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.

Previous Showing 21-23 of 23 results.

A365333 The number of exponentially odd coreful divisors of the largest square dividing n.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Sep 01 2023

Keywords

Comments

First differs from A043289, A053164, A063775, A203640 and A295658 at n = 64.
The number of squares dividing the largest exponentially odd divisor of n is A325837(n).
The sum of the exponentially odd divisors of the largest square dividing n is A365334(n). [corrected, Sep 08 2023]
The number of exponentially odd divisors of the largest square dividing n is the same as the number of squares dividing n, A046951(n). - Amiram Eldar, Sep 08 2023

Links

Crossrefs

Cf. A008833, A046100, A046951, A325837, A365334.
Cf. A043289, A053164, A063775, A203640, A295658.

Programs

  • Mathematica
    f[p_, e_] := Max[1, Floor[e/2]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> max(1, x\2), factor(n)[, 2]));

Formula

a(n) = A325837(A008833(n)).
a(n) = 1 if and only if n is a biquadratefree number (A046100).
Multiplicative with a(p^e) = max(1, floor(e/2)).
Dirichlet g.f.: zeta(s) * zeta(4*s) * zeta(6*s) / zeta(12*s).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 15015/(1382*Pi^2) = 1.100823... .

Extensions

Name corrected by Amiram Eldar, Sep 08 2023

A355265 Bicubeful numbers.

Original entry on oeis.org

64, 128, 192, 256, 320, 384, 448, 512, 576, 640, 704, 729, 768, 832, 896, 960, 1024, 1088, 1152, 1216, 1280, 1344, 1408, 1458, 1472, 1536, 1600, 1664, 1728, 1792, 1856, 1920, 1984, 2048, 2112, 2176, 2187, 2240, 2304, 2368, 2432, 2496, 2560, 2624, 2688, 2752
Offset: 1

Views

Author

Peter Luschny, Jul 12 2022

Keywords

Comments

Let lp(n, e) denote the largest positive integer b such that b^e divides n. For example for e = 1, 2, 3, 4 the sequences (lp(n, e), n >= 1) are A000027, A000188, A053150, and A053164. Let rad(n) = A007947(n) be the squarefree kernel of n. k is in this sequence if lp(n, 3) does not divide rad(n). The case e = 1 gives A013929, and the case e = 2 is A046101.
The asymptotic density of this sequence is 1 - 1/zeta(6) = 1 - 945/Pi^6 = 0.017047... . - Amiram Eldar, Jul 13 2022

Examples

			n = 512 = 2^9, rad(n) = 2, lp(n, 3) = 8 since n/8^3 = 1. But 8 does not divide 2.
n = 704 = 2^6*11, rad(n) = 22, lp(n, 3) = 4 since n/4^3 = 11. But 4 does not divide 22.

Links

Crossrefs

Cf. A007947, A000188, A053150, A053164, A013929, A046101 (biquadrateful).
Cf. A013664, A343359.

Programs

  • Maple
    with(NumberTheory):
isBicubeful := n -> irem(Radical(n), LargestNthPower(n, 3)) <> 0:
select(isBicubeful, [`$`(1..2752)]);
  • Mathematica
    bicubQ[n_] := AnyTrue[FactorInteger[n][[;; , 2]], # > 5 &]; Select[Range[3000], bicubQ] (* Amiram Eldar, Jul 13 2022 *)
  • Python
    from itertools import count, islice
from sympy import factorint
def A355265_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:any(map(lambda m:m>5,factorint(n).values())),count(max(startvalue,1)))
A355265_list = list(islice(A355265_gen(),30)) # Chai Wah Wu, Jul 12 2022

Formula

A number k is bicubeful iff it is divisible by the 6th power of an integer > 1.

A365490 The number of divisors of the largest 4th power dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Sep 05 2023

Keywords

Comments

The number of divisors of the 4th root of the largest 4th power dividing n, A053164(n), is A063775(n).

Links

Crossrefs

Cf. A000005, A000583, A046100, A008835, A053164, A063775.

Programs

  • Mathematica
    f[p_, e_] := 4*Floor[e/4] + 1; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> 4*(x\4) + 1, factor(n)[, 2]));
  • Python
    from math import prod
from sympy import factorint
def A365490(n): return prod(e&-4|1 for e in factorint(n).values()) # Chai Wah Wu, Aug 08 2024

Formula

a(n) = A000005(A008835(n)).
Multiplicative with a(p^e) = 4*floor(e/4) + 1.
a(n) = 1 if and only if n is a biquadratefree number (A046100).
a(n) <= A000005(n) with equality if and only if n is a fourth power (A000583).
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 + 3/p^(4*s)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 + 3/p^4) = 1.3414590511076... . In general, the asymptotic mean of the number of divisors of the largest k-th power dividing n is zeta(k) * Product_{p prime} (1 + (k-1)/p^k).
Previous Showing 21-23 of 23 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.