cp's OEIS Frontend

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

%I A289630 #25 Aug 01 2018 18:34:50
%S A289630 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,
%T A289630 30,16,9,33,34,15,9,19,20,15,15,41,18,29,33,15,46,47,15,15,50,51,15,
%U A289630 53,14,55,9,30,58,59,45,51,32,9,17,25,66,56,51,69,30,71
%N A289630 Number of modulo n residues among sums of two sixth powers.
%C A289630 This sequence appears to be multiplicative (verified through n = 10000).
%C A289630 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
%H A289630 Giovanni Resta, <a href="/A289630/b289630.txt">Table of n, a(n) for n = 1..10000</a>
%e A289630 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.:
%e A289630    (1^6 + 2^6) mod 5 = ( 1 + 64) mod 5 =  65 mod 5 = 0;
%e A289630    (0^6 + 1^6) mod 5 = ( 0 +  1) mod 5 =   1 mod 5 = 1;
%e A289630    (1^6 + 1^6) mod 5 = ( 1 +  1) mod 5 =   2 mod 5 = 2;
%e A289630    (2^6 + 2^6) mod 5 = (64 + 64) mod 5 = 128 mod 5 = 3;
%e A289630    (0^6 + 2^6) mod 5 = ( 0 + 64) mod 5 =  64 mod 5 = 4.
%e A289630 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.)
%o A289630 (PARI) a(n) = #Set(vector(n^2, i, ((i%n)^6 + (i\n)^6) % n)); \\ _Michel Marcus_, Jul 10 2017
%Y A289630 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).
%K A289630 nonn,mult
%O A289630 1,2
%A A289630 _Jon E. Schoenfield_, Jul 08 2017