A115964 -id:A115964 - cp's OEIS frontend

A362854 The sum of the divisors of n that are both bi-unitary and exponential.

1, 2, 3, 4, 5, 6, 7, 10, 9, 10, 11, 12, 13, 14, 15, 18, 17, 18, 19, 20, 21, 22, 23, 30, 25, 26, 30, 28, 29, 30, 31, 34, 33, 34, 35, 36, 37, 38, 39, 50, 41, 42, 43, 44, 45, 46, 47, 54, 49, 50, 51, 52, 53, 60, 55, 70, 57, 58, 59, 60, 61, 62, 63, 70, 65, 66, 67, 68

Offset: 1

Amiram Eldar, May 05 2023

The number of these divisors is A362852(n).

The indices of records of a(n)/n are the primorials (A002110) cubed, i.e., 1 and the terms of A115964.

			a(8) = 10 since 8 has 2 divisors that are both bi-unitary and exponential, 2 and 8, and 2 + 8 = 10.

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

Cf. A002110, A004709, A051377, A082020, A115964, A188999, A222266, A322791, A361810, A362852.

f[p_, e_] := DivisorSum[e, p^# &] - If[OddQ[e], 0, p^(e/2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

s(p, e) = sumdiv(e, d, p^d*(2*d != e));
a(n) = {my(f = factor(n)); prod(i = 1, #f~, s(f[i, 1], f[i, 2]));}

Multiplicative with a(p^e) = Sum_{d|e} p^d if e is odd, and (Sum_{d|e} p^d) - p^(e/2) if e is even.
a(n) >= n, with equality if and only if n is cubefree (A004709).
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, where c = (1/2) * Product_{p prime} ((1 - 1/p)*(1 + Sum_{e>=1} Sum_{d|e, d != e/2}, p^(d-2*e))) = 0.5124353304539905... .

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

Amiram Eldar, Jun 03 2025

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.

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

Cf. A002110, A005117, A051377, A072587, A082020, A115964, A268335, A331737, A368979, A374459, A384559.

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

f[p_, e_] := DivisorSum[e, p^# &, OddQ[#] &]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = {my(f = factor(n)); prod(i = 1, #f~, sumdiv(f[i,2], d, (d % 2) * f[i,1]^d));}

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... .

cp's OEIS Frontend

A362854 The sum of the divisors of n that are both bi-unitary and exponential.

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