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

1,2

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

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).

Sequence in context: A050976 A053447 A185026 * A023160 A085312 A046530

Adjacent sequences: A289627 A289628 A289629 * A289631 A289632 A289633

Keyword

nonn,mult

Author

Jon E. Schoenfield, Jul 08 2017

Status

approved