a(n) = (3*n*F(2n-1) + (3-n)*F(2n))/5 where F() = Fibonacci numbers A000045.

Terms

a(0) =0a(1) =1a(2) =3a(3) =9a(4) =27a(5) =80a(6) =234a(7) =677a(8) =1941a(9) =5523a(10) =15615a(11) =43906a(12) =122868a(13) =342409a(14) =950727a(15) =2631165a(16) =7260579a(17) =19982612a(18) =54865566a(19) =150316973a(20) =411015705a(21) =1121818311a(22) =3056773383a(23) =8316416134a(24) =22593883752a(25) =61301547025a(26) =166118284299a(27) =449639574897