cp's OEIS Frontend

A331546 a(n) = |{k^(k+1)+(k+1)^k (mod prime(n)): k = 0..prime(n)-1}|.

%I A331546 #21 Jan 22 2020 20:15:40
%S A331546 1,3,5,6,7,9,11,15,15,17,19,24,25,28,28,34,39,38,41,50,43,48,55,56,60,
%T A331546 62,70,68,70,76,76,83,83,78,88,106,95,98,105,110,117,106,114,126,114,
%U A331546 129,138,139,143,148,146,141,152,159,164,160,170,171,176,182,184,191,192,190,193,194,216,215,215,217
%N A331546 a(n) = |{k^(k+1)+(k+1)^k (mod prime(n)): k = 0..prime(n)-1}|.
%C A331546 Conjecture: |{k^(k+1)+(k+1)^k (mod p): k = 0..p-1}| = (1-1/e)*p + O(p^(1/2)), where p denotes a prime.
%H A331546 Robert Israel, <a href="/A331546/b331546.txt">Table of n, a(n) for n = 1..5000</a>
%H A331546 Zhi-Wei Sun, <a href="https://mathoverflow.net/questions/350784">Is it true that |{k^(k+1)+(k+1)^k (mod p): k = 0..p-1}| = (1-e^(-1))*p + O(sqrt(p))?</a> Question 350784 at MathOverflow, Jan. 20, 2020.
%e A331546 a(2) = 3 since {0^1+1^0, 1^2+2^1, 2^3+3^2} = {1, 3, 17} is a complete system of residues modulo the second prime 3.
%p A331546 f:= proc(p) local S,k;
%p A331546  nops({seq(k &^ (k+1) + (k+1) &^ k mod p, k=0..p-1)})
%p A331546 end proc:
%p A331546 seq(f(ithprime(i)),i=1..100); # _Robert Israel_, Jan 22 2020
%t A331546 p[n_]:=p[n]=Prime[n];
%t A331546 a[n_]:=a[n]=Length[Union[Table[Mod[PowerMod[k,k+1,p[n]]+PowerMod[k+1,k,p[n]],p[n]],{k,0,p[n]-1}]]];
%t A331546 Table[a[n],{n,1,70}]
%Y A331546 Cf. A000040, A001113, A051442.
%K A331546 nonn
%O A331546 1,2
%A A331546 _Zhi-Wei Sun_, Jan 19 2020