# Can a prime have arbitary many representations as a sum of two perfect powers?

Let $p$ be a prime number and $f(p)$ be the number of representations $$p=a+b$$ with perfect powers $0<a< b$

For example, $13$ has the only representation $$4+9=13$$ hence $f(13)=1$ $$41=9+32=16+25$$ has two representations, hence $f(41)=2$

The smallest primes that satisfy $$f(p)=1,2,3,4,5$$ are $$13,41,449,4481,93241$$ respective. I did not find this sequence in OEIS.

Is $f(p)$ a bounded function ? If not, is $f(p)$ surjective on the natural numbers ?

• There might be something at oeis.org/A206606 – Barry Cipra May 4 '18 at 21:51
• An observation, if $p=a_1^m+b_1^n$ then $\gcd(n,m)=2^k$, $k$ may be $0$ – rtybase May 4 '18 at 22:01
• $f(4297609)=6$ , not sure whether this is the smallest example. – Peter May 5 '18 at 16:03
• And $f(445\ 341\ 529)=7$. Also not sure whether it is the smallest. – Peter May 5 '18 at 19:08
• The smallest values for $8$ and $9$ representations are: $$f(4\:190\:216\:689)=8;$$ $$f(25\:140\:740\:257)=9.$$ – Oleg567 May 9 '18 at 20:49