

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 2adic valuation of y.


0



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
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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*n1) = a(12*n8) = A001511(3*n2); (iii) a(4*n2) = a(n).
Proof. Let n be a natural number. For part (i), we have F(4*4*n3) = (3*(16*n3)+1)/2^v(3*(16*n3)+1) = (48*n8)/2^v(48*n3) = 6*n1, hence a(4*n) = v(1 + (6*n1)) = v(6*n) = v(3*2*n) = v(2*n) = A001511(n); for part (ii), v(1+F(4*(2*n1)3)) = v(1+(24*n20)/2^v(24*n20)) = v(1+(6*n5)) = v(6*n4) and, similarly, v(1 + F(4*(12*n8)3)) = v(1+(144*n104)/2^v(144*n104)) = v(6*n4), so a(2*n1) = a(12*n8), as claimed, and finally note that v(6*n4) = v(2*(3*n2)) = A001511(3*n2); for part (iii), the claim follows from the fact that F(4*(4*n2)3) = (48*n32)/2^v(48*n32) = (3*n2)/2^v(3*n2) = F(4*n3). 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: A015137 A136148 A178252 * A325498 A224850 A269976
Adjacent sequences: A338196 A338197 A338198 * A338200 A338201 A338202


KEYWORD

nonn


AUTHOR

L. Edson Jeffery, Oct 17 2020


STATUS

approved



