A289630 Number of modulo n residues among sums of two sixth powers.

1, 2, 3, 3, 5, 6, 3, 3, 3, 10, 11, 9, 5, 6, 15, 5, 17, 6, 10, 15, 9, 22, 23, 9, 25, 10, 7, 9, 29, 30, 16, 9, 33, 34, 15, 9, 19, 20, 15, 15, 41, 18, 29, 33, 15, 46, 47, 15, 15, 50, 51, 15, 53, 14, 55, 9, 30, 58, 59, 45, 51, 32, 9, 17, 25, 66, 56, 51, 69, 30, 71

Offset: 1

Author

Jon E. Schoenfield, Jul 08 2017

Keywords

• nonn
• mult

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

Examples

			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.)
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A057760, A155918 (gives number of modulo n residues among sums of two squares), A289559 (Number of modulo n residues among sums of two fourth powers).

Programs

PARI

a(n) = #Set(vector(n^2, i, ((i%n)^6 + (i\n)^6) % n)); \\ Michel Marcus, Jul 10 2017