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%I #81 Dec 03 2023 09:41:35
%S 0,1,2,3,5,6,7,8,10,11,12,13,15,16,17,18,20,21,22,23,25,26,27,28,30,
%T 31,32,33,35,36,37,38,40,41,42,43,45,46,47,48,50,51,52,53,55,56,57,58,
%U 60,61,62,63,65,66,67,68,70,71,72,73,75,76,77,78,80,81,82,83,85,86,87,88
%N a(n) = floor(5*n/4), numbers that are congruent to {0, 1, 2, 3} mod 5.
%C From _M. F. Hasler_, Oct 21 2008: (Start)
%C Also, for n>0, the 4th term (after [0,n,3n]) in the continued fraction expansion of arctan(1/n). (Observation by V. Reshetnikov)
%C Proof:
%C arctan(1/n) = (1/n) / (1 + (1/n)^2/( 3 + (2/n)^2/( 5 + (3/n)^2/( 7 + ...)...)
%C = 1 / ( n + 1/( 3n + 4/( 5n + 9/( 7n + 25/(...)...)
%C = 1 / ( n + 1/( 3n + 1/( 5n/4 + (9/4)/( 7n + 25/(...)...),
%C and the term added to 5n/4, (9/4)/(7n+...) = (1/4)*9/(7n+...) is less than 1/4 for all n>=2. (End)
%H Vincenzo Librandi, <a href="/A001068/b001068.txt">Table of n, a(n) for n = 0..10000</a>
%H Paul Erdős, <a href="http://dx.doi.org/10.1016/S0012-365X(96)00044-1">Some recent problems and results in graph theory</a>, Discr. Math., Vol. 164, No. 1-3 (1997), pp. 81-85.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Arctan#Continued_fraction_for_arctangent">Continued fraction for arctangent</a>.
%H R. Witula, P. Lorenc, M. Rozanski and M. Szweda, <a href="http://yadda.icm.edu.pl/yadda/element/bwmeta1.element.baztech-142e1ad8-f56c-4c6e-b6e6-f71d1f31e76e">Sums of the rational powers of roots of cubic polynomials</a>, Zeszyty Naukowe Politechniki Slaskiej, Seria: Matematyka Stosowana z. 4, Nr. kol. 1920, (2014), pp. 17-34.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).
%F contfrac( arctan( 1/n )) = 0 + 1/( n + 1/( 3n + 1/( a(n) + 1/(...)))). - _M. F. Hasler_, Oct 21 2008
%F a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(1)=2 and b(k)=5*2^(k-2) for k>1. - _Philippe Deléham_, Oct 17 2011.
%F From _Bruno Berselli_, Oct 17 2011: (Start)
%F G.f.: x*(1+x+x^2+2*x^3)/((1+x)*(1-x)^2*(1+x^2)).
%F a(n) = (10*n+2*(-1)^((n-1)n/2)+(-1)^n-3)/8.
%F a(-n) = -A047203(n+1). (End)
%F From _Wesley Ivan Hurt_, Sep 17 2015: (Start)
%F a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.
%F a(n) = n + floor(n/4). (End)
%F a(n) = n + A002265(n). - _Robert Israel_, Sep 17 2015
%F E.g.f.: (sin(x) + cos(x) + (5*x - 2)*sinh(x) + (5*x - 1)*cosh(x))/4. - _Ilya Gutkovskiy_, May 06 2016
%F Sum_{n>=1} (-1)^(n+1)/a(n) = log(5)/4 + sqrt(5)*log(phi)/10 + sqrt(5-2*sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - _Amiram Eldar_, Dec 10 2021
%p A001068:=n->floor(5*n/4); seq(A001068(k), k=0..100); # _Wesley Ivan Hurt_, Nov 07 2013
%t Table[Floor[5*n/4],{n,0,120}] (* _Vladimir Joseph Stephan Orlovsky_, Jan 28 2012 *)
%t #+{0,1,2,3}&/@(5*Range[0,20])//Flatten (* or *) Complement[Range[0,103],5*Range[20]-1] (* _Harvey P. Dale_, Dec 03 2023 *)
%o (PARI) a(n)=5*n\4 /* or, cf. comment: */
%o a(n)=contfrac(atan(1/n))[4] \\ _M. F. Hasler_, Oct 21 2008
%o (Magma) [Floor(5*n/4): n in [0..80]]; // _Vincenzo Librandi_, Nov 13 2011
%Y Cf. A001622, A002265, A030308, A047203, A110255, A110256, A110257, A110258, A110259, A110260, A177704 (first differences), A249547.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
%E More terms from _James A. Sellers_, Sep 19 2000
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