cp's OEIS Frontend

A237719 Numbers n such that k(n) = (n(n+1)/2 mod n) = (antisigma(n) mod n) + (sigma(n) mod n).

%I A237719 #24 Sep 08 2022 08:46:06
%S A237719 1,2,6,12,18,20,24,28,30,40,42,54,56,66,70,78,80,88,100,102,104,112,
%T A237719 114,120,126,138,140,150,160,162,174,176,180,186,196,198,200,204,208,
%U A237719 220,222,224,228,234,240,246,258,260,272,276,282,294,304,306,308,318,320
%N A237719 Numbers n such that k(n) = (n(n+1)/2 mod n) = (antisigma(n) mod n) + (sigma(n) mod n).
%C A237719 Numbers n such that k(n) = A142150(n) = A229110(n) + A054024(n).
%C A237719 Numbers n such that k(n) = (A000217(n) mod n) = (A024816(n) mod n) + (A000203(n) mod n).
%C A237719 k(n) = 0 for odd n, k(n) = n/2 for even n.
%C A237719 If there are any odd multiply-perfect numbers, they are members of this sequence.
%C A237719 If there is no odd multiply-perfect number, then:
%C A237719 (1) the only odd number in this sequence is 1,
%C A237719 (2) corresponding sequence of numbers k(n): {0; a(n) / 2 for n > 1}.
%C A237719 Supersequence of A159907, A007691 and A000396.
%e A237719 12 is in the sequence because k(12) = (12*(12+1)/2) mod 12 = antisigma(12) mod 12 + sigma(12) mod 12; k(12) = 6 = 4 + 2 = n/2.
%o A237719 (Magma) [n: n in [1..320] | IsZero(n*(n+1)div 2 mod n - SumOfDivisors(n) mod n - (n*(n+1)div 2-SumOfDivisors(n)) mod n)]
%Y A237719 Cf. A000203, A000217, A024816, A054024, A142150, A229110.
%K A237719 nonn
%O A237719 1,2
%A A237719 _Jaroslav Krizek_, Mar 16 2014