Theorem: Sequence of Sum of Squares of Digits

For a positive integer $n$, let $\map f n$ be the integer created by adding the squares of digits of $n$.
Let $m \in \Z_{>0}$ be expressed in decimal notation.
Let $\sequence {S_m}_{n \mathop \in \Z_{>0} }$ be the sequence defined as follows:
:$n_k = \begin{cases} m & : n = 1 \\
\map f {n_{k - 1} } & : n > 1 \end{cases}$
Then eventually either $\sequence {S_m}$ sticks at $1$, or goes into the cycle:
:$\ldots, 4, 16, 37, 58, 89, 145, 42, 20, 4, \ldots$

Proof:
First note that:
:$1^2 + 9^2 + 9^2 = 163$
:$9^2 + 9^2 + 9^2 = 243$
and it can be seen that for a positive integer $m$ larger than $199$, $\map f m < m$.
Thus it is necessary to investigate numbers only up as far as that.
Starting from the bottom, we have that:
:$\map f 1 = 1^2 = 1$
and so $\sequence {S_1} = 1, 1, 1, \ldots$
We note the sequence:
{{begin-eqn}}
{{eqn | ll= \map f 4
      | lo= =
      | l = 4^2
      | r = 16
}}
{{eqn | ll= \map f {16}
      | lo= =
      | l = 1^2 + 6^2
      | r = 37
}}
{{eqn | ll= \map f {37}
      | lo= =
      | l = 3^2 + 7^2
      | r = 58
}}
{{eqn | ll= \map f {58}
      | lo= =
      | l = 5^2 + 8^2
      | r = 89
}}
{{eqn | ll= \map f {89}
      | lo= =
      | l = 8^2 + 9^2
      | r = 145
}}
{{eqn | ll= \map f {145}
      | lo= =
      | l = 1^2 + 4^2 + 5^2
      | r = 42
}}
{{eqn | ll= \map f {42}
      | lo= =
      | l = 4^2 + 2^2
      | r = 20
}}
{{eqn | ll= \map f {20}
      | lo= =
      | l = 2^2 + 0^2
      | r = 4
}}
{{end-eqn}}
Hence any $m$ in the set $\set {4, 16, 20, 37, 42, 58, 89, 145}$ stays within that sequence, which we will refer to as $\SS$.
It remains to test the rest.
Let $\TT$ be the set of integers $m$ for which $S_m$ ends up either in $\SS$ or $1$.
Then the elements of $\SS$ are all in $\TT$, as is $1$.
If $m \in \TT$, then any integer $m'$ whose non-zero digits are the same as those of $m$ is also in $\TT$.
Thus:
:$2, 10, 24, 40, 61, 73, 85, 98, 100, 106 \in \TT$
We use the notation:
:$a \to b \to c$
to denote:
:$\map f a = b, \map f b = c$
and work progressively through $\Z_{>0}$.
So:
{{begin-eqn}}
{{eqn | l = 3 \to 9 \to 81 \to 65 \to 61
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 18, 30, 56, 90
      | o = \in
      | r = \TT
}}
{{eqn | l = 5 \to 25 \to 29 \to 85
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 50, 52, 92
      | o = \in
      | r = \TT
}}
{{eqn | l = 6 \to 36 \to 45 \to 41 \to 17 \to 50
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 14, 54, 60, 63, 71
      | o = \in
      | r = \TT
}}
{{eqn | l = 7 \to 49 \to 97 \to 130 \to 10
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 13, 31, 70, 79, 94
      | o = \in
      | r = \TT
}}
{{eqn | l = 8 \to 64 \to 52
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 46, 80
      | o = \in
      | r = \TT
}}
{{eqn | l = 11 \to 2
      | o = \in
      | r = \TT
}}
{{eqn | l = 12 \to 5
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 21 
      | o = \in
      | r = \TT
}}
{{eqn | l = 15 \to 26 \to 40
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 51, 62
      | o = \in
      | r = \TT
}}
{{eqn | l = 19 \to 82 \to 68 \to 100
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 28, 86, 91
      | o = \in
      | r = \TT
}}
{{eqn | l = 22 \to 8
      | o = \in
      | r = \TT
}}
{{eqn | l = 23 \to 13
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 32
      | o = \in
      | r = \TT
}}
{{eqn | l = 27 \to 53 \to 34 \to 25
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 35, 43, 72
      | o = \in
      | r = \TT
}}
{{eqn | l = 33 \to 18
      | o = \in
      | r = \TT
}}
{{eqn | l = 38 \to 73
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 83
      | o = \in
      | r = \TT
}}
{{eqn | l = 39 \to 90
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 93
      | o = \in
      | r = \TT
}}
{{eqn | l = 44 \to 32
      | o = \in
      | r = \TT
}}
{{eqn | l = 47 \to 65
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 74
      | o = \in
      | r = \TT
}}
{{eqn | l = 48 \to 80
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 84
      | o = \in
      | r = \TT
}}
{{eqn | l = 55 \to 50
      | o = \in
      | r = \TT
}}
{{eqn | l = 57 \to 74
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 75
      | o = \in
      | r = \TT
}}
{{eqn | l = 59 \to 106
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 95
      | o = \in
      | r = \TT
}}
{{eqn | l = 66 \to 72
      | o = \in
      | r = \TT
}}
{{eqn | l = 67 \to 85
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 76
      | o = \in
      | r = \TT
}}
{{eqn | l = 69 \to 117 \to 51
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 96
      | o = \in
      | r = \TT
}}
{{eqn | l = 77 \to 98
      | o = \in
      | r = \TT
}}
{{eqn | l = 78 \to 113 \to 11
      | o = \in
      | r = \TT
}}
{{eqn | ll= \leadsto
      | l = 87
      | o = \in
      | r = \TT
}}
{{eqn | l = 88 \to 128 \to 69
      | o = \in
      | r = \TT
}}
{{eqn | l = 99 \to 162 \to 41
      | o = \in
      | r = \TT
}}
{{end-eqn}}
{{Finish|It's relatively easy to show that each starting number will decrease up to at least $170$ by considering that the squares of digits will be smaller than the number itself up to around that point (considering $9^2 + 9^2 {{=}} 162$). So manual work is needed "only" below that...<br/>Can it actually be shown that if $m$ is of 3 or more digits, then $\map f m < m$? If so then job done.<br/>Yes, by a series of bounds: $3 \cdot 9^2 {{=}} 243, 2^2 + 2 \cdot 9^2 {{=}} 166, 1 + 6^2 + 9^2 {{=}} 118, 1 + 1 + 9^2 < 100$. Not sure how to formulate nicely, though. For numbers of $n > 3$ digits, $n 9^2 < 10^{n - 1}$}}