A200219 Number of solutions of the equation x^n + (x+1)^n = (x+2)^n (mod n) for x = 0..n-1.

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

Offset: 1

Author

Michel Lagneau, Nov 14 2011

Keywords

• nonn

Comments

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

a(n) = 1 if n prime.

Examples

			a(6) = 2 because:
for x = 3,  3^6 + 4^6 == 1(mod 6) and 5^6 == 1(mod 6).
for x = 5,  5^6 + 6^6 == 1 (mod 6) and (7)^6 == 1 (mod 6).
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A195637, A200046.

Programs

Maple

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:

Mathematica

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