A006446 Numbers k such that floor(sqrt(k)) divides k.
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
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).
Links
- Harry J. Smith, Table of n, a(n) for n = 1..1000
- Benoit Cloitre, Some divisibility sequences.
- Curtis N. Cooper and Robert E. Kennedy, Chebyshev's inequality and natural density, Amer. Math. Monthly, Vol. 96, No. 2 (1989), pp. 118-124.
- S. W. Golomb, Problem E2491, Amer. Math. Monthly, 82 (1975), 854-855.
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
- Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).
Programs
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Haskell
a006446 n = a006446_list !! (n-1) a006446_list = filter (\x -> x `mod` a000196 x == 0) [1..] -- Reinhard Zumkeller, Mar 31 2011
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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
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Mathematica
Select[ Range[ 500 ], Mod[ #, Floor[ Sqrt[ # ]//N ] ]==0& ]
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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
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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
Comments