%I #14 Apr 01 2021 15:02:25

%S 1,1,3,1,1,1,2,2,1,3,5,1,1,1,2,3,1,1,3,1,1,1,2,2,1,2,4,1,1,2,2,4,1,1,

%T 3,1,1,3,2,2,1,5,7,1,1,1,2,3,1,1,3,1,1,1,2,2,1,2,4,1,1,3,2,5,1,1,3,1,

%U 1,1,2,2,1

%N a(n) = v(1 + F(4*n - 3)), where F(x) = (3*x + 1)/2^v(3*x + 1), x is any odd natural number, and v(y) is the 2-adic valuation of y.

%C This is a fractal sequence: removing all entries a(n) with indices n == 0,1 or 3 (mod 4) and reindexing yields the original sequence (see Thm 1 (iii)). This sequence also contains A001511 (the ruler sequence) as a subsequence (see Thm 1 (i)).

%C THEOREM 1. For all natural numbers n, the following hold: (i) a(4*n) = A001511(n); (ii) a(2*n-1) = a(12*n-8) = A001511(3*n-2); (iii) a(4*n-2) = a(n).

%C Proof. Let n be a natural number. For part (i), we have F(4*4*n-3) = (3*(16*n-3)+1)/2^v(3*(16*n-3)+1) = (48*n-8)/2^v(48*n-3) = 6*n-1, hence a(4*n) = v(1 + (6*n-1)) = v(6*n) = v(3*2*n) = v(2*n) = A001511(n); for part (ii), v(1+F(4*(2*n-1)-3)) = v(1+(24*n-20)/2^v(24*n-20)) = v(1+(6*n-5)) = v(6*n-4) and, similarly, v(1 + F(4*(12*n-8)-3)) = v(1+(144*n-104)/2^v(144*n-104)) = v(6*n-4), so a(2*n-1) = a(12*n-8), as claimed, and finally note that v(6*n-4) = v(2*(3*n-2)) = A001511(3*n-2); for part (iii), the claim follows from the fact that F(4*(4*n-2)-3) = (48*n-32)/2^v(48*n-32) = (3*n-2)/2^v(3*n-2) = F(4*n-3). QED

%t v[y_] := IntegerExponent[y, 2];f[x_] := (3*x + 1)/2^v[3*x + 1];Table[v[1 + f[4*k - 3]], {k, 73}]

%Y Cf. A001511, A254070, A257480.

%K nonn

%O 1,3

%A _L. Edson Jeffery_, Oct 17 2020