OFFSET
1,1
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.
LINKS
Jaroslav Krizek, Table of n, a(n) for n = 1..5000
EXAMPLE
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.
PROG
(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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, May 16 2014
STATUS
approved