A057760 -id:A057760 - cp's OEIS frontend

A057761 Least nonnegative cube root of 2 mod n for n in A057760.

0, 0, 2, 3, 2, 8, 7, 8, 8, 18, 16, 3, 26, 8, 4, 29, 8, 5, 20, 16, 21, 28, 8, 18, 18, 26, 38, 4, 62, 62, 49, 53, 46, 50, 8, 20, 26, 16, 20, 68, 26, 8, 18, 6, 57, 18, 81, 108, 38, 73, 5, 53, 32, 20, 54, 98, 62, 68, 120, 113, 70, 128, 38, 62, 71, 128, 50

Offset: 1

N. J. A. Sloane, Nov 01 2000

nonn

T. D. Noe, Table of n, a(n) for n=1..1000

Corrected by T. D. Noe, Apr 19 2007 [The errors were caused by the faulty Maple command "mroot"]

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

Original entry on oeis.org

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

Jon E. Schoenfield, Jul 08 2017

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

			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.)

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

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).

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

A259732 Numbers n > 1 that divide ((p-1)/2)^3 + 2 for some odd prime p.

Original entry on oeis.org

2, 3, 5, 10, 11, 17, 22, 23, 29, 31, 34, 41, 43, 46, 47, 53, 58, 59, 62, 71, 82, 83, 86, 89, 94, 101, 106, 107, 109, 113, 118, 121, 127, 131, 137, 142, 149, 157, 166, 167, 173, 178, 179, 187, 191, 197, 202

Offset: 1

Miguel Angel Velilla Mula, Jul 04 2015

nonn

n = 3,5,10 works only once, for p=3 (3-1)/2=1, then 1^3 + 2 = 3 and for p=5 (5-1)/2=2, then 2^3+2 = 10.
This sequence is a subset of A057760, where all elements that are multiples of 3 and 5 are excluded, except the three above (3,5,10).
"Mirror sequence" of this one, when n divides ((p+1)/2)^3 - 2, p = prime, produces a sequence very close to this one, the only differences being 10 (excluded), 25 (included for p=5 (p+1)/2=3 then 3^3-2 = 25) and 6 (included for p=3 (p+1)/2=2 then 2^3-2 = 6).
Analyzing ((p-1)/2)^3 + 2 = (p^3 - 3(p(p-1)-5))/8, every composite x (mod 3) trying to divide this one will fail.
To prove 5 can't divide ((p-1)/2)^3 + 2 = (p^3 - 3p^2 + 3p + 15)/8 we use the last digit of p, which can be 1,3,7 or 9. This leads the last digit of the formula to be (1,9,7 or 3) + 15, so it cannot be divided by 5, unless the last digit of p is 5. This happens just for the only prime divisible by 5, i.e., 5 itself, which occurs only once.
A179871 looks very similar to this sequence.

Cf. A057760, A040028.

Select[Rest[Union[Flatten[Divisors/@Rest[Table[((p-1)/2)^3+2,{p,Prime[Range[2000]]}]]]]],#<250&] (* Harvey P. Dale, Jun 01 2025 *)

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A057761 Least nonnegative cube root of 2 mod n for n in A057760.

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

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A259732 Numbers n > 1 that divide ((p-1)/2)^3 + 2 for some odd prime p.

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