|
%I #68 Apr 24 2024 04:14:24
%S 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,
%T 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,
%U 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
%N Denominator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of 4n, with the exponents of 2 being listed in descending order.
%C If the terms (n>0) are written as an array:
%C 1,
%C 1, 2,
%C 1, 2, 3, 3,
%C 1, 2, 3, 3, 4, 5, 4, 5,
%C 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8,
%C 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,
%C then the sum of the m-th 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,...).
%C If the rows are written in a right-aligned fashion:
%C 1,
%C 1, 2,
%C 1, 2, 3, 3,
%C 1, 2, 3, 3, 4, 5, 4, 5,
%C 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8,
%C ..., 7, 7, 8, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13,
%C 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^m-1 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
%H Paul D. Hanna, <a href="/A071766/b071766.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Fo#fraction_trees">Index entries for fraction trees</a>
%F a(n) = A071585(m), where m = n - floor(log_2(n));
%F a(0) = 1, a(2^k) = 1, a(2^k + 1) = 2.
%F a(2^k - 1) = Fibonacci(k+1) = A000045(k+1).
%F a(2^m+k) = A071585(k), m=0,1,2,..., k=0,1,2,...,2^m-1. - _Yosu Yurramendi_, Jun 23 2014
%F a(2^m-k) = F_k(m), k=0,1,2,..., m > log_2(k). F_k(m) is a Fibonacci sequence, where F_k(1) = a(2^(m_0(k))-1-k), F_k(2) = a(2^(m_0(k)+1)-1-k), m_0(k) = ceiling(log_2(k+1))+1 = A070941(k). - _Yosu Yurramendi_, Jun 23 2014
%F a(n) = A002487(A059893(A233279(n))) = A002487(1+A059893(A006068(n))), n > 0. - _Yosu Yurramendi_, Sep 29 2021
%e 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].
%e 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, ...
%t {1}~Join~Table[Denominator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2[NumberExpand[4 n, 2] /. 0 -> Nothing], {n, 120}] (* Version 11, or *)
%t {1}~Join~Table[Denominator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2@ DeleteCases[# Reverse[2^Range[0, Length@ # - 1]] &@ IntegerDigits[4 n, 2], k_ /; k == 0], {n, 120}] (* _Michael De Vlieger_, Aug 15 2016 *)
%o (R)
%o blocklevel <- 6 # arbitrary
%o a <- 1
%o for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
%o a[2^(m+1)+k] <- a[2^m+k]
%o a[2^(m+1)+2^(m-1)+k] <- a[2^m+2^(m-1)+k]
%o a[2^(m+1)+2^m+k] <- a[2^(m+1)+k] + a[2^(m+1)+2^(m-1)+k]
%o a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^(m+1)+2^m+k]
%o }
%o a
%o # _Yosu Yurramendi_, Jul 11 2014
%Y Cf. A071585, A086593.
%K easy,nonn,frac
%O 0,4
%A _Paul D. Hanna_, Jun 04 2002
|
