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A338199
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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.
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0
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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
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1,3
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COMMENTS
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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
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LINKS
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MATHEMATICA
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v[y_] := IntegerExponent[y, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; Table[v[1 + f[4*k - 3]], {k, 73}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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