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A242496 a(n)=sum_{j=0..n} sum_{i=0..j} F(i)*L(j), where F(n)=A000045(n) and L(n)=A000032(n).

%I A242496 #29 Oct 03 2020 15:17:43
%S A242496 0,1,7,23,72,204,564,1521,4059,10747,28336,74504,195576,512865,
%T A242496 1344063,3521007,9221688,24148468,63230860,165555665,433454835,
%U A242496 1134839091,2971111392,7778574288,20364739632,53315851969,139583151799,365434146311,956720165544
%N A242496 a(n)=sum_{j=0..n} sum_{i=0..j} F(i)*L(j), where F(n)=A000045(n) and L(n)=A000032(n).
%H A242496 Harvey P. Dale, <a href="/A242496/b242496.txt">Table of n, a(n) for n = 0..1000</a>
%H A242496 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-6,4,2,-1).
%F A242496 a(n) = A001519(n+2) - A000032(n+2) + A059841(n).
%F A242496 a(n) = L(n)*F(n+3) - L(n+2) + (1-3*(-1)^n)/2. - _Colin Barker_, May 18 2014
%F A242496 G.f.: -x*(3*x^2-3*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). - _Colin Barker_, May 16 2014
%e A242496 For n=5, 0*(2+1+3+4+7+11) + 1*(1+3+4+7+11) + 1*(3+4+7+11) + 2*(4+7+11) + 3*(7+11) + 5*11 = 204 = F(2*5+3) - L(n+2) + 0 = 233-29 = 204.
%p A242496 A242496 := proc(n)
%p A242496     add(add(A000045(i)*A000032(j),i=0..j),j=0..n) ;
%p A242496 end proc: # _R. J. Mathar_, May 17 2014
%t A242496 LinearRecurrence[{4,-2,-6,4,2,-1},{0,1,7,23,72,204},30] (* _Harvey P. Dale_, Oct 03 2020 *)
%o A242496 (PARI)
%o A242496 F(n) = fibonacci(n)
%o A242496 L(n) = if(n==0, 2, F(2*n)/F(n))
%o A242496 vector(30, n, sum(i=0, n-1, sum(j=i, n-1, F(i)*L(j)))) \\ _Colin Barker_, May 16 2014
%Y A242496 Cf. A190173, A000045, A000032, A242300.
%K A242496 nonn,easy
%O A242496 0,3
%A A242496 _J. M. Bergot_, May 16 2014
%E A242496 Two terms corrected, and more terms added by _Colin Barker_, May 16 2014
%E A242496 Formula corrected by _Colin Barker_, May 17 2014