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 1-7 of 7 results.

A062378 n divided by largest cubefree factor of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 2
Offset: 1

Views

Author

Henry Bottomley, Jun 18 2001

Keywords

Comments

Numerator of n/rad(n)^2, where rad is the squarefree kernel of n (A007947), denominator: A055231. - Reinhard Zumkeller, Dec 10 2002

Crossrefs

Cf. A000189, A000578, A007948, A008834, A019555, A048798, A050985, A053149, A053150, A056551, A056552. See A003557 for squares and A062379 for 4th powers.
Differs from A073753 for the first time at n=90, where a(90) = 1, while A073753(90) = 3.

Programs

Formula

a(n) = n / A007948(n).
a(n) = A003557(A003557(n)). - Antti Karttunen, Nov 28 2017
Multiplicative with a(p^e) = p^max(e-2, 0). - Amiram Eldar, Sep 07 2020
Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 - 1/p^(s-1) + 1/p^s - 1/p^(2*s-1) + 1/p^(2*s)). - Amiram Eldar, Dec 07 2023

A056192 a(n) = n divided by its characteristic cube divisor A056191.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 1, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 3, 25, 26, 1, 28, 29, 30, 31, 4, 33, 34, 35, 36, 37, 38, 39, 5, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 2, 55, 7, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 9, 73, 74, 75
Offset: 1

Views

Author

Labos Elemer, Aug 02 2000

Keywords

Comments

Different from A056552: e.g. a(16) = 16, while A056552(16) = 2.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[EvenQ[e], p^e, If[e == 1, p, p^(e - 3)]]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 06 2020 *)

Formula

a(n) = n/A055229(n)^3 = n/g^3=n/ggg and n=(LL)*(ggg)*f=L^2*g^3*f=LL*a(n)^3*f, so n=L^2*(g*3)*f, where L=A000188(n)/A055229(n), f=A055231(n), g=A055231(n).
Multiplicative with a(p^e)=p^e for even e, a(p)=p, a(p^e)=p^(e-3) for odd e>1. - Vladeta Jovovic, Apr 30 2002
Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^2/12) * Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4 + 1/p^6 - 1/p^7) = 0.4462648652... . - Amiram Eldar, Nov 13 2022

A056551 Smallest cube divisible by n divided by largest cube which divides n.

Original entry on oeis.org

1, 8, 27, 8, 125, 216, 343, 1, 27, 1000, 1331, 216, 2197, 2744, 3375, 8, 4913, 216, 6859, 1000, 9261, 10648, 12167, 27, 125, 17576, 1, 2744, 24389, 27000, 29791, 8, 35937, 39304, 42875, 216, 50653, 54872, 59319, 125, 68921, 74088, 79507, 10648
Offset: 1

Views

Author

Henry Bottomley, Jun 25 2000

Keywords

Examples

			a(16) = 8 since smallest cube divisible by 16 is 64 and smallest cube which divides 16 is 8 and 64/8 = 8.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^If[Divisible[e, 3], 0, 1]; a[n_] := (Times @@ (f @@@ FactorInteger[ n]))^3; Array[a, 100] (* Amiram Eldar, Aug 29 2019*)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%3, f[i,1], 1))^3; } \\ Amiram Eldar, Oct 28 2022

Formula

a(n) = A053149(n)/A008834(n) = A048798(n)*A050985(n) = A056552(n)^3.
From Amiram Eldar, Oct 28 2022: (Start)
Multiplicative with a(p^e) = 1 if e is divisible by 3, and a(p^e) = p^3 otherwise.
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(12)/(4*zeta(3))) * Product_{p prime} (1 - 1/p^2 + 1/p^3) = A013670 * A330596 / (4*A002117) = 0.1557163105... . (End)
Dirichlet g.f.: zeta(3*s) * Product_{p prime} (1 + 1/p^(s-3) + 1/p^(2*s-3)). - Amiram Eldar, Sep 16 2023

A056554 Powerfree kernel of 4th-powerfree part of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 1, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 3, 7, 10, 51, 26, 53, 6, 55, 14, 57, 58, 59, 30, 61, 62, 21, 2, 65, 66, 67, 34, 69, 70, 71, 6, 73, 74, 15, 38
Offset: 1

Views

Author

Henry Bottomley, Jun 25 2000

Keywords

Examples

			a(64) = 2 because 4th-power-free part of 64 is 4 and power-free kernel of 4 is 2.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] :=  p^If[Divisible[e, 4], 0, 1]; a[n_] := Times @@ (f @@@ FactorInteger[ n]); Array[a, 100] (* Amiram Eldar, Aug 29 2019 *)
  • PARI
    a(n) = my(f=factor(n)); for (k=1, #f~, if (frac(f[k,2]/4), f[k,2] = 1, f[k,2] = 0)); factorback(f); \\ Michel Marcus, Feb 28 2019

