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%I #64 Feb 13 2020 21:06:58
%S 1,1,1,3,5,11,21,43,85,171,341,683,1365,2731,5461,10923,21845,43691,
%T 87381,174763,349525,699051,1398101,2796203,5592405,11184811,22369621,
%U 44739243,89478485,178956971,357913941,715827883,1431655765,2863311531,5726623061,11453246123
%N a(n) = Product_{k=1..floor((n-1)/2)} (1 + 8*cos(k*Pi/n)^2) for n >= 0.
%C Apparently the same as A001045 after the first term. - _R. J. Mathar_, Nov 27 2008 [This conjecture is true. - _Peter Luschny_, Nov 28 2019]
%H Colin Barker, <a href="/A152046/b152046.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,2).
%F From _Sergei N. Gladkovskii_, May 22 2013 and Sep 09 2013: (Start)
%F G.f.: 1 + A(x) where A(x) is the g.f. of A001045.
%F G.f.: 1 + Q(0)/3, where Q(k)= 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction).
%F G.f.: 1+ Q(0)*x/2 , where Q(k) = 1 + 1/(1 - x*(2*k+1 + 2*x)/( x*(2*k+2 + 2*x) + 1/Q(k+1) )); (continued fraction). (End)
%F From _Colin Barker_, Nov 28 2019: (Start)
%F a(n) = a(n-1) + 2*a(n-2) for n>2.
%F a(n) = ((-1)^(1 + n) + 2^n)/ 3 for n>0. (End)
%F E.g.f.: (3 - exp(-x) + exp(2*x))/3. - _Stefano Spezia_, Feb 13 2020
%t a[n_] := Product[(1 + 8 Cos[k Pi/n]^2), {k, 1, Floor[(n - 1)/2]}];
%t Table[Round[a[n]], {n, 0, 20}]
%o (PARI) Vec((1 - 2*x^2) / ((1 + x)*(1 - 2*x)) + O(x^40)) \\ _Colin Barker_, Nov 28 2019
%o (PARI) apply( {A152046(n)=2^n\/3+!n}, [0..40]) \\ _M. F. Hasler_, Feb 13 2020
%Y Cf. A001045.
%K nonn,easy
%O 0,4
%A _Roger L. Bagula_, Nov 21 2008
%E Edited by _Peter Luschny_, Nov 28 2019
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