Let b(1)=x, b(2)=y, k*b(k)=(2k-1)*b(k-1) + 3(k+1)*b(k-2); then b(n)=a(n)*x+c(n)/3*y.

Terms

a(0) =1a(1) =0a(2) =4a(3) =7a(4) =27a(5) =74a(6) =230a(7) =681a(8) =2053a(9) =6148a(10) =18456a(11) =55355a(12) =166079a(13) =498222a(14) =1494682a(15) =4484029a(16) =13452105a(17) =40356296a(18) =121068908a(19) =363206703a(20) =1089620131a(21) =3268860370a(22) =9806581134a(23) =29419743377a(24) =88259230157a(25) =264777690444a(26) =794333071360