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.

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A065332 3-smooth numbers in their natural position, gaps filled with 0.

Original entry on oeis.org

1, 2, 3, 4, 0, 6, 0, 8, 9, 0, 0, 12, 0, 0, 0, 16, 0, 18, 0, 0, 0, 0, 0, 24, 0, 0, 27, 0, 0, 0, 0, 32, 0, 0, 0, 36, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 48, 0, 0, 0, 0, 0, 54, 0, 0, 0, 0, 0, 0, 0, 0, 0, 64, 0, 0, 0, 0, 0, 0, 0, 72, 0, 0, 0, 0, 0, 0, 0, 0, 81, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 96, 0, 0
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 29 2001

Keywords

Crossrefs

Programs

  • Mathematica
    smooth3Q[n_] := n == 2^IntegerExponent[n, 2]*3^IntegerExponent[n, 3];
    a[n_] := n Boole[smooth3Q[n]];
    Array[a, 100] (* Jean-François Alcover, Oct 17 2021 *)
  • PARI
    a(n) = if(n >> valuation(n, 2) == 3^valuation(n, 3), n, 0); \\ Amiram Eldar, Sep 16 2023

Formula

a(n) = if A065330(n) = 1 then n else 0.
a(n) = A065333(n) * n.
If a(k) > 0 then a(k) = (2^A065334(k)) * (3^A065335(k)).
From Amiram Eldar, Sep 16 2023: (Start)
Multiplicative with a(p^e) = p^e if p <= 3, and 0 otherwise.
Dirichlet g.f.: 6^s / ((2^s-2)*(3^s-3)).
Sum_{k=1..n} a(k) ~ (n/(log(2)*log(3))) * (log(n) + log(6)/2 - 1). (End)

A065335 3-exponents to represent 3-smooth numbers (A065332).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 29 2001

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := If[n/2^IntegerExponent[n, 2]/3^(e = IntegerExponent[n, 3]) == 1, e, 0]; Array[a, 100] (* Amiram Eldar, Feb 21 2021 *)

Formula

a(n) = A007949(n) * A065333(n).
a(n) = log_3(n / A006519(n)), where log_3 = A062153. For k > 0 with A065332(k) > 0: A065332(k) = (2^A065334(k)) * (3^a(k)).
Showing 1-2 of 2 results.