A242483 Numbers n such that A242481(n) = ((n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n) / n = 2.

4, 8, 10, 14, 16, 22, 26, 32, 34, 36, 38, 44, 46, 48, 50, 52, 58, 60, 62, 64, 68, 72, 74, 76, 82, 84, 86, 90, 92, 94, 96, 98, 106, 108, 110, 116, 118, 122, 124, 128, 130, 132, 134, 136, 142, 144, 146, 148, 152, 154, 156, 158, 164, 166, 168, 170, 172, 178, 182

Offset: 1

Author

Jaroslav Krizek, May 16 2014

Keywords

• nonn

Comments

Numbers n such that A242480(n) = (n*(n+1)/2) 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)) = 2n. 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 = 2.

Conjecture: with number 1 complement of A242482.

Examples

			8 is in sequence because [(8*(8+1)/2) mod 8 + sigma(8) mod 8 + antisigma(8) mod 8] / 8 = (36 mod 8 + 15 mod 8 + 21 mod 8) / 8 = (4 + 7 + 5 ) / 8 = 2.
		

Links

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

Crossrefs

Cf. A242480, A242481, A242482, A242484, A242485, A242486.

Programs

Magma

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