Wolfram Research, Number Theory Functions, http://functions.wolfram.com/NumberTheoryFunctions/DivisorSigma/29/
Pari/GP, http://pari.math.u-bordeaux.fr/download.html
Published by Jason Earls
Jason Earls is a writer, guitarist, and computational number theorist currently living in Texas with his wife, Christine. He is the author of Cocoon of Terror, Heartless Bast*rd In Ecstasy, Red Zen, How to B... View profile
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