Squaring is Daring (1)

Number Theory Level 4

The following sequences go by the rule:

$a_{n+1}$ = The sum of the squares of the individual digits of $a_{n}$

For example, 56 becomes 61, because $5^{2}$ + $6^{2}$ = $25 + 36$ = 61.

The following is an example of a bad sequence, because it reaches the number 1, and stays there.

{28, 68, 100, 1, 1, 1…}

The following is an example of a good sequence, because it repeats and is non-terminating:

{30, 3, 9, 81, 65, 61, 37 , 58, 89, 145, 42, 20, 2, 4, 16, 37 …}

If I create all sequences from 51 to 60 (inclusive), how many will be "good"?

Super Challenge: Can you find the general set of starting numbers that will make a sequence "good"/"bad"?

Ultimate Challenge: Prove It!

9 10 6 7 0 3 8 4

1 solution

Efren Medallo
Jun 9, 2015

The method I used is the crudest way. Forgive me.

Initially, we know that the sequence will only be bad if the sum of squares of the digits are 100 or 10. We know that every composite integer $n$ can be expressed as a sum of squares if every prime factor $p \equiv 3 (mod 4)$ has an even multiplicity (or zero multiplicity, which just means that no such factor exists for that number). For both 100 and 10, it can be both expressed as a sum of two squares.

For $100$ , the possible previous terms would be $68$ or $86$ . For $10$ , the possible previous terms would be $13$ or $31$ .

Out of these four numbers, $31$ and $86$ have prime factors $p \equiv 3 (mod 4)$ that have an odd multiplicity. Therefore these numbers can no longer be further expressed as a sum of squares. So, if these are the starting numbers of the sequence, your sequence will be bad.

Meanwhile, $13$ and $68$ can still be expressed as a sum of squares.

For $68$ , the possible previous terms would be $82$ or $28$ . For $13$ , the possible previous terms would be $23$ or $32$ .

Then we remove $28$ and $23$ , and again iterate on $82$ and $32$

For $82$ , the possible previous terms would be $91$ or $19$ . For $32$ , the possible previous term would just be $44$ .

Then we remove all $91$ , $44$ and $19$ , because they all have prime factors $p \equiv 3 (mod 4)$ that have an odd multiplicity.

In summary, we have the following starting numbers that will surely give a bad sequence.

$10, 13, 19, 23, 28, 31, 32, 44, 68, 82, 86, 91,$ and $100$ .

Thus, we can conclude that from 51-60, all sequences are good.

UPDATE! I found out that these numbers which start bad sequences are called "happy numbers".

I actually think this is good. It is very thorough. Do you think you could expand it to 3 digit numbers? How many 3 digit starting numbers become good sequences?

Ashwin Padaki - 6 years ago

For three digit numbers, it would mean just checking all of those "bad" starting numbers if they can be expressed as a sum of three squares. I purposefully did not include them in my solution because the problem was focused on how many from 51-60 start "good" sequences. :)

Efren Medallo - 6 years ago

Squaring is Daring (1)

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