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Sum of proper divisors of n of the form 4k+1.
4

%I #15 Nov 27 2023 03:04:33

%S 0,1,1,1,1,1,1,1,1,6,1,1,1,1,6,1,1,10,1,6,1,1,1,1,6,14,10,1,1,6,1,1,1,

%T 18,6,10,1,1,14,6,1,22,1,1,15,1,1,1,1,31,18,14,1,10,6,1,1,30,1,6,1,1,

%U 31,1,19,34,1,18,1,6,1,10,1,38,31,1,1,14,1,6,10,42,1,22,23,1,30,1,1,60,14,1,1,1,6,1,1,50,43,31,1,18,1,14,27

%N Sum of proper divisors of n of the form 4k+1.

%H Antti Karttunen, <a href="/A293901/b293901.txt">Table of n, a(n) for n = 1..16384</a>

%H <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>.

%F a(n) = Sum_{d|n, d<n} [1 == d mod 4]*d.

%F a(n) = A091570(n) - A293903(n).

%F G.f.: Sum_{k>=1} (4*k-3) * x^(8*k-6) / (1 - x^(4*k-3)). - _Ilya Gutkovskiy_, Apr 14 2021

%F Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/48 - 1/8 = 0.0806167... . - _Amiram Eldar_, Nov 27 2023

%t a[n_] := DivisorSum[n, # &, # < n && Mod[#, 4] == 1 &]; Array[a, 100] (* _Amiram Eldar_, Nov 27 2023 *)

%o (PARI) A293901(n) = sumdiv(n,d,(d<n)*(1==(d%4))*d);

%Y Cf. A050449, A091570, A293451, A293903.

%K nonn,easy

%O 1,10

%A _Antti Karttunen_, Oct 19 2017