A242482 Numbers n such that A242481(n) = ((n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n) / n = 1.

2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 33, 35, 37, 39, 40, 41, 42, 43, 45, 47, 49, 51, 53, 54, 55, 56, 57, 59, 61, 63, 65, 66, 67, 69, 70, 71, 73, 75, 77, 78, 79, 80, 81, 83, 85, 87, 88, 89, 91, 93, 95, 97, 99

Offset: 1

Author

Jaroslav Krizek, May 16 2014

Keywords

• nonn

Comments

Numbers n such that A242480(n) = (1/2*n*(n+1)) mod n + sigma(n) mod n + antisigma(n) mod n = (A142150(n) + A054024(n) + A229110(n)) = ((A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n)) = n. Numbers n such that A242481(n) = (A142150(n) + A054024(n) + A229110(n)) / n = ((A000217(n) mod n) + (A000203(n) mod n) + (A024816(n) mod n)) / n = 1.

Conjecture: with number 1 complement of A242483.

Supersequence of primes (A000040).

If there is no odd multiply-perfect number, then:

(1) a(n) = union of odd numbers >= 3 and even numbers from A239719.

(2) a(n) = supersequence of odd numbers (A005408).

Examples

			6 is in sequence because [(6*(6+1)/2) mod 6 + sigma(6) mod 6 + antisigma(6) mod 6] / 6 = (21 mod 6 + 12 mod 6 + 9 mod 6) / 6 = (3 + 0 + 3 ) / 6 = 1.
		

Links

• Jaroslav Krizek, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A242480, A242481, A242483, A242484, A242485, A242486.

Programs

Magma

[n: n in [1..1000] | n eq ((n*(n+1)div 2 mod n + SumOfDivisors(n) mod n + (n*(n+1)div 2-SumOfDivisors(n)) mod n))]