interesting

Let $N$ be a composite number such that $N=p_1\cdots p_m$, with $p_i$ being prime. Then the function is basically $1+\sum_{i=1}^m p_i$. Since $m$ is $\geq 2$ and $p_i\geq 2$ for all $i$, this is bounded by $f(N)\leq 1+2(m-1)+\frac{N}{2^{m-1}}$ (note that $\frac{N}{2^m}\geq 2$ since $N\geq 2^m$); this is basically equivalent to maximizing $x_1+x_2+\cdots +x_m$ given $x_1\cdots x_m=N$ with $x_i\in [2,+\infty)$ (and $N\geq 2^m$). Furthermore, we have $1+2(m-1)+\frac{N}{2^{m-1}}<N-1$ for all $m>2$, because $N(1-\frac {1}{2^{m-1}})\geq 2^m(1-\frac{1}{2^{m-1}})=2^m-2$ By an easy induction, this is greater than $2m$ for all $m> 3$, implying the result. For $m=2$ the wanted inequality is equivalent to $3+\frac{N}{2}<N\iff N>6$. For $m=3$ the inequality is equivalent to $5+\frac N4<N-1\iff N>8$. Furthermore, if $N$ is prime, $f(N)=N+1$ is composite, and so $f(f(N))=f(N+1)<(N+1)-1=N$ if $N+1>8\iff N>7$. Therefore, for all $N>8$ one of $f(N)$ and $f(f(N))$ is less than $N$. Therefore, every orbit eventually goes to a number which is at most $8$. Now, since $7\rightarrow 8\rightarrow 6,2\rightarrow 3\rightarrow 4\rightarrow 5\rightarrow 6\rightarrow 6$, it follows that the unique cycle (apart from $f(1)=1$, which anyway has period $1$) is $f(6)=1+2+3=6$, which has a fundamental period of $1$. Also,

cadaeibf wrote: Let $N$ be a composite number such that $N=p_1\cdots p_m$, with $p_i$ being prime. Then the function is basically $1+\sum_{i=1}^m p_i$. Since $m$ is $\geq 2$ and $p_i\geq 2$ for all $i$, this is bounded by $f(N)\leq 1+2(m-1)+\frac{N}{2^{m-1}}$ (note that $\frac{N}{2^m}\geq 2$ since $N\geq 2^m$); this is basically equivalent to maximizing $x_1+x_2+\cdots +x_m$ given $x_1\cdots x_m=N$ with $x_i\in [2,+\infty)$ (and $N\geq 2^m$). Furthermore, we have $1+2(m-1)+\frac{N}{2^{m-1}}<N-1$ for all $m>2$, because $N(1-\frac {1}{2^{m-1}})\geq 2^m(1-\frac{1}{2^{m-1}})=2^m-2$ By an easy induction, this is greater than $2m$ for all $m> 3$, implying the result. For $m=2$ the wanted inequality is equivalent to $3+\frac{N}{2}<N\iff N>6$. For $m=3$ the inequality is equivalent to $5+\frac N4<N-1\iff N>8$. Furthermore, if $N$ is prime, $f(N)=N+1$ is composite, and so $f(f(N))=f(N+1)<(N+1)-1=N$ if $N+1>8\iff N>7$. Therefore, for all $N>8$ one of $f(N)$ and $f(f(N))$ is less than $N$. Therefore, every orbit eventually goes to a number which is at most $8$. Now, since $7\rightarrow 8\rightarrow 6,2\rightarrow 3\rightarrow 4\rightarrow 5\rightarrow 6\rightarrow 6$, it follows that the unique cycle (apart from $f(1)=1$, which anyway has period $1$) is $f(6)=1+2+3=6$, which has a fundamental period of $1$. Also, how you found that $f(8)=6$ ?? $f(8)= 1+(2)(3)=7$

moom wrote: cadaeibf wrote: Let $N$ be a composite number such that $N=p_1\cdots p_m$, with $p_i$ being prime. Then the function is basically $1+\sum_{i=1}^m p_i$. Since $m$ is $\geq 2$ and $p_i\geq 2$ for all $i$, this is bounded by $f(N)\leq 1+2(m-1)+\frac{N}{2^{m-1}}$ (note that $\frac{N}{2^m}\geq 2$ since $N\geq 2^m$); this is basically equivalent to maximizing $x_1+x_2+\cdots +x_m$ given $x_1\cdots x_m=N$ with $x_i\in [2,+\infty)$ (and $N\geq 2^m$). Furthermore, we have $1+2(m-1)+\frac{N}{2^{m-1}}<N-1$ for all $m>2$, because $N(1-\frac {1}{2^{m-1}})\geq 2^m(1-\frac{1}{2^{m-1}})=2^m-2$ By an easy induction, this is greater than $2m$ for all $m> 3$, implying the result. For $m=2$ the wanted inequality is equivalent to $3+\frac{N}{2}<N\iff N>6$. For $m=3$ the inequality is equivalent to $5+\frac N4<N-1\iff N>8$. Furthermore, if $N$ is prime, $f(N)=N+1$ is composite, and so $f(f(N))=f(N+1)<(N+1)-1=N$ if $N+1>8\iff N>7$. Therefore, for all $N>8$ one of $f(N)$ and $f(f(N))$ is less than $N$. Therefore, every orbit eventually goes to a number which is at most $8$. Now, since $7\rightarrow 8\rightarrow 6,2\rightarrow 3\rightarrow 4\rightarrow 5\rightarrow 6\rightarrow 6$, it follows that the unique cycle (apart from $f(1)=1$, which anyway has period $1$) is $f(6)=1+2+3=6$, which has a fundamental period of $1$. Also, how you found that $f(8)=6$ ?? $f(8)= 1+(2)(3)=7$ I'm sorry I had a mistake. By correcting it we only have the two cycles $6\rightarrow 6$ and $7\rightarrow 8$, so the fundamental period is $1$ or $2$ depending on the starting numbers, so I guess we can say overall that the period is $2$.

