Undergrad Is there a happy semigroup of concatenated numbers?

[dkotschessaa]

I haven't slept in awhile and I might have just come up with a totally useless or vacuous concept. It could possibly be a cool example of something. For some reason, I really like happy numbers.

A happy number is a number such that when you separate the digits, square each, and add them back together, you get the number 1 in a finite number of steps. i.e.

13 --> 1^2 + 3^2 = 1 + 9 = 10
10 --> 1^2 + 0+2 = 1

It follows then that if you concatenate two happy numbers you'd get another happy number. So this set is closed under concatenation. Let $*$ be concatenation.Example:

13*10 = 1310 (which is clearly happy).

It's associative, and commutative. I suppose since I am using concatenation that the identity element is just the empty word. But the empty word isn't a number. It's certainly not a group since there is no inverse.

Without the identity it's at least a commutative semigroup. Kind of interesting. Or maybe not.

-Dave K

 

[fresh_42]

I don't get why $13\circ 10 =1310$ is happy. Shouldn't it be $1^2+3^2+1^2+0^2=11 \rightarrow 1^2+1^2=2$? I wouldn't see a problem with the unity, as you could simply define $\{\}$ to be happy and $a \circ \{\}=a\,.$ It's no group because left-concatenation is no bijection (I guess).

Edit: If you meant to continue: $2 \rightarrow 2^2=4 \rightarrow 16 \rightarrow 37 \rightarrow 58 \rightarrow 89 \rightarrow 145 \rightarrow 42 \rightarrow 20 \rightarrow 4$ which is a cycle without a $1$ in between.

 

[mfb]

I don't see an obvious closed operation between happy numbers, concatenation is not one.

13, 23 and 1323 are all happy: there are examples where it works, but in general it does not.

 

[dkotschessaa]

I don't get why $13\circ 10 =1310$ is happy. Shouldn't it be $1^2+3^2+1^2+0^2=11 \rightarrow 1^2+1^2=2$? I wouldn't see a problem with the unity, as you could simply define $\{\}$ to be happy and $a \circ \{\}=a\,.$ It's no group because left-concatenation is no bijection (I guess).

D'oh! Told you I didn't sleep. I had this funny feeling I would regret this post.

Edit: If you meant to continue: $2 \rightarrow 2^2=4 \rightarrow 16 \rightarrow 37 \rightarrow 58 \rightarrow 89 \rightarrow 145 \rightarrow 42 \rightarrow 20 \rightarrow 4$ which is a cycle without a $1$ in between.

Yeah, there might be something that works. But I should try again tomorrow.

I'm going to hide under a rock now.

 

[fresh_42]

I'm going to hide under a rock now.

Don't be square.