A071585 Numerator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of 4*n, with the exponents of 2 being listed in descending order.
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, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14, 19, 18, 21, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14, 19, 18, 21, 8, 13, 16, 17, 17, 22, 19, 23, 16, 23, 24, 27, 19
Offset: 0
Examples
a(37)=17 as it is the numerator 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].
Illustration of Sum_{m=0..2^(k-1)-1} a(2^k + m) = 3^k:
k=2: 3^2 = a(2^2) + a(2^2 + 1) = 4 + 5;
k=3: 3^3 = a(2^3) + a(2^3 + 1) + a(2^3 + 2) + a(2^3 + 3) = 5 + 7 + 7 + 8;
k=4: 3^4 = a(2^4) + a(2^4+1) + a(2^4+2) + a(2^4+3) + a(2^4+4) + a(2^4+5) + a(2^4+6) + a(2^4+7) = 6 + 9 + 10 + 11 + 9 + 12 + 11 + 13.
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, ...
From _Yosu Yurramendi_, Jun 27 2014: (Start)
a(0) = = 1;
a(1) = a(0) + a(0) = 2;
a(2) = a(0) + a(1) = 3;
a(3) = a(1) + a(0) = 3;
a(4) = a(0) + a(2) = 4;
a(5) = a(1) + a(3) = 5;
a(6) = a(2) + a(0) = 4;
a(7) = a(3) + a(1) = 5;
a(8) = a(0) + a(4) = 5;
a(9) = a(1) + a(5) = 7;
a(10) = a(2) + a(6) = 7;
a(11) = a(3) + a(7) = 8;
a(12) = a(4) + a(0) = 5;
a(13) = a(5) + a(1) = 7;
a(14) = a(6) + a(2) = 7;
a(15) = a(7) + a(3) = 8. (End)
Links
Crossrefs
Cf. A071766.
Programs
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Mathematica
ncf[n_]:=Module[{br=Reverse[Flatten[Position[Reverse[IntegerDigits[4 n,2]],1]-1]]}, Numerator[FromContinuedFraction[Flatten[Join[{Abs[ Differences[ br]],Last[br]}]]]]]; Join[{1},Array[ncf,80]] (* Harvey P. Dale, Jul 01 2012 *) {1}~Join~Table[Numerator@ FromContinuedFraction@ Append[Abs@ Differences@ #, Last@ #] &@ Log2[NumberExpand[4 n, 2] /. 0 -> Nothing], {n, 120}] (* Version 11, or *) {1}~Join~Table[Numerator@ 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 *) -
R
blocklevel <- 7 # arbitrary a <- c(1,2) for(k in 1:blocklevel) a <- c(a, a + c(a[((length(a)/2)+1):length(a)],a[1:(length(a)/2)])) a # Yosu Yurramendi, Jun 26 2014
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R
blocklevel <- 7 # arbitrary a <- c(1,2) for(p in 0:blocklevel) for(k in 1:2^(p+1)){ if (k <= 2^p) a[k + 2^(p+1)] = a[k] + a[k + 2^p] else a[k + 2^(p+1)] = a[k] + a[k - 2^p] } a # Yosu Yurramendi, Jun 27 2014
Formula
a(2^k + 2^j + m) = (k-j)*a(2^j + m) + a(m) when 2^k > 2^j > m >= 0.
a(0) = 1, a(2^k) = k + 2,
a(2^k + 1) = 2*k + 1 (k>0),
a(2^k + 2) = 3*k - 2 (k>1),
a(2^k + 3) = 3*k - 1 (k>1),
a(2^k + 4) = 4*k - 7 (k>2).
a(2^k - 1) = Fibonacci(k+2) = A000045(k+2).
Sum_{m=0..2^(k-1)-1} a(2^k + m) = 3^k (k>0).
From Yosu Yurramendi, Jun 27 2014: (Start)
Write n = k + 2^(m+1), k = 0,1,2,...,2^(m+1)-1, m = 0,1,2,...
if 0 <= k < 2^m, a(k + 2^(m+1)) = a(k) + a(k + 2^m).
if 2^m <= k < 2^(m+1), a(k + 2^(m+1)) = a(k) + a(k - 2^m).
with a(0)=1, a(1)=2. (End)
a(n) = A093873(2n) + A093873(2n+1), n > 0; a(n) = A093875(2n) = A093875(2n+1), n > 0. - Yosu Yurramendi, Jul 25 2016
a(n) = sqrt(A071766(2^(m+1)+n)*A229742(2^(m+1)+n) - A071766(2^m+n)*A229742(2^m+n)), for n > 0, where m = floor(log_2(n)+1). - Yosu Yurramendi, Jun 10 2019
Conjecture: a(n) = a(floor(n/2)) + Sum_{k=1..A000120(n)} a(b(n, k))*(-1)^(k-1) for n > 0 with a(0) = 1 where b(n, k) = A025480(b(n, k-1) - 1) for n > 0, k > 0 with b(n, 0) = n. - Mikhail Kurkov, Feb 20 2023
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