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A226605
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Irregular array read by rows of numerators in which row n has one numerator from each irreducible cycle of n rational numbers under iteration by the 3x+1 function. (See Comments for selection and order of numerators.)
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3
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-1, 0, 1, -5, 1, -19, 5, 1, -65, 19, 23, 5, 7, 1, -211, -65, -73, 19, 23, 31, 1, 7, 1, -665, -211, -227, 65, -251, 73, 89, 19, 85, 101, 23, 31, 47, 5, 37, 1, 11, 1, -2059, -665, -697, 211, -745, 227, 259, 13, 251, 283, 73, 331, 89, 121, 19, 319, 17, 101, 19, 23
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OFFSET
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1,4
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COMMENTS
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A cycle is irreducible if it is not a concatenation of copies of a shorter cycle.
The 3x+1 function T, on rational numbers in their lowest terms with a positive odd denominator, is defined by T(x) = x/2 if x's numerator is even, T(x) = (3x+1)/2 if x's numerator is odd.
Each numerator in a row is the first in the cyclic permutation with the lexicographically largest parity vector of numerators mod 2. The row lists these numerators in descending lexicographic order of the parity vectors.
The element with numerator a(n) has denominator A226606(n), as does every element in the same cycle.
a(n) is often the numerator with the least absolute value of the numerators in the cycle. a(20) and a(36) are the only exceptions in the first 7 rows.
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LINKS
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FORMULA
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If v(0) to v(m-1) are the bits of A102659(n), when 2's are replaced by 0's, then a(n) = N(n)/GCD(N(n),D(n)) where D(n) = 2^m - 3^(v(0)+...+v(m-1)) and N(n) = Sum_{j=0 to m-1} (2^j)(3^(v(j+1)+...+v(m-1)))v(j).
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EXAMPLE
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-1, 0, 1, -5, 1/5, -19/11, 5/7, 1/13, ... = A226605/A226606 for parity vectors 1, 0, 10, 110, 100, 1110, 1100, 1000, ... For example, the numerators of the rational cycle {-19/11,-23/11,-29/11,-38/11} have parity vector 1110.
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CROSSREFS
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There are A001037(n) terms in row n.
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KEYWORD
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sign,frac,tabf
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AUTHOR
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STATUS
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approved
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