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Somehow, I've done posts for 144 and 146, but not 145 . . .

In number theory, a happy number is a number that eventually reaches 1 when the number is replaced by the sum of the square of each digit.

For example, 13 is a happy number because

13 -> 1^2 + 3^2

= 10 -> 1^2 + 0^2

= 1

1 is itself a happy number because 1^2 = 1.

But there are numbers that are not happy. For example, 4 is an unhappy number because it gets stuck in an infinite loop (known as the Melancoil, whose name was popularized by Matt Parker):

4 -> 4^2

= 16 -> 1^2 + 6^2

= 37 -> 3^2 + 7^2

= 58 -> 5^2 + 8^2

= 89 -> 8^2 + 9^2

= 145 -> 1^2 + 4^2 + 5^2

= 42 -> 4^2 + 2^2

= 20 -> 2^2 + 0^2

= 4

145 is an interesting number because it is the only 3 digit number in the Melancoil.

All Numbers are Happy or Unhappy

I figured it was worth talking about how all numbers are happy or unhappy, and thus all numbers will eventually reach 1 or 145.

First note that large numbers always decrease. Suppose a number n has d digits. The largest possible value of f(n) occurs when every digit is 9, so f(n) <= 81d.

You can then compare 81d with the smallest d-digit number, 10^{d-1}.

For d >= 4, 81d <= 10^{d-1}

For example,

d = 4 : 324 < 1000 d = 5 : 405 < 10000

and it only gets more true as d grows.

Therefore, whenever n has at least 4 digits, f(n) < n. Thus, every sufficiently large number with at least four digits strictly decreases after one step.

If n has at most 3 digits, then

f(n) <= 3 * 81 = 243. Thus after at most a few iterations, every orbit enters the set

{1, 2, . . . , 243}

Once you're inside this set, every future value stays inside it because the sum of squares of three digits is still at most 243. Then you can use the fact that finite sets force cycles. If you keep iterating forever, eventually some value must repeat (this is essentially the pigeonhole principle).

Now the problem is finite: check the numbers 1, . . . , 243

A computer search shows there are exactly two cycles:

1 -> 1

4 -> 16 -> 37 -> 58 -> 89 -> 145 -> 42 -> 20 -> 4

Since every orbit must eventually enter a cycle, and these are the only two cycles, every positive integer eventually reaches either 1 or the Melancoil.

Factorion

As a final interesting fact, 145 is also a factorion, meaning that

145 = 1! + 4! + 5! = 1 + 24 + 120

It's one of only 4 factorions in base 10 (the others being 1, 2, and 40585).

Mathematicians have proven mathematically that no factorion can exist beyond seven digits. Any 8-digit number is at least 10,000,000, but the largest possible sum of the factorials of eight digits is only

8 * 9! = 2,903,040

Therefore, no 8-digit number can be a factorion. The same argument works even more strongly for numbers with more digits, so every factorion must have at most 7 digits.

Conclusion