

A071766


Denominator of the continued fraction expansion whose terms are the firstorder differences of exponents in the binary representation of 4*n, with the exponents of 2 being listed in descending order.


8



1, 1, 1, 2, 1, 2, 3, 3, 1, 2, 3, 3, 4, 5, 4, 5, 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13, 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13
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OFFSET

0,4


LINKS

Paul D. Hanna, Table of n, a(n) for n = 0..10000


FORMULA

a(n) = A071585(m), where m = n  floor(log(n)/log(2));
a(0) = 1, a(2^k) = 1, a(2^k + 1) = 2.
a(2^k  1) = Fibonacci(k+1) = A000045(k+1).


EXAMPLE

a(37)=5 as it is the denominator of 17/5 = 3 + 1/(2 + 1/2), which is a continued fraction that can be derived from the binary expansion of 4*37 = 2^7 + 2^4 + 2^2; the binary exponents are {7, 4, 2}, thus the differences of these exponents are {3, 2, 2}; giving the continued fraction expansion of 17/5=[3,2,2].
1, 2, 3, 3/2, 4, 5/2, 4/3, 5/3, 5, 7/2, 7/3, 8/3, 5/4, 7/5, 7/4, 8/5, 6, ...


PROG

(PARI) {a(n)=local(N=4*n, E=#binary(N)1, P=[E], CF); while(E>1, P=concat(P, E=#binary(N=N2^E)1)); CF=Vec(Ser(P)*(x1)); if(n==0, CF[1]=1, CF[1]=0); contfracpnqn(CF)[1, 1]} \\ Paul D. Hanna, Feb 22 2013
for(n=0, 256, print1(a(n), ", "))


CROSSREFS

Cf. A071585.
Sequence in context: A033803 A035531 A118977 * A007305 A112531 A100002
Adjacent sequences: A071763 A071764 A071765 * A071767 A071768 A071769


KEYWORD

easy,nonn,frac


AUTHOR

Paul D. Hanna, Jun 04 2002


STATUS

approved



