Iora wrote: For all $n>1$ let $f(n)$ be the sum of the smallest factor of $n$ that is not 1 and $n$ . The computer prints $f(2),f(3),f(4),...$ with order:$4,6,6,...$ ( Because $f(2)=2+2=4,f(3)=3+3=6,f(4)=4+2=6$ etc.). In this infinite sequence, how many times will be $ 2015$ and $ 2016$ written? (Explain your answer) why f(2)=2+2?

CROWmatician wrote: Iora wrote: For all $n>1$ let $f(n)$ be the sum of the smallest factor of $n$ that is not 1 and $n$ . The computer prints $f(2),f(3),f(4),...$ with order:$4,6,6,...$ ( Because $f(2)=2+2=4,f(3)=3+3=6,f(4)=4+2=6$ etc.). In this infinite sequence, how many times will be $ 2015$ and $ 2016$ written? (Explain your answer) why f(2)=2+2? 2's smallest factor expect 1 is 2, and n is 2, 2+2

Hint

The smallest divisor of a number is prime so 2015 cant be written 2016 has 3,7,2 prime divisors .So 2013+3=2009+7=2014+2=2016 so 2016 will be written 3 Times

Telman wrote: The smallest divisor of a number is prime so 2015 cant be written 2016 has 3,7,2 prime divisors .So 2013+3=2009+7=2014+2=2016 so 2016 will be written 3 Times I think your reasoning for $2015$ is wrong, you have to prove that $f(n)$ is never odd ( 2015 not being a smallest prime doesn't prove anything at all) ( look for hint above) And for 2016, your reasoning is wrong too, we have $$ d(n) +n= 2016$$where $$d(n)$$is smallest prime, you say that $$2,3,7+n=2016$$, but $d(n)$ and $2016$ are not connected Indeed your answers can be right, but I think that your reasonings and solutions are wrong , try to solve the problem with the hint above

Iora wrote: Telman wrote: The smallest divisor of a number is prime so 2015 cant be written 2016 has 3,7,2 prime divisors .So 2013+3=2009+7=2014+2=2016 so 2016 will be written 3 Times I think your reasoning for $2015$ is wrong, you have to prove that $f(n)$ is never odd ( 2015 not being a smallest prime doesn't prove anything at all) ( look for hint above) And for 2016, your reasoning is wrong too, we have $$ d(n) +n= 2016$$where $$d(n)$$is smallest prime, you say that $$2,3,7+n=2016$$, but $d(n)$ and $2016$ are not connected Indeed your answers can be right, but I think that your reasonings and solutions are wrong , try to solve the problem with the hint above i think people could understand what i wrote thats because i wrote shorter than you)))

Telman you are wrong

KhayalAliyev wrote: Telman you are wrong Why Khayal Aliyev?İf İ have any mistake you can say me

i think 2015 is written 2 times bcs f(2015) = 2015 + 5 and there is no other 2015 in that sequence.

the solution Telman proposed is right, you just have to understand it a bit deeper.

if n is even, $f(n) = n + 2$, so it outputs an evem number. if n is odd, it's smallest divisor is a prime $p$ other than 2, so $f(n)$ is again, even. From this, $f(n)$ is always even, so 2015 cannot be write as 2015 is an odd number. If $n = 2014$, $f(n) = 2016$, so we have one soltuion. Because $2016 = 2^5 * 3^2 * 7$, $n = 2016 - 3 = 2013$ and $n = 2016 - 7 = 2009$ are also solutions. We can conclude that 2016 will be written 3 time on the board.

Telman wrote: The smallest divisor of a number is prime so 2015 cant be written 2016 has 3,7,2 prime divisors .So 2013+3=2009+7=2014+2=2016 so 2016 will be written 3 Times Your soltuion is correct, but the first part is badly written. The smallest divisor being a prime matters if $n$ is odd, as the divisor will also be odd.

If $n$ is even, then the number will be $n+2$, which is even. If $n$ is odd, then the number will be odd+odd=even. Hence, f(n) is always even, so $2015$ is impossible, $2016=2^5 \cdot 3^2 \cdot 7$, so $n=2014, n=2013, n=2009$, work and are the only solutions, so we have $\boxed{3}.$

Let m be the smallest positive divisor of n that is not equal to 1 Claim:f(n) can't compute an odd number Proof:If n is even,then f(n)=n+2. If n is odd,then$$f(n)=n+m\equiv 1+1\equiv0\mod 2$$So,f(n) can't be equal to 2015 Claim:f(n) is divisible by m Proof:$$f(n)=n+m=m\cdot n/m + m=m(n/m+1)$$Prime divisors of 2016 is 2,3 and 7 f(2014)=2014+2=2016 f(2013)=2013+3=2016 f(2009)=2009+7=2016

Let $d(n)$ be the smallest divisor of $n$ here we have two cases; 1.If $n$ is odd then $d(n)$ is also odd so we get that $f(n)=n+d(n)$ since $odd+odd=even$ we cant ever get an odd number. 2.Same for the even numbers.$f(n)=n+2$ which is always an even number. Here we can say that $\boxed{f(n) \neq 2015}$ Lets look for $2016$ $2016=n+d(n)$ since $RHS$ is $0(\bmod d(n)$ we can say that $d(n)|2016$. Prime divisors of $2016$ are $2,3$ and $7$. 3.1 $2016=n+7 \rightarrow n=2009$ 3.2 $2016=n+2 \rightarrow n=2014$ 3.3$2016=n+3\rightarrow n=2013$ (Number of 2016)=3 (Number of 2015)=0