3. The Digit Sum Process.

In this section we are going to study the process proposed by the authors.

Definition 3.1.   We define the dsf function by
, where $\left\{n_1,n_2,\cdots ,n_m \right\}$ is the list of the digits of an integer n.

Remark 3.1.   In mathematics we usually define $0^0=1$, and hence in Definition 3.1 $n_i^{n_i} = 1$ if $n_i = 0$.

we start with any non-negative integer n, and repeatedly apply the function dsf, then we can generate a sequence .

Example 3.1.   If we start with the number 134 and apply dsf function for 5 times, then we have
$134 \to 1^1+3^3+4^4 = 284 \to 2^2+8^8+4^4 = 16777476 \to ...$,
and hence we have {134, 284, 16777476, 3387741, 18424613, 16824417 }.

Example 3.2.   $dsf(99999999999)=4261625379 <99999999999$. This is an example that for a sufficiently large natural number n $dsf(n)<n$.

Our next aim is to find a natural number N such that $dsf(n)<n$ for any natural number $n \geq N$.

Theorem 3.   We have $dsf(n)>n$ for $n=2999999999$, and we have $dsf(n)<n$ for $n\geq3000000000.$

We are going to prove this theorem by the following 5 Lemmas.

Lemma 3.1.   For $n=2999999999$ we have $dsf(n)>n$.

Proof.   $dsf(n)=9\times9^9+2^2=3486784405>n.$

Lemma 3.2.   For a natural number $n$ such that $3000000000 \leq n \leq 3999999999$ we have $dsf(n)<n$.

Proof.   Let $\{n_1=3,n_2,...,n_{10} \}$ be the list of the digits of the integer $n$ .
$(1)$ When the number of 9 in the list is exactly 9,then $n = 3999999999$. $dsf(n) = dsf(3999999999)=3^3+9\times9^9=3486784428<3999999999 = n$.
$(2)$ Suppose that the number of 9 in the list is exactly 8. The remainning 2 numbers are $n_1$=3 and an integer $m$ such that $m \leq 8$.
$(\alpha)$ When $m =0$ , then $dsf(n) = 8\times9^9+3^3+0^0 = 3099363940 <3099999999 \leq n$.
$\beta$ When $1 \leq m \leq 8$, then $dsf(n) =8\times9^9+3^3+m^m \leq 8\times9^9+3^3+8^8 = 3116141155 < 3199999999 \leq n$.
$(3)$ When the number of 9 in the list is less than 8, we have $dsf(n)<3^3+7\times9^9+2\times8^8=2745497882<3000000000 \leq n$.

Lemma 3.3.   For a natural number n such that
$4000000000\leq n \leq 9999999999$ we have $dsf(n)<n$.

Proof.   If $4000000000\leq n \leq 9999999999$, $dsf(n) \leq 10\times9^9=3874204890<n$.

Lemma 3.4.   For any natural number k such that $k \geq 11$ we have

$$k \times 9^9 < 10^{k-1}.$$ (3.1)

Proof.   Let

$$f(x) = 10^{x-1} - 9^9x.$$ (3.2)

Then

$$f^{\prime}(x) = (log_e10)10^{x-1}-9^9.$$ (3.3)

When $x \geq 11$,

$$f^{\prime}(x) = (log_e10)10^{x-1}-9^9 \geq (log_e10)10^{10}-9^9 > 0.$$ (3.4)

Since

$$f(11) =10^{10} - 9^9\times 11 > 0,$$ (3.5)

by 3.4 we have 3.1.

Lemma 3.5.   For a natural number n such that $10000000000 \leq n$ we have $dsf(n)<n$.

Proof.   If $\{n_1, n_2, ...,n_m \}$ is the list of the digits of the integer n , then $m \geq$ 11 and by Lemma 3.4 dsf(n)= $n_1^{n_1}+n_2^{n_2}+...+n_m^{n_m} \leq m \times 9^9$ $<10^{m-1} \leq n$.

Theorem 4.   For any natural number $n \geq 3000000000$ we have $dsf(n)<n$.

Proof.   This is direct from Lemma 3.2, Lemma3.3 and Lemma3.5.

Theorem 5.   For any natural number $n$ the sequence $\{(dsf^m)(n),m=1,2,...\}$ eventually will enter into a loop. In other words there exist numbers p and q such that $(dsf^{p+kq+r})(n)= (dsf^{p+r})(n)$ for any natural number k and any integer r such that $0 \leq r \leq q-1$, where we call q the length of the loop. The sequence $(dsf^m)(n)$ enters into a loop when m = p.

Proof.   If we start with a natural number n, we have a sequence
$\{(dsf^m)(n),m=1,2,...\}$. When $(dsf^m)(n)\geq 3000000000$, by Theorem 3 $(dsf^{m+1})(n) = dsf((dsf^m)(n)) < (dsf^m)(n)$. If $(dsf^{m+1})(n) \geq3000000000$, by Theorem 3 we have $(dsf^{m+2})(n)<(dsf^{m+1})(n)$. In this way this sequence decreases while the value of the sequence is bigger or equal to 3000000000. Therefore there exists a natural number t such that $(dsf^{m+t})(n)<3000000000$. In this way we can prove that there are infinite number of m such that $(dsf^m)(n)<3000000000$, and hence there is a non-negative integer u such that $(dsf^m)(n) = u$ for at least two values of m. Let v be the smallest integer such that $(dsf^v)(n) = u$ and let w be the second smallest integer such that $(dsf^w)(n) = u$ .
Then clearly
$\{(dsf^v)(n),(dsf^{v+1})(n),(dsf^{v+2})(n)$,..., $(dsf^{w-1})(n)\}$ is a loop.

Example 3.3.   If we apply dsf function to the number 132 for 114 times, then we get the following sequence.

{132,32,31,28,16777220,2517295,388250291,420978083,421798753,
405848204,33558331,16783575,18477236,18471246,
17647933,389114542,404201375,827214,17601024,870463,
17647699,776581633,18520794,405024635,53451,6534,50064,
50039,387423643,17648012,17647678,19341414,387421287,
35201810,16780376,18517643,17650825,17653671,1743552,830081,
33554462,53476,873607,18470986,421845378,34381644,16824695,
404294403,387421546,17651084,17650799,776537847,20121452,
3396,387467199,793312220,388244100,33554978,405027808,
34381363,16824237,17647707,3341086,16824184,33601606,140025,
3388,33554486,16830688,50424989,791621836,405114593,
387427281,35201810,16780376,18517643,17650825,17653671,
1743552,830081,33554462,53476,873607,18470986,421845378,
34381644,16824695,404294403,387421546,17651084,17650799,
776537847,20121452,3396,387467199,793312220,388244100,
33554978,405027808,34381363,16824237,17647707,3341086,
16824184,33601606,140025,3388,33554486,16830688,50424989,
791621836,405114593,387427281,35201810,16780376}.

From this sequence we get Graph 3.1.

Graph 3.1.  

If you look at the Graph3.1, you will find that the same pattern occurs more than once.
You will also find the same number appears twice in the sequence if you look at the sequence carefully.