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Numbers n such that A242481(n) = ((n*(n+1)/2) mod n + sigma(n) mod n + antisigma(n) mod n) / n = 1.
7

%I #6 Sep 08 2022 08:46:08

%S 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,

%T 37,39,40,41,42,43,45,47,49,51,53,54,55,56,57,59,61,63,65,66,67,69,70,

%U 71,73,75,77,78,79,80,81,83,85,87,88,89,91,93,95,97,99

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

%C 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.

%C Conjecture: with number 1 complement of A242483.

%C Supersequence of primes (A000040).

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

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

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

%H Jaroslav Krizek, <a href="/A242482/b242482.txt">Table of n, a(n) for n = 1..5000</a>

%e 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.

%o (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))]

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

%K nonn

%O 1,1

%A _Jaroslav Krizek_, May 16 2014