Define $$f(n):=\sum_{j=1}^n j^{j+1}=1+2^3+3^4+\cdots +n^{n+1}$$
Does for every integer $p\ge 2$ exist a positive integer $n$ with $p\mid f(n)$ ? In other words, can every positive integer be a divisor of $f(n)$ ?
I used this simple sieve in PARI/GP :
? x=[2..10^4];n=1;s=1;while(length(x)>0,n=n+1;s=s+n^(n+1);y=select(a->Mod(s,a)==0,x);if(length(y)>0,x=setminus(x,y);print(n," ",length(x))))
to determine upto which $\ n\ $ we must continue to cover all divisors upto $\ p=10^4\ $
Last line in the output is
69696 0
Hence every positive integer $p\le 10^4$ can be a divisor and we need not go beyond $\ n=69696\ $. The toughest divisor is $\ 8549\ $