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The number of unitary divisors of the exponentially odd numbers ( A268335).
+10
6
1, 2, 2, 2, 4, 2, 2, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 4, 4, 2, 2, 8, 2, 2, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 2, 4, 2, 4, 4, 4, 4, 4, 2, 2, 4, 4, 8, 2, 4, 8, 2, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 2, 8, 2, 4, 8, 4, 2, 2, 8, 4, 2, 8, 4, 4, 4, 8, 4
MATHEMATICA
f[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[AllTrue[e, OddQ], 2^Length[e], Nothing]]; f[1] = 1; Array[f, 150]
PROG
(PARI) lista(max) = for(k = 1, max, my(e = factor(k)[, 2], isexpodd = 1); for(i = 1, #e, if(!(e[i] % 2), isexpodd = 0; break)); if(isexpodd, print1(2^(#e), ", ")));
The sum of unitary divisors of the cubefree numbers ( A004709).
+10
5
1, 3, 4, 5, 6, 12, 8, 10, 18, 12, 20, 14, 24, 24, 18, 30, 20, 30, 32, 36, 24, 26, 42, 40, 30, 72, 32, 48, 54, 48, 50, 38, 60, 56, 42, 96, 44, 60, 60, 72, 48, 50, 78, 72, 70, 54, 72, 80, 90, 60, 120, 62, 96, 80, 84, 144, 68, 90, 96, 144, 72, 74, 114, 104, 100, 96
FORMULA
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(3)^2 * Product_{p prime} (1 + 1/p^2 - 2/p^3 + 1/p^4 - 1/p^5) = 1.665430860774244601005... .
The asymptotic mean of the unitary abundancy index of the cubefree numbers: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/ A004709(k) = c / zeta(3) = 1.38548421160152785073... .
MATHEMATICA
s[n_] := Module[{f = FactorInteger[n], e}, e = f[[;; , 2]]; If[AllTrue[e, # < 3 &], Times @@ (1 + Power @@@ f), Nothing]]; s[1] = 1; Array[s, 100]
PROG
(PARI) lista(max) = for(k = 1, max, my(f = factor(k), e = f[, 2], iscubefree = 1); for(i = 1, #e, if(e[i] > 2, iscubefree = 0; break)); if(iscubefree, print1(prod(i = 1, #e, 1 + f[i, 1]^e[i]), ", ")));
(Python)
from sympy.ntheory.factor_ import udivisor_sigma
from sympy import mobius, integer_nthroot
def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x, 3)[0]+1))
m, k = n, f(n)
while m != k:
m, k = k, f(k)
The number of unitary divisors of the exponentially 2^n-numbers ( A138302).
+10
5
1, 2, 2, 2, 2, 4, 2, 2, 4, 2, 4, 2, 4, 4, 2, 2, 4, 2, 4, 4, 4, 2, 2, 4, 4, 2, 8, 2, 4, 4, 4, 4, 2, 4, 4, 2, 8, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 8, 2, 4, 4, 8, 2, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 2, 8, 4, 4, 4, 2, 8, 4, 4, 4, 4, 4, 2, 4
COMMENTS
Also, the number of infinitary divisors of the terms of A138302, since A138302 is also the sequence of numbers whose sets of unitary divisors ( A077610) and infinitary divisors ( A077609) coincide.
MATHEMATICA
f[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[AllTrue[e, # == 2^IntegerExponent[#, 2] &], 2^Length[e], Nothing]]; f[1] = 1; Array[f, 150]
PROG
(PARI) lista(max) = for(k = 1, max, my(e = factor(k)[, 2], is = 1); for(i = 1, #e, if(e[i] >> valuation(e[i], 2) > 1, is = 0; break)); if(is, print1(2^#e, ", ")));
The sum of squarefree divisors of the cubefree numbers.
+10
1
1, 3, 4, 3, 6, 12, 8, 4, 18, 12, 12, 14, 24, 24, 18, 12, 20, 18, 32, 36, 24, 6, 42, 24, 30, 72, 32, 48, 54, 48, 12, 38, 60, 56, 42, 96, 44, 36, 24, 72, 48, 8, 18, 72, 42, 54, 72, 80, 90, 60, 72, 62, 96, 32, 84, 144, 68, 54, 96, 144, 72, 74, 114, 24, 60, 96, 168
COMMENTS
The number of squarefree divisors of the n-th cubefree number is A366536(n).
FORMULA
Sum_{j=1..n} a(j) ~ c * n^2, where c = zeta(3)^2/(2*zeta(5)) = 0.6967413068... .
In general, the formula holds for the sum of squarefree divisors of the k-free numbers with c = zeta(k)^2/(2*zeta(2*k-1))..., for k >= 2.
MATHEMATICA
f[p_, e_] := p + 1; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; cubefreeQ[n_] := Max[FactorInteger[n][[;; , 2]]] < 3; s /@ Select[Range[100], cubefreeQ]
(* or *)
f[p_, e_] := If[e > 2, 0, p + 1]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Array[s, 100], # > 0 &]
PROG
(PARI) lista(kmax) = {my(f, s, p, e); for(k = 1, kmax, f = factor(k); s = prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; if(e < 3, p + 1, 0)); if(s > 0, print1(s, ", "))); }
(Python)
from math import prod
from sympy import mobius, integer_nthroot, primefactors
def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x, 3)[0]+1))
m, k = n, f(n)
while m != k:
m, k = k, f(k)
return prod(p+1 for p in primefactors(m)) # Chai Wah Wu, Aug 12 2024
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