# Is there any other number that has similar properties as $21$?

It's my observation.

Let $$n=p_1×p_2×p_3×\dots×p_r$$ where $$p_i$$ are prime factors and $$f$$ and $$g$$ are the functions $$f(n)=1+2+\dots+n$$ And $$g(n)=p_1+p_2+\dots+p_r$$ If we put $$n=21$$ then $$g(f(21))=g(231)=21.$$ I checked it upto $$n=10000$$, I did not find another number with this property $$g(f(n))=n$$.

Can we prove that other such numbers do not exist?

• To clarify, is $g(12)=5$ or $=7$? – Hagen von Eitzen Mar 1 '19 at 7:22
• 12=3*2*2 then g(12)=2+2+3=7 – Pruthviraj Mar 1 '19 at 7:35
• @PruthvirajHajari Did you come up with this problem by yourself? If so, please state that in your question and include the program code used to run it. – TheSimpliFire Mar 1 '19 at 7:47
• I encourage you to accept my answer by clicking the tick you see beside it. – Parcly Taxel Mar 1 '19 at 13:37
• @PruthvirajHajari sure it is. – Parcly Taxel Mar 1 '19 at 14:01

$$\newcommand{sopfr}{\operatorname{sopfr}}$$ $$f(n)=\frac{n(n+1)}2$$ and $$g(n)=\sopfr(n)$$, the sum of prime factors of $$n$$ with repeats (OEIS A001414). We want $$n$$ such that $$g(f(n))=n$$ or $$\sopfr\left(\frac{n(n+1)}2\right)=n\tag1$$ which can be split into two cases due to the property $$\sopfr(ab)=\sopfr(a)+\sopfr(b)$$.
• If $$n$$ is even, then $$\sopfr\left(\frac n2\right)+\sopfr(n+1)=n$$. We know that $$\sopfr(n)\le n$$, so $$\sopfr\left(\frac n2\right)\le\frac n2$$ and consequently $$\sopfr(n+1)\ge\frac n2$$. Either $$n+1$$ is a prime, in which case the LHS of $$(1)$$ is greater than $$n$$ and so the equality cannot hold, or $$n+1$$ is odd composite and so has a least prime factor at least 3*, yielding $$\sopfr(n+1)\le3+\frac{n+1}3$$ and thus $$\frac n2\le3+\frac{n+1}3$$ which is only true for $$n\le20$$. Checking those $$n$$ reveals no solutions to $$(1)$$.
• If $$n$$ is odd, the reasoning is similar: $$\sopfr\left(\frac{n+1}2\right)+\sopfr(n)=n$$, where $$\sopfr\left(\frac{n+1}2\right)\le\frac{n+1}2$$ and so $$\sopfr(n)\ge\frac{n-1}2$$. Since $$n$$ is odd, either it is prime and the LHS of $$(1)$$ is greater than $$n$$, or it has a least prime factor at least 3* and $$\sopfr(n)\le3+\frac n3$$, giving $$\frac{n-1}2\le3+\frac n3$$ which only holds for $$n\le21$$. 21 is the solution to $$(1)$$ pointed out in the original question; we have just shown it is the only one.
*Technically we have to repeat the argument for other possible least prime factors $$k$$ of $$n$$ or $$n+1$$ – and the upper bound $$N_k$$ of the solution to the inequalities in $$n$$ increases accordingly, each 3 replaced with $$k$$. However, the least composite number with least prime factor $$k$$ is $$k^2$$, and this increases much faster than $$N_k$$ (which is $$\sim\frac k2$$). Indeed, $$5^2$$ already exceeds $$N_5$$ for both inequalities.
• To elaborate on the footnote: If $n$ is composite and $p$ its minimal prime factor, then $\operatorname{sopfr}(n)=p+\operatorname{sopfr}(n/p)\le p+\frac np$ and $x\mapsto x+\frac nx$ is a decreasing function on $(0,\sqrt n)$, hence $n$ odd and composite implies $\operatorname{sopfr}(n)\le 3+\frac n3$. – Hagen von Eitzen Mar 1 '19 at 15:47