A093873 Numerators in Kepler's tree of harmonic fractions.
1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 3, 1, 3, 2, 3, 1, 4, 3, 4, 2, 5, 3, 5, 1, 4, 3, 4, 2, 5, 3, 5, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 5, 4, 5, 3, 7, 4, 7, 2, 7, 5, 7, 3, 8, 5, 8, 1, 6, 5, 6, 4, 9, 5, 9, 3, 10, 7, 10, 4, 11, 7, 11, 2, 9, 7, 9, 5, 12, 7, 12, 3, 11, 8, 11, 5, 13, 8, 13, 1, 6
Offset: 1
Examples
The first few fractions are: 1 1 1 1 2 1 2 1 3 2 3 1 3 2 3 1 4 3 4 2 5 3 5 1 4 3 4 2 5 3 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ... 1 2 2 3 3 3 3 4 4 5 5 4 4 5 5 5 5 7 7 7 7 8 8 5 5 7 7 7 7 8 8
Links
- R. Zumkeller, Table of n, a(n) for n = 1..10000
- Johannes Kepler, Harmonices Mundi, Liber III, see p. 27.
- Index entries for fraction trees
- Index entries for sequences related to music
Crossrefs
Programs
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Haskell
{-# LANGUAGE ViewPatterns #-} import Data.Ratio((%), numerator, denominator) rat :: Rational -> (Integer,Integer) rat r = (numerator r, denominator r) data Harmony = Harmony Harmony Rational Harmony rows :: Harmony -> [[Rational]] rows (Harmony hL r hR) = [r] : zipWith (++) (rows hL) (rows hR) kepler :: Rational -> Harmony kepler r = Harmony (kepler (i%(i+j))) r (kepler (j%(i+j))) where (rat -> (i,j)) = r -- Full tree of Kepler's harmonic fractions: k = rows $ kepler 1 :: [[Rational]] -- as list of lists h = concat k :: [Rational] -- flattened a093873 n = numerator $ h !! (n - 1) a093875 n = denominator $ h !! (n - 1) a011782 n = numerator $ (map sum k) !! n -- denominator == 1 -- length (k !! n) == 2^n -- numerator $ (map last k) !! n == fibonacci (n + 1) -- denominator $ (map last k) !! n == fibonacci (n + 2) -- numerator $ (map maximum k) !! n == n -- denominator $ (map maximum k) !! n == n + 1 -- eop. -- Reinhard Zumkeller, Oct 17 2010 -
Maple
M:= 8: # to get a(1) .. a(2^M-1) gen[1]:= [1]; for n from 2 to M do gen[n]:= map(t -> (numer(t)/(numer(t)+denom(t)), denom(t)/(numer(t)+denom(t))), gen[n-1]); od: seq(op(map(numer,gen[i])),i=1..M): # Robert Israel, Jan 11 2016 -
Mathematica
num[1] = num[2] = 1; den[1] = 1; den[2] = 2; num[n_?EvenQ] := num[n] = num[n/2]; den[n_?EvenQ] := den[n] = num[n/2] + den[n/2]; num[n_?OddQ] := num[n] = den[(n-1)/2]; den[n_?OddQ] := den[n] = num[(n-1)/2] + den[(n-1)/2]; A093873 = Table[num[n], {n, 1, 97}] (* Jean-François Alcover, Dec 16 2011 *)
Formula
a(n) = a([n/2])*(1 - n mod 2) + A093875([n/2])*(n mod 2).
a(1) = 1. For all n>0 a(2n) = a(n), a(2n+1) = A093875(n). - Yosu Yurramendi, Jan 09 2016
a(4n+3) = a(4n+1), a(4n+2) = a(4n+1) - a(4n), a(4n+1) = A071585(n). - Yosu Yurramendi, Jan 11 2016
G.f. G(x) satisfies G(x) = x + (1+x) G(x^2) + Sum_{k>=2} x (1+x^(2^(k-1))) G(x^(2^k)). - Robert Israel, Jan 11 2016
a(2^(m+1)+k) = a(2^(m+1)+2^m+k) = A020651(2^m+k), m>=0, 0<=k<2^m. - Yosu Yurramendi, May 18 2016
a(2^(m+1)+k) - a(2^m+k) = a(k) , m >=0, 0 <= k < 2^m. For k=0 a(0)=0 is needed. - Yosu Yurramendi, Jul 22 2016
a(2^(m+2)-1-k) = a(2^(m+1)-1-k) + a(2^m-1-k), m >= 1, 0 <= k < 2^m. - Yosu Yurramendi, Jul 22 2016
a(2^m-1-(2^r -1)) = A000045(m-r), m >= 1, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
a(2^m+2^r) = m-r, , m >= 1, 0 <= r <= m-1 ; a(2^m+2^r+2^(r-1)) = m-(r-1), m >= 2, 0 <= r <= m-1. - Yosu Yurramendi, Jul 22 2016
A093875(2n) - a(2n) = A093875(n), n > 0; A093875(2n+1) - a(2n+1) = a(n), n > 0. - Yosu Yurramendi, Jul 23 2016
Comments