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.

A006446 Numbers k such that floor(sqrt(k)) divides k.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 12, 15, 16, 20, 24, 25, 30, 35, 36, 42, 48, 49, 56, 63, 64, 72, 80, 81, 90, 99, 100, 110, 120, 121, 132, 143, 144, 156, 168, 169, 182, 195, 196, 210, 224, 225, 240, 255, 256, 272, 288, 289, 306, 323, 324, 342, 360, 361, 380, 399, 400, 420
Offset: 1

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Keywords

Comments

Numbers of the form k^2, k*(k+1), or k*(k+2). Nonsquare elements of this sequence are given by A035106. - Max Alekseyev, Nov 27 2006
Union of A000290, A002378, and A005563. - Fred Daniel Kline, Feb 06 2016
The asymptotic density of this sequence is 0 (Cooper and Kennedy, 1989). - Amiram Eldar, Jul 10 2020

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 21.
  • Jeffrey Shallit, personal communication.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a006446 n = a006446_list !! (n-1)
    a006446_list = filter (\x -> x `mod` a000196 x == 0) [1..]
    -- Reinhard Zumkeller, Mar 31 2011
  • Maple
    A006446:=(-1-z-z**2+z**3)/(z**2+z+1)**2/(z-1)**3; # conjectured by Simon Plouffe in his 1992 dissertation
    A006446:=n->`if`(type(n/floor(sqrt(n)), integer), n, NULL); seq(A006446(n), n=1..100); # Wesley Ivan Hurt, Feb 11 2014
  • Mathematica
    Select[ Range[ 500 ], Mod[ #, Floor[ Sqrt[ # ]//N ] ]==0& ]
  • PARI
    { n=0; for (m=1, 10^9, if (m%floor(sqrt(m)) == 0, write("b006446.txt", n++, " ", m); if (n==1000, return)) ) } \\ Harry J. Smith, Feb 12 2010
    
  • PARI
    a(n)=my(k=n--\3+1);k*(k+n%3) \\ Charles R Greathouse IV, Jul 07 2011
    

Formula

For k>=1 a(3*k-2) = k^2, a(3*k-1) = k*(k+1) and a(3*k) = k*(k+2). - Benoit Cloitre, Jan 14 2012
a(n) mod A000196(a(n)) = 0. - Reinhard Zumkeller, Apr 12 2013
a(n) = floor((n+1)/3)*(floor(n/3) + 1) + floor((n+2)/3). - Ridouane Oudra, Nov 21 2020
a(n) = a(n-1) + 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7) for n > 7. - Chai Wah Wu, Apr 05 2021
Sum_{n>=1} 1/a(n) = 7/4 + Pi^2/6. - Amiram Eldar, Sep 24 2022