cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A384558 The sum of the exponential divisors of n that are exponentially odd numbers (A268335).

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 10, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 30, 5, 26, 30, 14, 29, 30, 31, 34, 33, 34, 35, 6, 37, 38, 39, 50, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, 51, 26, 53, 60, 55, 70, 57, 58, 59, 30, 61, 62, 21, 10, 65, 66, 67, 34, 69
Offset: 1

Views

Author

Amiram Eldar, Jun 03 2025

Keywords

Comments

First differs from A384559 at n = 512: a(512) = 522, while A384559(512) = 514.
The number of these divisors is A368979(n), and the largest of them is A331737(n).
The indices of records of a(n)/n are the primorial numbers (A002110) cubed, i.e., 1 and the terms of A115964.

Crossrefs

Programs

  • Maple
    A384558:=proc(n)
        local a, pe,p,e,af,d;
        a := 1;
        for pe in ifactors(n)[2] do
            p := op(1,pe) ;
            e := op(2,pe) ;
            af := 0 ;
            for d in numtheory[divisors](e) do
                if type(d,'odd') then
                    af := af+p^d ;
                end if;
            end do:
            a := a*af ;
        end do;
        a
    end proc:
    seq(A384558(n), n=1..100); # R. J. Mathar, Jun 04 2025
  • Mathematica
    f[p_, e_] := DivisorSum[e, p^# &, OddQ[#] &]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, sumdiv(f[i,2], d, (d % 2) * f[i,1]^d));}

Formula

Multiplicative with a(p^e) = Sum_{d|e, d odd} p^d.
a(n) = n if and only if n is squarefree (A005117).
a(n) < n if and only if n is in A072587.
a(n) > n if and only if n is in A374459.
limsup_{n->oo} a(n)/n = Product_{p prime} (1 + 1/p^2) = 15/Pi^2 (A082020).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} (1 + 1/(p*(p^2-1)) - 1/(p^2-1) + (1-1/p) * Sum_{k>=1} p^(2*k+1)/(p^(4*k+2)-1)) = 0.80824764393216997768... .