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A109794 a(2n) = A001906(n+1), a(2n+1) = A002878(n).

%I A109794 #43 Mar 13 2024 21:51:48
%S A109794 1,1,3,4,8,11,21,29,55,76,144,199,377,521,987,1364,2584,3571,6765,
%T A109794 9349,17711,24476,46368,64079,121393,167761,317811,439204,832040,
%U A109794 1149851,2178309,3010349,5702887,7881196,14930352,20633239,39088169
%N A109794 a(2n) = A001906(n+1), a(2n+1) = A002878(n).
%C A109794 Sequence relates bisections of Lucas and Fibonacci numbers (see also A098149).
%C A109794 Floretion Algebra Multiplication Program, FAMP code: 4jesleftforsumseq[ + .25'i + .25i' + .25'ii' + .25'jj' + .25'kk' + .25'jk' + .25'kj' + .25e], vesleftforsumseq = A000045, sumtype: (Y[15], *, inty*sum) (internal program code)
%H A109794 Alois P. Heinz, <a href="/A109794/b109794.txt">Table of n, a(n) for n = 0..1000</a>
%H A109794 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-1)
%F A109794 G.f.: (1+x+x^3)/((1+x-x^2)*(1-x-x^2)).
%F A109794 a(n) = ((3/20)*sqrt(5) + 3/4)*(1/2 + (1/2)*sqrt(5))^n + (-(3/20)*sqrt(5) + 3/4)*(1/2 - (1/2)*sqrt(5))^n + (-(3/20)*sqrt(5) - 1/4)*(-1/2 + (1/2)*sqrt(5))^n + ((3/20)*sqrt(5) - 1/4) *(-1/2 - (1/2)*sqrt(5))^n.
%F A109794 a(n) = 3*a(n-2) - a(n-4), n >= 4; a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 4. - _Daniel Forgues_, May 07 2011
%p A109794 a:= n-> (<<0|1>, <-1|3>>^iquo(n, 2, 'r'). <<1, 3+r>>)[1, 1]:
%p A109794 seq(a(n), n=0..50);  # _Alois P. Heinz_, May 02 2011
%t A109794 LinearRecurrence[{0, 3, 0, -1}, {1, 1, 3, 4}, 40] (* _Robert G. Wilson v_, Aug 06 2018 *)
%t A109794 CoefficientList[Series[(1+x+x^3)/((1+x-x^2)(1-x-x^2)),{x,0,40}],x] (* _Harvey P. Dale_, Aug 10 2021 *)
%o A109794 (GAP) a:=[1,1,3,4];; for n in [5..40] do a[n]:=3*a[n-2]-a[n-4]; od; a; # _Muniru A Asiru_, Aug 09 2018
%Y A109794 Cf. A001906, A002878, A098149, A000045, A189761.
%K A109794 nonn,easy
%O A109794 0,3
%A A109794 _Creighton Dement_, Aug 14 2005