Question

1,7,13,19,...
are Happy numbers
the rule is like this ( for example 13)
1^2+3^2=1+9=10
1^2+0^2=1 then 13 is Happy

Answer 1

under which operation the happy numbers are closed
multiple , addition , none

Answer 2

how is 7 a happy number ?

Answer 3

7^2=49
4^2+9^2=16+81=97
9^2+7^2=81+49=130---> see same as 13
1^2+3^2=1+9=10
1^2+0^2=1

Answer 4

Oh they need to end up at 1 after doing a sequence of squaring and adding ok got it

Answer 5

yosh
sad number end up with cycle of no happy number >.<

Answer 6

by that definition 10 is also a happy number right ?

Answer 7

Here are the happy numbers below 100 :

```
1
7
10
13
19
23
28
31
32
44
49
68
70
79
82
86
91
94
97
100
```

Answer 8

they are not closed under addition because 1+7 = 8 is not happy

Answer 9

they seem to be closed under multiplication
need to prove it mathematically yeah

Answer 10

it seems just like a sequence that has:
~ \(\large\color{black}{ ~a_{_1}=8 }\)
~ \(\large\color{black}{ ~d=6 }\) \(\LARGE\color{white}{ \rm \left| \right| }\)

Answer 11

I mean if you just look at the first row of the question (in the light blue box).

Answer 12

In base two all numbers are happy numbers! =D

Answer 13

As an optimist and a programmer, I will offer this not very rigorous proof that all numbers are happy in base 2.

All digits in base 2 are 1, so the square of all the digits are 1 as well. So we can simply sum the digits as they are. But of course the sum of digits is less than the number itself, so if we do this an arbitrary number of times, then eventually we will get a happy number!

Not only that, but it only takes a single step to show that all powers of 2 are happy, how cute!

Answer 14

They are not closed under multiplication either :
13*13 = 169 is not happy as it leads to 4-loop.

Answer 15

I think 0 is the only number thats not happy in base2 :)

Answer 16

we can make base2 also interesting by stopping the process right after reaching a 2 digit number. that way we will have four options to look for

Answer 17

https://oeis.org/A007770/list

Answer 18

Suppose I offer to give you a jar with a number of cookies in it. Then you open the jar and find out it's empty. Then I laugh and say, "Zero is a number!"

So that's my argument for defending that all numbers in base two are happy, since 0 isn't a number haha. XD

Answer 19

lol i would say im convinced xD

Answer 20

`Not only that, but it only takes a single step to show that all powers of 2 are happy, how cute!`

are you still refering to base 2 here kai ?

Answer 21

Yeah, I suppose analogously we know that "normally" all powers of 10 can be shown to be happy numbers in a single step as well.

Answer 22

I see

Answer 23

Are there more happy numbers than unhappy numbers, and are there only so many finitely many "trees", meaning 7 and 13 are part of the same tree like in ikram's example earlier.

Answer 24

Interesting. There cannot be infinitely many trees because the value of number is greater than the sum of squares of digits for all 4+ digit numbers. Below can be shown trivially :

\[A_3A_2A_1A_0 \gt A_3^2 + A_2^2 + A_1^2+A_0^2\]

Answer 25

so a loose upperbound for number of trees can be 999

Answer 26

Wow interesting, very cool find! But I think I have found a counterexample! We can look at any power of 10 to be a direct path in a single step that bypasses all the other "routes" and funnels you straight to happiness while skipping any other trees.

So by themselves, we have 1, 10, 100, 1000, 10000, 100000000000000, etc as independent "roots" of an infinite number of independent happy trees. Now as an added bonus, we can always find a sum of squares of large numbers that funnel to these as well, and one formula to find these is: \[\Large x81+y=10^n\] So for instance to find a number that funnels into 10,000 we have: 123 remainder 37 so any way we can arrange these digits:

111111111111111111111111111999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

xD

Answer 27

100 reduces to 1 so it sits on the same tree

Answer 28

Hmm but if we look at it like that, then that trivializes it since all happy numbers would be on the same tree?

Answer 29

yes, we are looking for other bad trees right ?

Answer 30

for example : 1 and 4 are different trees

Answer 31

Ohhh I think I see where we are misunderstanding each other.

Answer 32

I was imagining there being a sort of set of independent routes to 1 with other independent cycles that fail like this maybe: