%I #23 Apr 12 2024 03:27:37
%S 1,1,2,3,3,4,4,6,6,7,7,6,10,10,11,11,9,11,13,15,16,14,16,15,15,17,17,
%T 22,22,22,20,18,20,24,24,25,27,27,27,26,28,26,30,30,29,31,31,37,35,35,
%U 35,31,35,35,40,40,38,40,40,41,41,41,43,47,47,46,42,44,46,50,48,46,52,52,52,54,52,55,55,53,55,53,59,58,56,58
%N a(n) = Sum_{i=1..n} d(8*i+1) - Sum_{i=1..n} d(2*i+1), where d(n) = A000005(n) is the number of divisors of n.
%C Cimadevilla proved that a(n) >= 0. That is surprising because d(8*i+1) - d(2*i+1) < 0 for i = 12, 17, 22, 24, 31, 32, 40, 42, 45, 49, 52, 57, 66, 67, 71, 72, 77, 80, 82, 84, 85, ...
%H Amiram Eldar, <a href="/A230476/b230476.txt">Table of n, a(n) for n = 1..10000</a>
%H Jorge Luis Cimadevilla Villacorta, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.120.09.832">Certain inequalities associated with the divisor function</a>, Amer. Math. Monthly, 120 (2013), 832-837. See inequalities (1.5).
%F a(n) = Sum_{i=1..n} (d(8*i+1) - d(2*i+1)) = A230293(n) + A230294(n).
%F a(n) = log(2) * n + O(n^(1/3)*log(n)). - _Amiram Eldar_, Apr 12 2024
%e The divisors of 8*1 + 1 = 9 are 1, 3, 9 and those of 2*1 + 1 = 3 are 1, 3, so a(1) = d(9) - d(3) = 3 - 2 = 1.
%t Table[Sum[ DivisorSigma[0, 8 i + 1] - DivisorSigma[0, 2 i + 1], {i, n}], {n, 100}]
%o (PARI) a(n) = sum(i=1, n, numdiv(8*i+1) - numdiv(2*i+1)); \\ _Michel Marcus_, Jun 19 2015
%Y Cf. A000005, A002162, A230290, A230291, A230292, A230293, A230295, A230296.
%K nonn
%O 1,3
%A _Jonathan Sondow_, Oct 20 2013