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 1-3 of 3 results.

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

A056555 Smallest number k (k>0) such that n*k is a perfect 4th power.

Original entry on oeis.org

1, 8, 27, 4, 125, 216, 343, 2, 9, 1000, 1331, 108, 2197, 2744, 3375, 1, 4913, 72, 6859, 500, 9261, 10648, 12167, 54, 25, 17576, 3, 1372, 24389, 27000, 29791, 8, 35937, 39304, 42875, 36, 50653, 54872, 59319, 250, 68921, 74088, 79507, 5324, 1125
Offset: 1

Views

Author

Henry Bottomley, Jun 25 2000

Keywords

Examples

			a(64) = 4 because the smallest 4th power divisible by 64 is 256 and 64*4 = 256.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^Mod[4 - e, 4]; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 08 2020 *)
  • PARI
    a(n,f=factor(n))=f[,2]=-f[,2]%4; factorback(f) \\ Charles R Greathouse IV, Apr 24 2020

Formula

a(n) = A053167(n)/n = n^3/A000190(n)^4 = A056553(n)/A053165(n).
Multiplicative with a(p^e) = p^((4 - e) mod 4). - Amiram Eldar, Sep 08 2020
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(16)/(4*zeta(4))) * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^7 + 1/p^8) = 0.1537848996... . - Amiram Eldar, Oct 27 2022

A062379 n divided by largest 4th-power-free factor of 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, 4, 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, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Henry Bottomley, Jun 18 2001

Keywords

Crossrefs

Programs

Formula

a(n) = n/A058035(n).
Multiplicative with a(p^e) = p^max(e-3, 0). - Amiram Eldar, Sep 07 2020
Showing 1-3 of 3 results.