

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.


17



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


AUTHOR

Paul D. Hanna, Jun 04 2002


STATUS

approved



