A242483 - OEIS

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

%S 4,8,10,14,16,22,26,32,34,36,38,44,46,48,50,52,58,60,62,64,68,72,74,

%T 76,82,84,86,90,92,94,96,98,106,108,110,116,118,122,124,128,130,132,

%U 134,136,142,144,146,148,152,154,156,158,164,166,168,170,172,178,182

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

%C 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.

%C Conjecture: with number 1 complement of A242482.

%H Jaroslav Krizek, <a href="/A242483/b242483.txt">Table of n, a(n) for n = 1..5000</a>

%e 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.

%o (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]

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

%K nonn

%O 1,1

%A _Jaroslav Krizek_, May 16 2014