Let b(1)=b(2)=1, b(k) = (2^b(k-1)+2^b(k-2)) (mod k); sequence gives values of n such that b(n)=0.

Terms

a(0) =4a(1) =10a(2) =16a(3) =20a(4) =24a(5) =40a(6) =64a(7) =80a(8) =128a(9) =144a(10) =192a(11) =256a(12) =358a(13) =528a(14) =1152a(15) =1536a(16) =1600a(17) =1672a(18) =2048a(19) =2052a(20) =2056a(21) =2176a(22) =2260a(23) =2560a(24) =2804a(25) =3072a(26) =3898a(27) =4528a(28) =5120a(29) =5139