login
a(n) = floor(2*n/5).
23

%I #58 Sep 30 2022 07:47:25

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

%T 12,12,13,13,14,14,14,15,15,16,16,16,17,17,18,18,18,19,19,20,20,20,21,

%U 21,22,22,22,23,23,24,24,24,25,25,26,26,26,27,27,28,28,28,29,29,30,30

%N a(n) = floor(2*n/5).

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

%C The sequence a(n) can be used in determining confidence intervals for the median of a population. Let Y(i) denote the i-th smallest datum in a random sample of size n from any population of values. When estimating the population median with a symmetric interval [Y(r), Y(n-r+1)], the exact confidence coefficient c for the interval is given by c = Sum_{k=r..n-r} C(n, k)(1/2)^n. If r = a(n-4), then the confidence coefficient will be (i) at least 0.90 for all n>=7, (ii) at least 0.95 for all n>=35, and (iii) at least 0.99 for all n>=115. To use the sequence, for example, decide on the minimum level of confidence desired, say 95%. Hence use a sample size of 35 or greater, say n=40. We then find a(n-4)=a(36)=14, and thus the 14th smallest and 14th largest values in the sample will form the bounds for the confidence interval. If the exact confidence coefficient c is needed, calculate c = Sum_{k=14..26} C(40,k)(1/2)^40, which is 0.9615226917. - _Dennis P. Walsh_, Nov 28 2011

%C a(n+2) is also the domination number of the n-antiprism graph. - _Eric W. Weisstein_, Apr 09 2016

%C Equals partial sums of 0 together with 0, 0, 1, 0, 1, ... (repeated). - _Bruno Berselli_, Dec 06 2016

%C Euler transform of length 5 sequence [1, 1, 0, -1, 1]. - _Michael Somos_, Dec 06 2016

%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 Colin Barker, <a href="/A057354/b057354.txt">Table of n, a(n) for n = 0..1000</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 Dennis Walsh, <a href="http://frank.mtsu.edu/~dwalsh/4370/MEDIANC5.pdf">Median estimation with the point-four-n-minus-two rule</a>.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AntiprismGraph.html">Antiprism Graph</a>.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DominationNumber.html">Domination Number</a>.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).

%F G.f.: x^3*(1 + x^2) / ((x^4 + x^3 + x^2 + x + 1)*(x - 1)^2). - Numerator corrected by _R. J. Mathar_, Feb 20 2011

%F a(n) = a(n-1) + a(n-5) - a(n-6) for n>5. - _Colin Barker_, Dec 06 2016

%F a(n) = -a(2-n) for all n in Z. - _Michael Somos_, Dec 06 2016

%F a(n) = A002266(2*n). - _R. J. Mathar_, Jul 21 2020

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

%e G.f. = x^3 + x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 3*x^8 + 3*x^9 + 4*x^10 + 4*x^11 + ...

%t Table[Floor[2 n/5], {n, 0, 80}] (* _Bruno Berselli_, Dec 06 2016 *)

%t a[ n_] := Quotient[2 n, 5]; (* _Michael Somos_, Dec 06 2016 *)

%o (PARI) a(n)=2*n\5 \\ _Charles R Greathouse IV_, Nov 28 2011

%o (PARI) concat(vector(3), Vec(x^3*(1 + x^2) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^80))) \\ _Colin Barker_, Dec 06 2016.

%o (Python) [int(2*n/5) for n in range(80)] # _Bruno Berselli_, Dec 06 2016

%o (Sage) [floor(2*n/5) for n in range(80)] # _Bruno Berselli_, Dec 06 2016

%o (Magma) [2*n div 5: n in [0..80]]; // _Bruno Berselli_, Dec 06 2016

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

%Y Cf. A002266.

%K nonn,easy

%O 0,6

%A _Mitch Harris_