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a(n) = floor(3*n/7).
18

%I #36 Dec 30 2023 23:52:34

%S 0,0,0,1,1,2,2,3,3,3,4,4,5,5,6,6,6,7,7,8,8,9,9,9,10,10,11,11,12,12,12,

%T 13,13,14,14,15,15,15,16,16,17,17,18,18,18,19,19,20,20,21,21,21,22,22,

%U 23,23,24,24,24,25,25,26,26,27,27,27,28,28,29,29,30,30,30,31,31,32,32

%N a(n) = floor(3*n/7).

%C The cyclic pattern (and numerator of the gf) is computed using Euclid's algorithm for GCD.

%C This sequence relates to 3/7 = 0.42857142... (essentially given by A020806). It differs from the Beatty sequence A308358 for sqrt(3)/4 = 0.43301270... = A120011.

%D N. Dershowitz and E. M. Reingold, Calendrical Calculations, Cambridge University Press, 1997.

%D R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, NY, 1994.

%H G. C. Greubel, <a href="/A057357/b057357.txt">Table of n, a(n) for n = 0..5000</a>

%H N. Dershowitz and E. M. Reingold, <a href="http://emr.cs.iit.edu/home/reingold/calendar-book/first-edition/">Calendrical Calculations Web Site</a>.

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,1,-1).

%H <a href="/index/Be#Beatty">Index to sequences related to Beatty sequences</a>.

%F G.f.: (1+x^2+x^4)*x^3/((1-x)*(1-x^7)) - Bruce Corrigan (scentman(AT)myfamily.com), Jul 03 2002

%F for all m>=0 a(7m)=0 mod 3; a(7m+1)=0 mod 3; a(7m+2)= 0 mod 3; a(7m+3) = 1 mod 3; a(5m+4) = 1 mod 3; a(7m+5) = 2 mod 3; a(7m+6) = 2 mod 3 - Bruce Corrigan (scentman(AT)myfamily.com), Jul 03 2002

%F Sum_{n>=3} (-1)^(n+1)/a(n) = log(2)/3 (A193535). - _Amiram Eldar_, Sep 30 2022

%t Floor[(3*Range[0,80])/7] (* _Harvey P. Dale_, Aug 22 2013 *)

%o (PARI) a(n)=3*n\7 \\ _Charles R Greathouse IV_, Sep 02 2015

%o (Magma) [Floor(3*n/7): n in [0..50]]; // _G. C. Greubel_, Nov 02 2017

%Y Floors of other ratios: A004526, A002264, A002265, A004523, A057353, A057354, A057355, A057356, A057357, A057358, A057359, A057360, A057361, A057362, A057363, A057364, A057365, A057366, A057367.

%Y Cf. A020806, A193535, A308358.

%K nonn,easy

%O 0,6

%A _Mitch Harris_