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 1-5 of 5 results.

A372380 The number of divisors of n that are numbers whose number of divisors is a power of 2.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Apr 29 2024

Keywords

Comments

First differs from A061389 and A322483 at n = 32.
First differs from A380922 at n = 128. - Vaclav Kotesovec, Apr 22 2025

Links

Crossrefs

Cf. A000005, A005117, A036537, A372379.
Cf. A061389, A322483, A380922.

Programs

  • Mathematica
    f[p_, e_] := Floor[Log2[e + 1]] + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> exponent(x+1)+1, factor(n)[, 2]));
  • Python
    from math import prod
from sympy import factorint
def A372380(n): return prod((e+1).bit_length() for e in factorint(n).values()) # Chai Wah Wu, Apr 30 2024

Formula

Multiplicative with a(p^e) = floor(log_2(e+1)) + 1.
a(n) = A000005(n) if and only if n is squarefree (A005117).

A372604 The maximal exponent in the prime factorization of the largest divisor of n whose number of divisors is a power of 2.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, May 07 2024

Keywords

Comments

First differs from A331273 at n = 32.
Differs from A368247 at n = 1, 128, 216, 256, 384, 432, 512, ... .
All the terms are of the form 2^k-1 (A000225).

Examples

			4 has 3 divisors, 1, 2 and 4. The number of divisors of 4 is 3, which is not a power of 2. The number of divisors of 2 is 2, which is a power of 2. Therefore, A372379(4) = 2 and a(4) = A051903(2) = 1.

Links

Crossrefs

Cf. A000225, A051903, A092323, A372379, A372466.
Cf. A331273, A368247.
Similar sequences: A007424, A368781, A372601, A372602, A372603.

Programs

  • Mathematica
    f[n_] := 2^Floor[Log2[n + 1]] - 1; a[n_] := f[Max[FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100]
  • PARI
    s(n) = 2^exponent(n+1) - 1;
a(n) = if(n>1, s(vecmax(factor(n)[,2])), 0);

Formula

a(n) = A051903(A372379(n)).
a(n) = A092323(A051903(n)+1).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{i>=1} 2^i * (1 - 1/zeta(2^(i+1)-1)) = 1.36955053734097783559... .

A282940 Largest non-infinitary divisor of A162643(n) having no non-infinitary divisors.

Original entry on oeis.org

2, 3, 6, 8, 6, 10, 5, 14, 8, 6, 22, 15, 24, 7, 10, 26, 30, 21, 8, 34, 24, 15, 38, 40, 27, 42, 30, 46, 24, 14, 33, 10, 54, 56, 58, 39, 11, 62, 42, 66, 70, 24, 21, 74, 30, 51, 78, 40, 54, 82, 13, 57, 86, 35, 88, 30, 94, 24, 14, 66, 40, 102, 69, 104, 106, 110, 56
Offset: 1

Views

Author

Vladimir Shevelev, Feb 25 2017

Keywords

Comments

Or largest term of A036537 dividing A162643(n) (cf. our comment in A036537).

Links

Crossrefs

Cf. A000005, A036537, A162643, A282900, A327839, A372379.

Programs

  • Mathematica
    f[p_, e_] := p^(2^Floor[Log2[e + 1]] - 1); s[1] = Nothing; s[n_] := Module[{v = Times @@ f @@@ FactorInteger[n]}, If[v == n, Nothing, v]]; Array[s, 300] (* Amiram Eldar, Apr 29 2024 *)

Formula

a(n) = A036537(m), where m = max{k: A036537(k)|A162643(n)}.
From Amiram Eldar, Apr 29 2024: (Start)
a(n) = A372379(A162643(n)).
Sum_{k=1..n} a(k) ~ ((c-d)/(1-d)^2) * n^2 / 2, where d = A327839 and c = 0.7907361848... is the constant in the asymptotic formula in A372379. (End)

Extensions

More terms from Peter J. C. Moses, Feb 25 2017

A372381 The number of divisors of the largest divisor of n whose number of divisors is a power of 2.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 8, 4, 4, 2, 8, 4, 4, 4
Offset: 1

Views

Author

Amiram Eldar, Apr 29 2024

Keywords

Comments

First differs from A286324 at n = 32, and from A331109 at n = 64.
Also, the number of infinitary divisors of the largest divisor of n whose number of divisors is a power of 2.

Links

Crossrefs

Cf. A000005, A004709, A036537, A037445, A372379, A372380.
Cf. A286324, A331109.

Programs

  • Mathematica
    f[p_, e_] := 2^Floor[Log2[e + 1]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> 2^exponent(x+1), factor(n)[, 2]));
  • Python
    from math import prod
from sympy import factorint
def A372381(n): return prod(1<<(e+1).bit_length()-1 for e in factorint(n).values()) # Chai Wah Wu, Apr 30 2024

Formula

Multiplicative with a(p^e) = 2^floor(log_2(e+1)).
a(n) = A000005(A372379(n)).
a(n) = A037445(A372379(n)).
a(n) = A000005(n) if and only if n is in A036537.
a(n) <= A372380(n), with equality if and only if n is cubefree (A004709).

A382788 The sum of divisors of n that are numbers whose number of divisors is a power of 2.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Apr 05 2025

Keywords

Comments

First differs from A033634 at n = 32.
The sum of the terms of A036537 that divide n.
The number of these divisors is A372380(n) and the largest of them is A372379(n).

Links

Crossrefs

Cf. A000523, A033634, A036537, A372379, A353900, A372380.

Programs

  • Mathematica
    f[p_, e_] := Sum[p^(2^k-1), {k, 0, Floor[Log2[e + 1]]}]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = my(f = factor(n), p = f[,1], e = f[,2]); prod(i = 1, #p, sum(k = 0, exponent(e[i]+1), p[i]^(2^k-1)));

Formula

Multiplicative with a(p^e) = Sum_{k = 0..floor(log_2(e+1))} p^(2^k-1).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} (1 + Sum_{k>=1} a(p^k)/p^(2*k)) = 1.13143029377358401678... .
Showing 1-5 of 5 results.

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