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A135511
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Number of Pierce-Engel hybrid expansions of 3/b, b>=3.
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0
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1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3, 1, 3, 2, 1, 2, 3
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OFFSET
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3,2
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COMMENTS
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Gives the number of representations of 3/b (for b>=3) as a sum of fractions 3/b= a_1/q_1 + a_2/(q_1 q_2) + a_3/(q_1 q_2 q_3) + ... a_n/(q_1 q_2 ... q_n), where each a_i is either 1 or -1 and the q_i are chosen greedily.
Equivalently, the q_i can be found by taking r_1 = 3 and applying either b=r_i q_i + r_(i+1) or b=r_i q_i - r(i-1), where 0<=r_(i-1)<r_i. (When the first equation is used to find q_i, then a_(i+1) will be of opposite sign than a_i. If the second is used, a_(i+1) will be of the same sign as a_i.) The process terminates when some r_(n+1)=0.
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LINKS
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FORMULA
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h(n) = h(n mod 6), for n, (n mod 6) >= 3.
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EXAMPLE
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5 = 3(1) + 2 -> 2(2) + 1 -> 1(5) + 0 or
5 = 3(1) + 2 -> 2(3) - 1 -> 1(5) + 0 or
5 = 3(2) - 1 -> 1(5) + 0
So 3/5 = 1 - 1/2 + 1/10, 1 - 1/3 - 1/15, 1/2 + 1/10: thus h(5)=3.
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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A. Sutyak (asutyak(AT)gmail.com), Feb 09 2008
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STATUS
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approved
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