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%I #42 Aug 09 2022 12:04:58
%S 3,3,11,9,27,19,51,33,83,51,123,73,171,99,227,129,291,163,363,201,443,
%T 243,531,289,627,339,731,393,843,451,963,513,1091,579,1227,649,1371,
%U 723,1523,801,1683,883,1851,969,2027,1059,2211,1153
%N a(2n) = 4*n*(n+1) + 3; a(2n+1) = 2*n*(n+2) + 3.
%C a(n) = largest odd divisor of A059100(n+1). Proof: Observe that a(2n) = A059100(2n+1) and a(2n+1) = (A059100(2n+2))/2 and note that (A059100(m))/2 is odd for even m. - _Jeremy Gardiner_, Aug 25 2013
%C a(n) is also the denominator of the (n+1)-st largest circle in a special case of the Pappus chain inspired by the Yin-Yang symbol. See illustration in the links. - _Kival Ngaokrajang_, Jun 20 2015
%H Vincenzo Librandi, <a href="/A164900/b164900.txt">Table of n, a(n) for n = 0..10000</a>
%H Kival Ngaokrajang, <a href="/A164900/a164900.pdf">Illustration of initial terms</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-3,0,1)
%F a(2n) = A164897(n); a(2n+1) = A058331(n+1).
%F a(n) = A164845(n-1)/A026741(n), n>0.
%F G.f.: ( -3-3*x-2*x^2-3*x^4-x^5 ) / ( (x-1)^3*(1+x)^3 ). - _R. J. Mathar_, Jan 21 2011
%F a(n) = ((-1)^n+3)*(n^2+2*n+3)/4. - _Bruno Berselli_, Jan 21 2011
%F From _Amiram Eldar_, Aug 09 2022: (Start)
%F a(n) = numerator(((n+1)^2 + 2)/2).
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(2))*Pi/sqrt(2) + tanh(Pi/sqrt(2))*Pi/(2*sqrt(2)) - 1)/2. (End)
%t LinearRecurrence[{0, 3, 0, -3, 0, 1}, {3, 3, 11, 9, 27, 19}, 50] (* _Amiram Eldar_, Aug 09 2022 *)
%o (Magma) [((-1)^n+3)*(n^2+2*n+3)/4: n in [0..50]]; // _Vincenzo Librandi_, Aug 07 2011
%o (PARI) vector(100,n,n--;(1/4)*((-1)^n+3)*(n^2+2*n+3)) \\ _Derek Orr_, Jun 27 2015
%Y Cf. A026741, A058331, A059100, A164845, A164897.
%K nonn,easy
%O 0,1
%A _Paul Curtz_, Aug 30 2009
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