A200219 - OEIS

%I #12 Mar 30 2012 18:36:00

%S 1,1,1,2,1,2,1,4,3,2,1,4,1,2,0,8,1,6,1,4,1,2,1,8,0,2,9,2,1,4,1,16,0,2,

%T 0,12,1,2,0,8,1,4,1,2,3,2,1,16,7,10,2,2,1,18,0,8,0,2,1,8,1,2,3,32,2,4,

%U 1,4,0,2,1,24,1,2,0,4,6,4,1,16,27,2,1,8

%N Number of solutions of the equation x^n + (x+1)^n = (x+2)^n  (mod n) for x = 0..n-1.

%C a(n) = 0 for n = 15, 25, 33, 35, 39, 55, 57,… (see A200046).

%C a(n) = 1 if n prime.

%H Michel Lagneau, <a href="/A200219/b200219.txt">Table of n, a(n) for n = 1..1000</a>

%e a(6) = 2 because:

%e for x = 3,  3^6 + 4^6 == 1(mod 6) and 5^6 == 1(mod 6).

%e for x = 5,  5^6 + 6^6 == 1 (mod 6) and (7)^6 == 1 (mod 6).

%p for n from 1 to 100 do:ii:=0:for x from 0 to n-1 do:if x^n+(x+1)^n -(x+2)^n mod n=0 then ii:=ii+1:else fi:od: printf(`%d, `,ii):od:

%t Array[Function[n,Count[Array[Mod[#^n+(#+1)^n-(#+2)^n,n]&,n,0],0]],84]

%Y Cf. A195637, A200046.

%K nonn

%O 1,4

%A _Michel Lagneau_, Nov 14 2011