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.

A056553 Smallest 4th-power divisible by n divided by largest 4th-power which divides n.

Original entry on oeis.org

1, 16, 81, 16, 625, 1296, 2401, 16, 81, 10000, 14641, 1296, 28561, 38416, 50625, 1, 83521, 1296, 130321, 10000, 194481, 234256, 279841, 1296, 625, 456976, 81, 38416, 707281, 810000, 923521, 16, 1185921, 1336336, 1500625, 1296, 1874161, 2085136
Offset: 1

Views

Author

Henry Bottomley, Jun 25 2000

Keywords

Examples

			a(64) = 16 because smallest 4th power divisible by 64 is 256 and largest 4th power which divides 64 is 16 and 256/16 = 16.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^If[Divisible[e, 4], 0, 1]; a[n_] := (Times @@ (f @@@ FactorInteger[ n]))^4; Array[a, 100] (* Amiram Eldar, Aug 29 2019*)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%4, f[i,1], 1))^4; } \\ Amiram Eldar, Oct 27 2022

Formula

a(n) = A053167(n)/A008835(n) = A056556(n)*A053165(n) = A056554(n)^4.
From Amiram Eldar, Oct 27 2022: (Start)
Multiplicative with a(p^e) = 1 if e is divisible by 4, and a(p^e) = p^4 otherwise.
Sum_{k=1..n} a(k) ~ c * n^5, where c = (zeta(20)/(5*zeta(4))) * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^7 + 1/p^8) = 0.123026157003... . (End)

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.