# «The Collatz Scale» by henklass

on 08 Jan'14 08:37 inSome time ago a did a thing on a series of numbers, that, as I learned later, was officially called the Collatz problem and it can be found on the Online Encyclopedia of Integer Sequences: http://oeis.org/A070165 . One day I was playing with those numbers on a keyboard and I had the impression, that some notes occur more frequent than others. This implies some kind of tonality. So I rewrote the code to use notenumbers in stead of frequencies. And here is the result. The fourth step, which coincidentally is a C, is dominant and the 10th, F# follows. There is quite some A, B flat, C#, E and G and very little E flat. A flat, B, D, and F are almost absent. So I guess the Collatz scale is: C, C#, E, F#, G, A, B flat, C and you can use E flat as some kind of Blue note or something.

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//series of numbers:Collatz-scale: http://oeis.org/A070165//from1to100//while number>1//if even: divide by2, else multiply by3and add1//remember highest value, remember longest series, both with their startnumber//startnumber->cutoff, number-> frequency, pan mid//startnumber->cutoff, highest value -> frequency, pan right//startnumber->cutoff, longest series -> cutoff, pan left//Thereseemd to be some tonality in the numbers when used asMIDI-notenumbers//Tofind that out you can count the occurrences of all the numbers wrapped between0en11Server.default.boot; (SynthDef(\beep, { |freq=440, cutoff=1000, pan=0|Out.ar(0, (Pan2.ar(LPF.ar(VarSaw.ar(freq, width:0.01), cutoff), pan))*0.4)}).send(s); ) (varhighest=1, longest=1, starthighest=1, startlongest=1;varnumberbeep, highestbeep, longestbeep; u=Array.new(12);//array to store the numbers, wrapped to{for(0..110,12, {u.add(0)})}.value; numberbeep=Synth(\beep, [\freq,200,\cutoff,200,\pan,0]); longestbeep=Synth(\beep, [\freq,200,\cutoff,200,\pan, -1]); highestbeep=Synth(\beep, [\freq,200,\cutoff,200,\pan,1]);Routine({ for (1,100, {argstartnumber;varnumber, length=1, maximum; number=startnumber; maximum=startnumber; u[startnumber%12]=u[startnumber%12]+1; (number.asString+" ").post; numberbeep.set(\cutoff,100*startnumber); while ( {number>1},{ if (number.asInteger.even, {number=(number/2)}, {number=3*number+1}); length=length+1; if (number>maximum, {maximum=number}); numberbeep.set(\freq, (32+(number%96)).midicps);//frequency aslongestbeep.set(MIDI-notenumber between32and128, that is8octaves\freq, (32+(length%96)).midicps); highestbeep.set(\freq, (32+(maximum%96)).midicps); u[number%12]=u[number%12]+1;//count the occurrences ofthisnumber wrapped between0and110.1.wait; }); if(length>longest, {longest=length; startlongest=startnumber}); if (maximum>highest, {highest=maximum; starthighest=startnumber}); longestbeep.set(\cutoff,100*longest); highestbeep.set(\cutoff, highest*2); ("length: " + length.asString).post; (" maximum: " + maximum.asString).postln; }); postf("longest series % at %\n", longest, startlongest); postf("highest value % at %\n", highest, starthighest); numberbeep.set(\freq,33.midicps,\cutoff,10000); longestbeep.set(\freq, ((32+longest)%96).midicps,\cutoff,100*startlongest); highestbeep.set(\freq, ((32+highest)%128).midicps,\cutoff,100*starthighest);10.wait; numberbeep.free; highestbeep.free; longestbeep.free; }).play; ) (/*u.plot(bounds:Thisshows theCollatz-scale:Thefourth step, which coincidentally is a C, is dominant and the10th, F# follows.Thereis quite some A, B flat, C#, E and G and very little E flat. A flat, B, D, and F are almost absent.SoI guess theCollatzscale is: C, C#, E, F#, G, A, B flat, C and you can use E flat as some kind ofBluenote or something.*/Rect(10,100,400,400));//show as graphfor(0,12, {argi;//show as listif( (u[i] >0), {(i.asString+" ").post; u[i].postln}) } ); ) numberbeep.set(\freq, (32+(0%96)).midicps); {VarSaw.ar(440, [0.1,1],0.05,1,0)}.scope