A390372 The sum of the bi-unitary divisors of n that are exponentially odd numbers (A268335).

1, 3, 4, 1, 6, 12, 8, 11, 1, 18, 12, 4, 14, 24, 24, 11, 18, 3, 20, 6, 32, 36, 24, 44, 1, 42, 31, 8, 30, 72, 32, 43, 48, 54, 48, 1, 38, 60, 56, 66, 42, 96, 44, 12, 6, 72, 48, 44, 1, 3, 72, 14, 54, 93, 72, 88, 80, 90, 60, 24, 62, 96, 8, 35, 84, 144, 68, 18, 96, 144

Offset

1,2

Links

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

Formula

Multiplicative with a(p^e) = 1 + (p^(2*floor((e-1)/2)+2) - 1)*p / (p^2-1) - [e == 2 (mod 4)] * p^(e/2), where [] is the Iverson bracket.

a(n) = 1 if and only if n is the square of a squarefree number (A062503).

a(n) = sigma(n) = A000203(n) if and only if n is squarefree (A005117).

a(n) = A033634(n) if and only if all the exponents in the prime factorization of n are in A042965.

Dirichlet g.f.: zeta(s) * zeta(2*s-2) * zeta(4*s-2) * Product_{p prime} (1 + 1/p^(s-1) - 1/p^(2*s-2) - 1/p^(2*s-1) + 1/p^(3*s-1) + 1/p^(4*s-3) - 1/p^(4*s-2) - 2/p^(5*s-3) + 1/p^(6*s-4)).

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(2) * zeta(6) * Product_{p prime} (1 - 1/p^2 + 2/p^5 - 3/p^6 + 1/p^7) = 1.05515925301614983992... .

Mathematica

f[p_, e_] := 1 + (p^(2*Floor[(e-1)/2]+2) - 1)*p / (p^2-1) - If[Mod[e, 4] == 2, p^(e/2), 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

Prog

(PARI) a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; 1 + (p^(2*((e-1)\2)+2) - 1)*p / (p^2-1) - if(e % 4 == 2, p^(e/2))); }

Crossrefs

Cf. A000203, A005117, A033634, A042965, A062503, A188999, A268335, A286324, A358346, A384558, A384559, A390370, A390371.

Cf. A013661, A013664.

Sequence in context: A388994 A390370 A368471 * A351569 A163762 A347084

Adjacent sequences: A390369 A390370 A390371 * A390373 A390374 A390375

Keyword

nonn,easy,mult

Author

Amiram Eldar, Nov 03 2025

Status

approved