A133269 - OEIS

%I #30 Aug 27 2017 23:40:14

%S 1,5,8,12,5,9,12,4,8,12,3,7,12,4,7,11,5,9,12,4,9,1,4,8,12,4,7,11,4,8,

%T 11,3,8,12,3,7,12,4,7,11,3,7,10,2,7,11,2,6,12,4,7,11,4,8,11,3,7,11,2,

%U 6,11,3,6,10,5,9,12,4,9,1,4,8,12,4,7,11,4,8,11,3,9,1,4,8,1,5,8,12,4,8,11,3,8

%N Fractal sequence consisting of quadruples, each of which contains the tones comprising a major 7th chord (i.e., root, major third, fifth, and major 7th), with the tones in an octave assigned to the numbers 1..12, and with the n-th quadruple using a(n) as its root.

%C A major 7th chord consists of its root tone and the tones that lie 4, 7, and 11 semitones (a.k.a. half tones or half steps) above the root.

%C Starting from root tones at C, C#, D, D#, E, F, etc., wrapping around the tones higher than an octave (i.e., 13->1, 14->2, etc.), and assigning numbers 1 to 12 to the half tones within the octave, the major 7th chords (4 tones each) are represented as 1-> {1, 5, 8, 12}, 2-> {2, 6, 9, 1}, 3-> {3, 7, 10, 2}, 4-> {4, 8, 11, 3}, 5-> {5, 9, 12, 4}, 6-> {6, 10, 1, 5}, 7-> {7, 11, 2, 6}, 8-> {8, 12, 3, 7}, 9-> {9, 1, 4, 8}, 10-> {10, 2, 5, 9}, 11-> {11, 3, 6, 10}, 12-> {12, 4, 7, 11}.

%C The sequence starts with the chord at root 1, which defines a(1)..a(4), and then appends iteratively in groups of 4 the chords at base a(k), k = 2, 3, 4, ... to the sequence.

%C Note that construction of a chord in terms of the intervals from its root to each of the chord's constituent notes is simpler if we instead number the notes in an octave from 0 through 11, rather than from 1 through 12 (and a sequence built this way would also have a simpler formula, not requiring us to add 1 after taking modulo-12 residues). (See the link to the Wikipedia article, which says that a major 7th chord "can be represented by the integer notation {0, 4, 7, 11}.) - _Jon E. Schoenfield_, Aug 27 2017

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Major_seventh_chord">Major seventh chord</a>

%F From _Jon E. Schoenfield_, Aug 27 2017: (Start)

%F a(1) = 1, a(2) = 5, a(3) = 8, a(4) = 12;

%F for k > 1,

%F    a(4k - 3) =  a(k);

%F    a(4k - 2) = (a(k) +  3) mod 12 + 1;

%F    a(4k - 1) = (a(k) +  6) mod 12 + 1;

%F    a(4k)     = (a(k) + 10) mod 12 + 1. (End)

%e From _Jon E. Schoenfield_, Aug 27 2017: (Start)

%e If we assign the number 1 to the note C in some octave (e.g., middle C) and assign the numbers 2, 3, 4, ..., to the notes reached as we move up the scale by half tones, then the C notes at successive octaves will be assigned the numbers 1, 13, 25, ... (see the illustration below, which is intended to represent a portion of a standard musical keyboard).

%e   |  |C| |D|  |  |F| |G| |A|  |  |C| |D|  |  |F| |G| |A|  |

%e   |  |#| |#|  |  |#| |#| |#|  |  |#| |#|  |  |#| |#| |#|  |

%e   |  |/| |/|  |  |/| |/| |/|  |  |/| |/|  |  |/| |/| |/|  |

%e   |  |D| |E|  |  |G| |A| |B|  |  |D| |E|  |  |G| |A| |B|  |

%e   |  |b| |b|  |  |b| |b| |b|  |  |b| |b|  |  |b| |b| |b|  |

%e   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |

%e   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |

%e   | C | D | E | F | G | A | B | C | D | E | F | G | A | B | C

%e    --- --- --- --- --- --- --- --- --- --- --- --- --- --- --

%e                         1 1 1   1 1 1 1 1   1 1 2 2 2 2 2   2

%e     1 2 3 4 5   6 7 8 9 0 1 2   3 4 5 6 7   8 9 0 1 2 3 4   5

%e .

%e Then the sequence consists of quadruples constructed as follows:

%e - the values of this sequence's first quadruple, a(1)..a(4) (i.e., 1, 5, 8, and 12) give a C major 7th chord (the tones C, E, G, B);

%e - the sequence's 2nd quadruple, a(5)..a(8), starts from a(5) = a(2) = 5 and thus gives an E major 7th chord (i.e., 5, 9, 12, 16, but the 16 is wrapped down an octave to 16-12 = 4);

%e - the sequence's 3rd quadruple, a(9)..a(12), starts from a(9) = a(3) = 8 and thus gives a G major 7th chord (8, 12, 15->3, 19->7);

%e - the sequence's 4th quadruple, a(13)..a(16), starts from a(13) = a(4) = 12, giving a B major 7th chord (12, 16->4, 19->7, 23->11);

%e - in general, the sequence's k-th quadruple, a(4k-3)..a(4k), starts with a(4k-3) = a(k), and its three remaining terms are a(k)+4, a(k)+7, and a(k)+11 (with any that exceed 12 decreased by 12, i.e., by one octave).

%e Thus, the sequence is fractal: keeping a(1) and every 4th term after it, and discarding the rest, yields the original sequence. (End)

%t Clear[s, p] s[i_] = {i, If[i + 4 > 12,i - 8, i + 4], If[i + 7 > 12, i - 5, i + 7], If[i + 11 > 12, i - 1, i + 11]}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]]; p[4]

%Y See A133270 for a companion sequence which provides more explanation.

%K nonn,easy,tabf

%O 1,2

%A _Roger L. Bagula_, Oct 16 2007

%E Comments inserted for clarification - The Assoc. Eds. of the OEIS - Aug 29 2010

%E Further edits from _Jon E. Schoenfield_, Aug 27 2017