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Number of Pierce-Engel hybrid expansions of 5/b, b>=5.
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%I #2 Feb 27 2009 03:00:00

%S 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,

%T 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,

%U 5,3,1,2,5,2,5,1,3,3

%N Number of Pierce-Engel hybrid expansions of 5/b, b>=5.

%C 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.

%H Weisstein, Eric W., <a href="http://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>.

%H Weisstein, Eric W., <a href="http://mathworld.wolfram.com/EngelExpansion.html">Engel Expansion</a>.

%F h(n)=h(n mod 60), for n, (n mod 60) >= 5

%e 14 = 5(2) + 4 -> 4(3) + 2 -> 2(7) + 0 or

%e 14 = 5(2) + 4 -> 4(4) - 2 -> 2(7) + 0 or

%e 14 = 5(3) - 1 -> 1(14) + 0

%e So 5/14 = 1/2 - 1/6 + 1/42, 1/2 - 1/8 - 1/56, 1/3 + 1/42: thus h(14) = 3.

%K easy,nonn

%O 5,2

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