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A135513
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Number of Pierce-Engel hybrid expansions of 4/b, b>=4.
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0
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1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1, 4, 2, 2, 1, 4, 2, 4, 1, 2, 2, 4, 1
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OFFSET
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4,2
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COMMENTS
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Gives the number of representations of 4/b (for b>=4) as a sum of fractions 4/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 = 4 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 12), for n, (n mod 12) >= 4.
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EXAMPLE
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13 = 4(3) + 1 -> 1(13) + 0 or
13 = 4(4) - 3 -> 3(4) + 1 -> 1(13) + 0 or
13 = 4(4) - 3 -> 3(5) - 2 -> 2(6) + 1 -> 1(13) + 0 or
13 = 4(4) - 3 -> 3(5) - 2 -> 2(7) - 1 -> 1(13) + 0
So 4/13 = 1/3 - 1/39, 1/4 + 1/16 - 1/208, 1/4 + 1/20 + 1/120 - 1/1560, 1/4 + 1/20 + 1/140 + 1/1820: thus h(13) = 4
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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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