------------------------------------------------------------------------------ http://www.research.att.com/~njas/sequences/a086592.txt ------------------------------------------------------------------------------ An excerpt from the Book III of "The Harmony of the World by Johannes Kepler" (an English translation of "Harmonices Mundi", latin original published at Linz, 1619) translated with an Introduction and Notes by E.J. Aiton, A.M. Duncan, and J.V. Field, published by American Philosophical Society; April 1997, (Memoirs of the American Philosophical Society, Vol 209). On page 163, chapter II of the book III, we have: Corollaries I. The harmonic divisions of a single string are seven in number, not more. II. The expansion of the numbers which are characteristic of divisions occurs in the following manner. To begin with, the whole is expressed in the form of a fraction, that is to say with unity above as numerator, and unity below for denominator. Then each number separately is put as a numerator, and the sum of the two is put as a denominator in each case. Hence from any given fraction two branches arise, until from the sum occurs the number which indicates an unconstructible figure. __ 1/6 ..... 7 1/5 __ .. 5/6 ..... 11 1/4 / .. / 4/5 ............ 9 1/3 \ / \ / 3/4 ................... 7 / 1/1 -- 1/2 . \ . \ 2/5 ................... 7 . \ / . / . 2/3 . \ 3/8 ............ 11 . \ .. . 3/5 . .. . 5/8 ............ 13 . . . . . -- 1/2 Same I found these seven divisions of the string first with hearing as guide, in other words the same number as there are harmonies not greater than a single diapason (= octave). After that I dug out the causes both of the individual divisions and of the number of the total, not without toil, from the deepest fountains of geometry. Let the diligent reader read what I wrote about these divisions 22 years ago in The Secret of the Universe (Mysterium cosmographicum), Chapter XII (*), and ponder how in that passage I was under a delusion about the causes of the divisions and the harmonies, mistakenly striving to deduce their number and the reasons from the number of the five regular solid bodies; whereas the truth is rather that both the five solid figures and the musical harmonies and divisions of the string have a common origin in the regular plane figures. The footnote 58 of the translators: The seven harmonic divisions were illustrated in exactly the same way as here in the Mysterium cosmographicum, Chapter 12. At that time, as he relates, Kepler attempted to derive the harmonies from the regular solids but later found the causes of the harmonies in the constructible polygons. Evidently he had expected to find inspiration in Ptolemy's Harmonica but when he was eventually able to read the work, he found that his own theory of the causes of the harmonies was wholly original. On page 161, Chapter II of the Book III, the footnote 55 of the translators: The foregoing propositions established divisions of the string which produce all the consonances; that is, unison (1:1), octave (1:2), fourth (3:4), fifth (2:3), major third (4:5), minor third (5:6), major sixth (3:5), and minor sixth (5:8). The following propositions will show that there are no further harmonic divisions of the string. This is essential to Kepler's theory, for such further divisions would introduce dissonances. On page 178: "This therefore, is the origin of the dissonant melodic intervals, to which we shall give their names a little later on." Associated footnote 72 by translators: The reader may find it more convenient to have the names at this point. They are major tone (8:9), minor tone (9:10), semitone (15:16), and diesis (24:25). ------------------------------------------------------------------------------ Antti Karttunen's notes: In OEIS-terms we proceed each branch of the fraction-tree A020651/A086592, until we encounter a denominator which belongs into A004169: 7,9,11,13,14,18,19,21,..., "Regular polygons not constructible with ruler and compass". Thus we get 14 fractions in all: allfracs := [1/1,1/2,1/3,2/3,1/4,3/4,2/5,3/5,1/5,4/5,3/8,5/8,1/6,5/6]; We can "coerce" them to the range [1,2[ with the following two functions (written as Maple procedures): invert := r -> 1/r; coerce1st_octave := proc(r) option remember; if(r < 1) then coerce1st_octave(2*r); else if(r >= 2) then coerce1st_octave(r/2); else (r); fi; fi; end; Calling invert, coerce1st_octave and invert in succession, and discarding the duplicates: convert(map(invert,sort(map(coerce1st_octave,map(invert,allfracs)))),set); we get the following set of seven intervals (see the footnote 55 above): {1, 2/3, 3/4, 3/5, 4/5, 5/8, 5/6} --------------------------------- To see that A020651 really gives the numerators, and A086592 which is a bisection of A020650 (or equally: the termwise sum of A020650 and A020651) gives the denominators, we notice that the numerators and denominators for the above tree are given by the following mutually recursive formulae. (We scan its levels from 1/2 onward in top-down fashion, so that the upper edge 1/2, 1/3, 1/4, 1/5, 1/6, ... gets the indices 1,2,4,8,16,...). num(1) = 1 num(2n) = num(n) num(2n+1) = den(n) den(1) = 2 den(2n) = den(n)+num(n) den(2n+1) = den(n)+num(n) On the other hand, David Wilson defines A020650 and A020651 as Numerators and Denominators in recursive bijection from positive integers to positive rationals; where the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1). Here one has to assume that the numerators and denominators are given in their reduced form, i.e. that A020650(n) and A020651(n) are ALWAYS relatively prime. (The terms given in OEIS fulfill this condition, and their definition does not give any reason to suspect otherwise). Then f(n) = A020650(n)/A020651(n) [by definition] f(2n) = A020650(n)/A020651(n) + A020651(n)/A020651(n) f(2n+1) = A020651(2n)/A020650(2n), i.e. A020650(2n+1) = A020651(2n) and A020651(2n+1) = A020650(2n) and because gcd(A020650(n),A020651(n)) is always 1, we have also A020650(2n) = A020650(n)+A020651(n) and A020651(2n) = A020651(n). Thus, summa summarum, the recurrences for A020650 and A020651 are: A020650(1) = 1 A020650(2n) = A020650(n)+A020651(n) A020650(2n+1) = A020651(2n) = A020651(n) A020651(1) = 1 A020651(2n) = A020651(n) A020651(2n+1) = A020650(2n) Furthermore, if we mark f(2n) = X/Y, (with gcd(X,Y) = 1) then f(2n+1) = Y/X and f(4n) = (X+Y)/(Y+Y) and f(4n+2) = f(2(2n+1)) = (Y+X)/(X+X) implying that A020650(4n) = A020650(4n+2) for all n >= 1. Then, let's make an induction hypothesis that den(n) = A020650(2n) up to some value k of n. This implies immediately that num(n) = A020651(n) (up to that same k) and furthermore, we have: den(2n) = den(n)+num(n) [by den's definition] = A020650(2n)+A020651(n) [by the ind. hypoth. and its implication] = A020650(2n)+A020651(2n) [by A020651's definition] = A020650(4n) [by A020650's definition] den(2n+1) = den(n)+num(n) [by den's definition] = A020650(4n) [... as before] = A020650(4n+2) [proved above] Thus if we know that our hypothesis is true for all n's in range [1,k] then it will be true also for all n's in range [k+1,2k+1], and this observation completes the induction. Q.E.D. by Antti Karttunen, August 27 2003. ------------------------------------------------------------------------------