A338199 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.

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, 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, 1, 1, 2, 2, 1

Offset

1,3

Comments

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)).

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).

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

Links

Table of n, a(n) for n=1..73.

Mathematica

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

Crossrefs

Cf. A001511, A254070, A257480.

Sequence in context: A368202 A136148 A178252 * A325498 A224850 A269976

Adjacent sequences: A338196 A338197 A338198 * A338200 A338201 A338202

Keyword

nonn

Author

L. Edson Jeffery, Oct 17 2020

Status

approved