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.

Showing 1-4 of 4 results.

A366076 The number of prime factors of the largest divisor of n that is a cubefull number (A036966), counted with multiplicity.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 3, 0, 3, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 4, 4, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Amiram Eldar, Sep 28 2023

Keywords

Comments

The sum of exponents larger than 2 in the prime factorization of n.
The number of distinct prime factors of the largest divisor of n that is a cubefull number is A295659(n).

Links

Crossrefs

Cf. A001222, A004709, A036966, A046100, A085541, A152441, A295659, A360540, A366077.
Similar sequence: A275812 (number of prime factors of the powerful part).

Programs

  • Mathematica
    f[p_, e_] := If[e < 3, 0, e]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecsum(apply(x -> if(x < 3, 0, x), factor(n)[, 2]));

Formula

a(n) = A001222(A360540(n)).
a(n) = A001222(n) - A366077(n).
Additive with a(p^e) = 0 if e <= 2, and a(p^e) = e for e >= 3.
a(n) >= 0, with equality if and only if n is cubefree (A004709).
a(n) <= A001222(n), with equality if and only if n is cubefull (A036966).
a(n) >= 3*A295659(n), with equality if and only if n is a biquadratefree number (A046100).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} (2/p^3 + 1/(p^2*(p-1))) = 2 * A085541 + A152441 = 0.67043452760761670220... .

A366078 The number of distinct prime factors of the cubefree part of n (A360539).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 2, 2, 0, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 0, 2, 1, 3, 1, 0, 2, 2, 2, 2, 1, 2, 2, 1, 1, 3, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 3, 1, 2, 2, 0, 2, 3, 1, 2, 2, 3, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 0, 2, 1, 3, 2, 2, 2
Offset: 1

Views

Author

Amiram Eldar, Sep 28 2023

Keywords

Comments

The number of exponents smaller than 3 in the prime factorization of n.
The number of prime factors of the cubefree part of n (A360539), counted with multiplicity is A366077(n).

Links

Crossrefs

Cf. A001221, A004709, A005117, A036966, A077761, A085541, A295659, A360539, A366077.

Programs

  • Mathematica
    f[p_, e_] := If[e < 3, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecsum(apply(x -> if(x < 3, 1, 0), factor(n)[, 2]));

Formula

a(n) = A001221(A360539(n)).
a(n) = A001221(n) - A295659(n).
Additive with a(p^e) = 1 if e <= 2, and a(p^e) = 0 for e >= 3.
a(n) >= 0, with equality if and only if n is cubefull (A036966).
a(n) <= A001221(n), with equality if and only if n is cubefree (A004709).
a(n) <= A366077(n), with equality if and only if n is squarefree (A005117).
Sum_{k=1..m} a(k) = n * (log(log(n)) + B - C) + O(n/log(n)), where B is Mertens's constant (A077761) and C = Sum_{p prime} 1/p^3 = 0.174762... (A085541).

A366147 The number of divisors of the cubefree part of n (A360539).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 1, 3, 4, 2, 6, 2, 4, 4, 1, 2, 6, 2, 6, 4, 4, 2, 2, 3, 4, 1, 6, 2, 8, 2, 1, 4, 4, 4, 9, 2, 4, 4, 2, 2, 8, 2, 6, 6, 4, 2, 2, 3, 6, 4, 6, 2, 2, 4, 2, 4, 4, 2, 12, 2, 4, 6, 1, 4, 8, 2, 6, 4, 8, 2, 3, 2, 4, 6, 6, 4, 8, 2, 2, 1, 4, 2, 12, 4, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Oct 01 2023

Keywords

Links

Crossrefs

Cf. A000005, A004709, A036966, A056671, A360539, A366077, A366078, A366145, A366148.

Programs

  • Mathematica
    f[p_, e_] := If[e < 3, e+1, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> if(x < 3, x+1, 1), factor(n)[, 2]));

Formula

a(n) = A000005(A360539(n)).
a(n) = A000005(n)/A366145(n).
Multiplicative with a(p^e) = e+1 if e <= 2, and 1 otherwise.
a(n) >= 1, with equality if and only if n is cubefull (A036966).
a(n) <= A000005(n), with equality if and only if n is cubefree (A004709).
Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^s + 1/p^(2*s) - 2/p^(3*s)).
From Vaclav Kotesovec, Oct 01 2023: (Start)
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 3/p^(3*s) + 2/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 3/p^(3*s) + 2/p^(4*s)).
Sum_{k=1..n} a(k) ~ f(1) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 3/p^3 + 2/p^4) = 0.66219033176371496870504912254207846719824904470940603905284774924086...,
f'(1) = f(1) * Sum_{p prime} (9*p - 8) * log(p) / (p^4 - 3*p + 2) = f(1) * 1.04316863044761953555286128194165251303791613504188623828521117799260...
and gamma is the Euler-Mascheroni constant A001620. (End)

A366148 The sum of divisors of the cubefree part of n (A360539).

Original entry on oeis.org

1, 3, 4, 7, 6, 12, 8, 1, 13, 18, 12, 28, 14, 24, 24, 1, 18, 39, 20, 42, 32, 36, 24, 4, 31, 42, 1, 56, 30, 72, 32, 1, 48, 54, 48, 91, 38, 60, 56, 6, 42, 96, 44, 84, 78, 72, 48, 4, 57, 93, 72, 98, 54, 3, 72, 8, 80, 90, 60, 168, 62, 96, 104, 1, 84, 144, 68, 126, 96
Offset: 1

Views

Author

Amiram Eldar, Oct 01 2023

Keywords

Links

Crossrefs

Cf. A000203, A004709, A036966, A092261, A360539, A366077, A366078, A366146, A366147.

Programs

  • Mathematica
    f[p_, e_] := If[e < 3, (p^(e+1)-1)/(p-1), 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2] < 3, (f[i,1]^(f[i,2]+1)-1)/(f[i,1]-1), 1));}

Formula

a(n) = A000203(A360539(n)).
a(n) = A000203(n)/A366146(n).
Multiplicative with a(p^e) = (p^(e+1)-1)/(p-1) if e <= 2, and 1 otherwise.
a(n) >= 1, with equality if and only if n is cubefull (A036966).
a(n) <= A000203(n), with equality if and only if n is cubefree (A004709).
Dirichlet g.f.: zeta(s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-2) - 1/p^(3*s-2) - 1/p^(3*s-1)). CHECK
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} (p/(p+1) + 1/p - 1/p^4) = 0.63884633697952950095... .
Showing 1-4 of 4 results.

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