

A135514


Number of PierceEngel 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
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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(i1), where 0<=r_(i1)<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

Table of n, a(n) for n=5..82.
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: A101778 A292570 A161670 * A251754 A225581 A275391
Adjacent sequences: A135511 A135512 A135513 * A135515 A135516 A135517


KEYWORD

easy,nonn


AUTHOR

A. Sutyak (asutyak(AT)gmail.com), Feb 09 2008


STATUS

approved