first we can found that $f(1)= 2$$ f(2)=3$ $f(3)=4$ $f(4)=5$ $f(5)=6$ $f(6)=6$ $f(7)=8$ $f(8)=7$ we can see that $f(p)=p+1$ claim 1: for all prime $p_i$ and positive integer $a_i$ we have $ p_i^{a_i} \ge p_ia_i $ proof : its easy by induction claim 2 : for all coprime number$ n $\ge$ 9$we have $n-2$ $\ge$ $f(n)$ proof : we discuse tow case case1) $ n= p_i ^ {a_i} $ and when n is coprime then $a_i\ge 2$ so here the proof is easy by induction case2) n has at least tow different primer factor from claim 1 we have $2+p_1a_1 +........+p_ka_k$ $\ge$ $2+p_1 ^ {a_1}+....p_k ^ {a_k}$ let $s_i =p_i ^{ a_i}$ we want to prove that $s_1+s_2+s_3+s_4 +.......s_k$ $\le $ $s_1s_2..s_k$ for all $k\ge 2$ the proof is also by induction k=2 $2+s_1+s_2$ when $n \ge 10$ we found that one of $s_i$ is bigger than three and wlog $s_2 \ge s_1$ if $s_1=2$ we find that $s_1s_2 = s_2 +s_2(s_1-1)\ge s_2 \ge s_1+2=4 $ is true because $n \ge 10 $ if $s_1\ge 3$then $s_1s_2 = s_2 +s_2(s_1-1)\ge 2s_2 \ge 2s_1 \ge s_1+2 $ so k=2 is true suppose is true for $k$ we want to prove is true for $k+1$ wlog $s_1=min{s_1,s_2,...,s_(k+1) } $ so $s_1s_2s_3...s_(k+1)= s_2s_3..s_(k+1)(s_1-1) +s_2s_3..s_(k+1) \ge 2+s_1+s_2+......s_(k+1)$ so we are done now we can found that f(n) is increas when n is prime and equal n+1 abd this isn't prime so it will be less than n so the function will be degreas until many steps he will take the value 1 ,2 ,...8 so fundamental period is $1$ or $2$ depending on the starting numbers

cadaeibf wrote: moom wrote: cadaeibf wrote: Let $N$ be a composite number such that $N=p_1\cdots p_m$, with $p_i$ being prime. Then the function is basically $1+\sum_{i=1}^m p_i$. Since $m$ is $\geq 2$ and $p_i\geq 2$ for all $i$, this is bounded by $f(N)\leq 1+2(m-1)+\frac{N}{2^{m-1}}$ (note that $\frac{N}{2^m}\geq 2$ since $N\geq 2^m$); this is basically equivalent to maximizing $x_1+x_2+\cdots +x_m$ given $x_1\cdots x_m=N$ with $x_i\in [2,+\infty)$ (and $N\geq 2^m$). Furthermore, we have $1+2(m-1)+\frac{N}{2^{m-1}}<N-1$ for all $m>2$, because $N(1-\frac {1}{2^{m-1}})\geq 2^m(1-\frac{1}{2^{m-1}})=2^m-2$ By an easy induction, this is greater than $2m$ for all $m> 3$, implying the result. For $m=2$ the wanted inequality is equivalent to $3+\frac{N}{2}<N\iff N>6$. For $m=3$ the inequality is equivalent to $5+\frac N4<N-1\iff N>8$. Furthermore, if $N$ is prime, $f(N)=N+1$ is composite, and so $f(f(N))=f(N+1)<(N+1)-1=N$ if $N+1>8\iff N>7$. Therefore, for all $N>8$ one of $f(N)$ and $f(f(N))$ is less than $N$. Therefore, every orbit eventually goes to a number which is at most $8$. Now, since $7\rightarrow 8\rightarrow 6,2\rightarrow 3\rightarrow 4\rightarrow 5\rightarrow 6\rightarrow 6$, it follows that the unique cycle (apart from $f(1)=1$, which anyway has period $1$) is $f(6)=1+2+3=6$, which has a fundamental period of $1$. Also, how you found that $f(8)=6$ ?? $f(8)= 1+(2)(3)=7$ I'm sorry I had a mistake. By correcting it we only have the two cycles $6\rightarrow 6$ and $7\rightarrow 8$, so the fundamental period is $1$ or $2$ depending on the starting numbers, so I guess we can say overall that the period is $2$. not a problem , very nice solution