%I #20 Dec 31 2022 15:16:24
%S 1,1,2,1,2,2,3,1,2,2,3,2,3,2,3,1,2,2,3,2,3,2,3,2,3,2,3,2,3,2,3,1,2,2,
%T 3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,1,2,2,3,2,
%U 3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2,3,2
%N a(1) = 1; for n >= 1, a(2*n) = A000005(a(n)), a(2*n + 1) = A000005(a(n)) + 1.
%F Sum_{i = 2^k..2^(k + 1) - 1} a(i) = 5*2^(k - 1) - 2, for k >= 1.
%F a(2^k) = 1.
%e a(1) = 1;
%e a(2) = A000005(a(1)) = 1;
%e a(3) = A000005(a(1)) + 1 = 2;
%e a(4) = A000005(a(2)) = 1;
%e a(5) = A000005(a(2)) + 1 = 2;
%e and so on.
%t a[1] = 1; a[n_] := a[n] = If[EvenQ[n], DivisorSigma[0, a[n/2]], DivisorSigma[0, a[(n - 1)/2]] + 1]; Array[a, 100] (* _Amiram Eldar_, Dec 25 2022 *)
%Y Cf. A000005, A131051.
%K nonn
%O 1,3
%A _Ctibor O. Zizka_, Dec 25 2022