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A385130 The sum of divisors of n whose maximum exponent in their prime factorization is even.

%I A385130 #10 Jun 24 2025 06:15:13
%S A385130 1,1,1,5,1,1,1,5,10,1,1,17,1,1,1,21,1,28,1,25,1,1,1,17,26,1,10,33,1,1,
%T A385130 1,21,1,1,1,80,1,1,1,25,1,1,1,49,55,1,1,81,50,76,1,57,1,28,1,33,1,1,1,
%U A385130 97,1,1,73,85,1,1,1,73,1,1,1,80,1,1,101,81,1,1
%N A385130 The sum of divisors of n whose maximum exponent in their prime factorization is even.
%C A385130 The sum of terms in A368714 that divide n.
%C A385130 The number of these divisors is A385128(n).
%H A385130 Amiram Eldar, <a href="/A385130/b385130.txt">Table of n, a(n) for n = 1..10000</a>
%F A385130 a(n) = Sum_{d|n} (d * (1 - A051903(d) mod 2)).
%F A385130 a(n) = A000203(n) - A385131(n).
%F A385130 a(n) = Sum_{k=1..kmax(n)} (-1)^(k+1) * Product_{i=1..r} (p_i^(min(e_i, k-1) + 1)-1)/(p_i-1), for n >= 2; if n = Product_{i=1..r} p_i^e_i, r = omega(n) = A001221(n), then emax(n) = max(e_i) = A051903(n), and kmax(n) = emax(n) if emax(n) is odd, and emax(n)+1 otherwise.
%F A385130 Sum_{k=1..n} a(k) ~ c * zeta(2) * n^2 / 2, where c = Sum_{k>=2} (-1)^k * (1-1/zeta(k)) = 0.27591672059822700769...,
%t A385130 q[n_] := EvenQ[Max[FactorInteger[n][[;; , 2]]]]; q[1] = True; a[n_] := DivisorSum[n, # &, q[#] &]; Array[a, 100]		
%t A385130 (* second program: *)
%t A385130 a[n_] := Module[{f = FactorInteger[n], p, e, emax, kmax}, p = f[[;;,1]]; e = f[[;;,2]]; emax = Max[e]; kmax = emax + 1 - Mod[emax, 2]; Sum[(-1)^(k+1) * Product[(p[[i]]^(Min[e[[i]], k-1]+1)-1)/(p[[i]]-1), {i, 1, Length[e]}], {k, 1, kmax}]]; a[1] = 1; Array[a, 100]
%o A385130 (PARI) q(n) = if(n == 1, 1, !(vecmax(factor(n)[,2]) % 2));
%o A385130 a(n) = sumdiv(n, d, d*q(d));
%o A385130 (PARI) a(n) = if(n == 1, 1, my(f = factor(n), p = f[,1], e = f[,2], emax = vecmax(e), kmax = emax + 1 - emax % 2); sum(k = 1, kmax, (-1)^(k+1) * prod(i = 1, #e, (p[i]^(min(e[i], k-1)+1)-1)/(p[i]-1))));
%Y A385130 Cf. A000203, A001221, A013661, A051903, A368714, A383156, A385128, A385131.
%K A385130 nonn,easy
%O A385130 1,4
%A A385130 _Amiram Eldar_, Jun 24 2025