Let n be a positive integer and let s(n) be the digit's sum of n. Is there a prime number of the form $n^n+s(n)^{s(n)}$ ? after the trivial one $1^1+1^1$ ?, I've checked n up to $2500$ without finding a prime anymore (!). I found that numbers of such form often divisible by small primes below $100$, and occasionally they have large least prime factor, but after that they become divisible by small primes again (very often!). Is there a prime number of such form after the trivial one 2 ?
Asked
Active
Viewed 172 times
1
-
Interesting question, but I'm sure you meant to put this on MATH.stackexchange.com... – rasher May 08 '17 at 04:51
-
If s(n)=n+1, dp you mean $n^n+(n+1)^{n+1}$? – marty cohen May 08 '17 at 05:48
-
@martycohen For what positive integer $n$ can the digit sum of $n$ exceed $n$? – Erick Wong May 08 '17 at 06:06
-
This problem is very weird. I cannot find by computation a single prime in any odd-numbered base. Base 10 I cannot find and 16 is quite stubborn, too. – law-of-fives May 08 '17 at 06:55
-
1I found no prime for $1<n\le 5700$ – Peter May 10 '17 at 18:06
-
I don't see why there should be no other solutions. Are there infinitely many primes of the form $n^n+m^m$? It wouldn't be surprising. They're probably just way too large to find. – Bart Michels May 12 '17 at 22:29
-
Looking at $n^{s(n)}+s(n)^n$ might be fun. You get at least one more prime. (@barto, primes of the form $n^m+m^n$ seem to be plentiful, early on at least.) – Barry Cipra May 12 '17 at 22:38
-
@BarryCipra In this case, the numbers $n$ leading to a prime upto $n=1000$ are $0,1,10,100$ – Peter May 14 '17 at 12:26
1 Answers
0
I'm just stating a few trivial facts:
If $n$ is divisible by $3$ then clearly the number is not prime as it's divisible by 3 and greater than $3$.
If $n$ and $s_n$ are both even then also it's not prime.
For a general proof, I believe a mathematical proof is way very difficult and wouldn't be elementary.
Arpan1729
- 3,414
-
2In place of your second point, you might say that if $n$ and $s_n$ have the same parity, then it's not prime (except for $n=1$). – Barry Cipra May 12 '17 at 22:22