OFFSET
1,2
COMMENTS
This sequence appears to be multiplicative (verified through n = 10000).
This sequence is multiplicative. In general, by the Chinese remainder theorem, the number of distinct residues modulo n among the values of any multivariate polynomial with integer coefficients will be multiplicative. - Andrew Howroyd, Aug 01 2018
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..10000
EXAMPLE
a(5) = 5 because (j^6 + k^6) mod 5, where j and k are integers, can take on all 5 values 0..4; e.g.:
(1^6 + 2^6) mod 5 = ( 1 + 64) mod 5 = 65 mod 5 = 0;
(0^6 + 1^6) mod 5 = ( 0 + 1) mod 5 = 1 mod 5 = 1;
(1^6 + 1^6) mod 5 = ( 1 + 1) mod 5 = 2 mod 5 = 2;
(2^6 + 2^6) mod 5 = (64 + 64) mod 5 = 128 mod 5 = 3;
(0^6 + 2^6) mod 5 = ( 0 + 64) mod 5 = 64 mod 5 = 4.
a(7) = 3 because (j^6 + k^6) mod 7 can take on only the three values 0, 1, and 2. (This is because j^6 mod 7 = 0 for all j divisible by 7, 1 otherwise.)
PROG
(PARI) a(n) = #Set(vector(n^2, i, ((i%n)^6 + (i\n)^6) % n)); \\ Michel Marcus, Jul 10 2017
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
Jon E. Schoenfield, Jul 08 2017
STATUS
approved