Formula

a(n) = A007947(A053165(n)) = A053166(A053165(n)) = n/(A053164(n)*A000190(n)) = A053166(n)/A053164(n) = A056553(n)^(1/4).
If n = Product_{j} Pj^Ej then a(n) = Product_{j} Pj^Fj, where Fj = 0 if Ej is 0 or a multiple of 4 and Fj = 1 otherwise.
Multiplicative with a(p^e) = p^(if 4|e, then 0, else 1). - Mitch Harris, Apr 19 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(8)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6) = 0.3513111135... . - Amiram Eldar, Oct 27 2022

A055491 Smallest square divisible by n divided by largest square which divides n.

Original entry on oeis.org

1, 4, 9, 1, 25, 36, 49, 4, 1, 100, 121, 9, 169, 196, 225, 1, 289, 4, 361, 25, 441, 484, 529, 36, 1, 676, 9, 49, 841, 900, 961, 4, 1089, 1156, 1225, 1, 1369, 1444, 1521, 100, 1681, 1764, 1849, 121, 25, 2116, 2209, 9, 1, 4, 2601, 169, 2809, 36, 3025, 196, 3249, 3364
Offset: 1

Views

Author

Henry Bottomley, Jun 28 2000

Keywords

Examples

			a(12) = 36/4 = 9.
		

Crossrefs

Programs

  • Haskell
    a055491 = (^ 2) . a007913  -- Reinhard Zumkeller, Jul 23 2014
    
  • Mathematica
    With[{sqs=Range[100]^2},Table[SelectFirst[sqs,Divisible[#,n]&]/ SelectFirst[ Reverse[sqs],Divisible[n,#]&],{n,60}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Feb 18 2018 *)
    f[p_, e_] := p^(2 * Mod[e, 2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(2*(f[i,2]%2)));} \\ Amiram Eldar, Oct 27 2022

Formula

If n is written as Product(Pj^Ej) then a(n) = Product(Pj^(2*(Ej mod 2))).
a(n) = A053143(n)/A008833(n) = A007913(n)^2 = (A019554(n)/A000188(n))^2 = A000290(n)/A008833(n)^2.
Sum_{k=1..n} a(k) ~ c * n^3, where c = (zeta(6)/(3*zeta(2))) = 2*Pi^4/945 = 0.206156... . - Amiram Eldar, Oct 27 2022
Dirichlet g.f.: zeta(s-2) * zeta(2*s) / zeta(2*s-4). - Amiram Eldar, Sep 16 2023

A056194 Characteristic cube divisor of n!: a(n) = A056191(n!).

Original entry on oeis.org

1, 1, 1, 8, 8, 1, 1, 8, 8, 1, 1, 27, 27, 216, 1000, 1000, 1000, 125, 125, 1, 9261, 74088, 74088, 343, 343, 2744, 74088, 216, 216, 125, 125, 1000, 35937000, 4492125, 12326391, 12326391, 12326391, 98611128, 8024024008, 125375375125
Offset: 1

Views

Author

Labos Elemer, Aug 02 2000

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e] && e > 1, p^3, 1]; a[n_] := Times @@ f @@@ FactorInteger[n!]; Array[a, 40] (* Amiram Eldar, Sep 06 2020 *)

Formula

a(n) = A056191(A000142(n)). - Amiram Eldar, Sep 06 2020

A383717 Dirichlet g.f.: Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-1)).

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 0, 3, 10, 11, 6, 13, 14, 15, 0, 17, 6, 19, 10, 21, 22, 23, 0, 5, 26, 0, 14, 29, 30, 31, 0, 33, 34, 35, 6, 37, 38, 39, 0, 41, 42, 43, 22, 15, 46, 47, 0, 7, 10, 51, 26, 53, 0, 55, 0, 57, 58, 59, 30, 61, 62, 21, 0, 65, 66, 67, 34, 69, 70, 71, 0, 73
Offset: 1

Views

Author

Vaclav Kotesovec, May 07 2025

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e < 3, p, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 07 2025 *)
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 + X*p + X^2*p))[n], ", "))

Formula

Sum_{k=1..n} a(k) ~ c * n^2/2, where c = Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4) = 0.684286924186862318141968725791218083472312736723163777284618226290055...
Multiplicative with a(p^e) = p is e <= 2, and 0 otherwise. - Amiram Eldar, May 07 2025
Showing 1-7 of 7 results.