

A071766


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


9



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


COMMENTS

If the terms (n>0) are written as an array:
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,
then the sum of the mth row is 3^m (m = 0,1,2,3,...), each column is constant and the terms are from A071585 ( a(2^m+k) = A071585(k) , k = 0,1,2,...).
If the rows are written in a rightaligned fashion:
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,
..., 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13,
then each column is a Fibonacci sequence ( a(2^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k), m = 0,1,2,..., k = 0,1,2,...,2^m1 with a_k(1) = A071766(k) and a_k(2) = A086593(k) being the first two terms of each column sequence).  Yosu Yurramendi, Jun 23 2014


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).
a(2^m+k) = A071585(k), m=0,1,2,..., k=0,1,2,...,2^m1.  Yosu Yurramendi, Jun 23 2014
a(2^mk) = F_k(m), k=0,1,2,..., m > log2(k). F_k(m) is a Fibonacci sequence, where F_k(1) = a(2^(m_0(k))1k), F_k(2) = a(2^(m_0(k)+1)1k), m_0(k) = ceiling(log_2(k+1))+1 = A070941(k).  Yosu Yurramendi, Jun 23 2014


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), ", "))
(R)
blocklevel < 6 # arbitrary
a < 1
for(m in 0:blocklevel) for(k in 0:(2^(m1)1)){
a[2^(m+1)+k] < a[2^m+k]
a[2^(m+1)+2^(m1)+k] < a[2^m+2^(m1)+k]
a[2^(m+1)+2^m+k] < a[2^(m+1)+k] + a[2^(m+1)+2^(m1)+k]
a[2^(m+1)+2^m+2^(m1)+k] < a[2^(m+1)+2^m+k]
}
a
# Yosu Yurramendi, Jul 11 2014


CROSSREFS

Cf. A071585, A086593.
Sequence in context: A033803 A035531 A118977 * A007305 A112531 A100002
Adjacent sequences: A071763 A071764 A071765 * A071767 A071768 A071769


KEYWORD

easy,nonn,frac,changed


AUTHOR

Paul D. Hanna, Jun 04 2002


STATUS

approved



