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A135514
Number of Pierce-Engel hybrid expansions of 5/b, b>=5.
0
1, 3, 5, 3, 3, 1, 5, 2, 5, 3, 1, 2, 5, 2, 5, 1, 3, 3, 5, 2, 1, 3, 3, 3, 5, 1, 5, 3, 3, 3, 1, 2, 5, 3, 3, 1, 5, 2, 5, 2, 1, 3, 5, 2, 5, 1, 3, 3, 5, 3, 1, 2, 3, 3, 5, 1, 5, 3, 3, 2, 1, 3, 5, 3, 3, 1, 5, 2, 5, 3, 1, 2, 5, 2, 5, 1, 3, 3
OFFSET
5,2
COMMENTS
Gives the number of representations of 5/b (for b>=5) as a sum of fractions 5/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 = 5 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.
LINKS
Weisstein, Eric W., Pierce Expansion.
Weisstein, Eric W., Engel Expansion.
FORMULA
h(n)=h(n mod 60), for n, (n mod 60) >= 5
EXAMPLE
14 = 5(2) + 4 -> 4(3) + 2 -> 2(7) + 0 or
14 = 5(2) + 4 -> 4(4) - 2 -> 2(7) + 0 or
14 = 5(3) - 1 -> 1(14) + 0
So 5/14 = 1/2 - 1/6 + 1/42, 1/2 - 1/8 - 1/56, 1/3 + 1/42: thus h(14) = 3.
CROSSREFS
Sequence in context: A292570 A372801 A161670 * A251754 A225581 A275391
KEYWORD
easy,nonn
AUTHOR
A. Sutyak (asutyak(AT)gmail.com), Feb 09 2008
STATUS
approved