A385128 The number of divisors of n whose maximum exponent in their prime factorization is even.

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 3, 2, 1, 2, 3, 1, 1, 1, 3, 1, 1, 1, 6, 1, 1, 1, 3, 1, 1, 1, 3, 3, 1, 1, 5, 2, 3, 1, 3, 1, 3, 1, 3, 1, 1, 1, 5, 1, 1, 3, 4, 1, 1, 1, 3, 1, 1, 1, 6, 1, 1, 3, 3, 1, 1, 1, 5, 3, 1, 1, 5, 1, 1, 1

Offset

1,4

Comments

The number of terms in A368714 that divide n.

The sum of these divisors is A385130(n).

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

a(n) = Sum_{d|n} (1 - A051903(d) mod 2).

a(n) = A000005(n) - A385129(n).

a(n) = Sum_{k=1..kmax(n)} (-1)^(k+1) * Product_{i=1..r} (min(e_i, k-1) + 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.

Sum_{k=1..n} a(k) ~ c1 * n * (log(n) + 2*gamma - 1) + c2 * n, where gamma is Euler's constant (A001620), c1 = Sum_{k>=2} (-1)^k * (1-1/zeta(k)) = 0.27591672059822700769..., and c2 = 1 + Sum_{k>=2} (-1)^k * k * zeta'(k)/zeta(k)^2 = 0.56812633046434345687... .

Mathematica

q[n_] := EvenQ[Max[FactorInteger[n][[;; , 2]]]]; q[1] = True; a[n_] := DivisorSum[n, 1 &, q[#] &]; Array[a, 100]
(* second program: *)
a[n_] := Module[{e = FactorInteger[n][[;; , 2]], emax, kmax}, emax = Max[e]; kmax = emax + 1 - Mod[emax, 2]; Sum[(-1)^(k+1) * Product[Min[e[[i]], k-1] + 1, {i, 1, Length[e]}], {k, 1, kmax}]]; Array[a, 100]

Prog

(PARI) q(n) = if(n == 1, 1, !(vecmax(factor(n)[, 2]) % 2));
a(n) = sumdiv(n, d, q(d));
(PARI) a(n) = if(n == 1, 1, my(e = factor(n)[, 2], emax = vecmax(e), kmax = emax + 1 - emax %2); sum(k = 1, kmax, (-1)^(k+1) * prod(i = 1, #e, min(e[i], k-1)+1)));

Crossrefs

Cf. A000005, A001221, A001620, A051903, A368714, A383156, A385129, A385130.

Sequence in context: A379018 A265991 A353841 * A322871 A322870 A170979

Adjacent sequences: A385125 A385126 A385127 * A385129 A385130 A385131

Keyword

nonn,easy

Author

Amiram Eldar, Jun 24 2025

Status

approved