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a(2*n) = A001045(n+2); a(2*n+1) = A001045(n+1).
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%I #11 May 30 2018 13:44:08

%S 1,1,3,1,5,3,11,5,21,11,43,21,85,43,171,85,341,171,683,341,1365,683,

%T 2731,1365,5461,2731,10923,5461,21845,10923,43691,21845,87381,43691,

%U 174763,87381,349525,174763,699051,349525,1398101,699051,2796203,1398101,5592405

%N a(2*n) = A001045(n+2); a(2*n+1) = A001045(n+1).

%C Consider the Harmonacci sequence: H(1)=x, H(2)=y, H(3)=2xy/(x+y), H(4)=4xy/(3x+y)...; H(m) is the harmonic mean of H(m-1) and H(m-2). a(2n) and a(2n+1) are the denominator coefficients of H(n+3).

%H Colin Barker, <a href="/A112447/b112447.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,2).

%F a(n) = (a(n-1)+1)/2 for n=2, 6, 10...

%F a(n) = 4*a(n-1)-1 for n=3, 7, 11...

%F a(n) = (a(n-1)-1)/2 for n=4, 8, 12...

%F a(n) = 4*a(n-1)+1 for n=5, 9, 13....

%F From _Colin Barker_, Dec 15 2017: (Start)

%F G.f.: (1 + x + 2*x^2) / ((1 + x^2)*(1 - 2*x^2)).

%F a(n) = a(n-2) + 2*a(n-4) for n>3.

%F (End)

%t LinearRecurrence[{0,1,0,2},{1,1,3,1},50] (* _Harvey P. Dale_, May 30 2018 *)

%o (PARI) Vec((1 + x + 2*x^2) / ((1 + x^2)*(1 - 2*x^2)) + O(x^60)) \\ _Colin Barker_, Dec 15 2017

%Y Cf. A001045.

%K nonn,easy

%O 0,3

%A _Edwin F. Sampang_, Dec 12 2005

%E Edited by _Don Reble_, Jan 25 2006