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Message: 826 - Contents - Hide Contents Date: Thu, 23 Aug 2001 18:15:49 Subject: Re: Tetrachordal alterations From: Paul Erlich --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:>> I might not call these tetrachordal. According to my paper, the >> disjunction is supposed to contain some pattern of scale steps found >> with the tetrachords themselves. But in this case, the 5/72 interval >> isn't found in the tetrachords. >> >> Perhaps we can call these "weakly tetrachordal". >> Hmm. So in the case where the disjunction consists of a single step, > you insist that that step (always an approx 8:9) appears in the > tetrachords, before calling a scale tetrachordal? Right. > > Isn't this just an additional desirable property because it tends to > make the scale tetrachordal in more rotations. No. > > "Weakly tetrachordal" is still tetrachordal, right? Weakly. > Wouldn't it be > better if we simply said "tetrachordal in n rotations". Of course the > minimum for n is 3 (other than 0). And when n is the number of notes > in the scale it is omnitetrachordal.This can all still be "weakly" or "strongly".> > I'll abbreviate both tetrachord and tetrachordal as "Tc" in future I'm > getting tired of typing it. > > I suppose one could have the disjunction containing only intervals > from the Tc but still only have 3 Tc rotations.Yes, like Mohajira.> But is the important > thing about this the minimising of different step sizes, rather than > having the steps in the disjunction the same as in the Tc?I think the disjuction should smoothly connect the tetrachords, rather than sounding "different".
Message: 827 - Contents - Hide Contents Date: Thu, 23 Aug 2001 18:20:10 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote: >>> What makes the apparent relationship between MOS and >> tempered periodicity blocks hard to fathom ... >> Well, tastes differ. I certainly hope your definition is correct,It's almost correct. There can be an interval of repetition which is 1/P of the interval of equivalence, with P an integer.> because: > > (1) It makes sense > > (2) It certainly seems like a condition we would want a scale with > one generator in the octave to satisfy > > (3) It explains why Paul's hypothesis is true. > > For the last, note if for instance we want a q-step scale generated > by interating the p-th step, where p/q is approximately log_2(3/2), > so that fifths are represented by p steps in the q-division, then if > x is log_2 of our approximation for the fifth, we want p/q to be a > semiconvergent of x. > > Now, if memory serves, if |x - p/q| < 1/q^2 then p/q is a > semiconvergent for x, and the converse is "almost" true. The details > don't much matter, unless you wanted to write this up for > publication, because it is clear that if x is anything in a certain > interval around p/q, then p/q will be a semiconvergent for it. If p/q > is a good enough approximation for log_2(3/2), say one that satisfies > |log_2(3/2) - p/q| < 1/(2*q^2), then anything close to log_2(3/2) > will be close enough to p/q so that p/q must be a semiconvergent for > it. Since the approximation x ought to be better than that given by > p/q, we would expect to find (and will find, if we wrote this up with > the details worked out as if for publication) that it gives a MOS.Lost me there. Did you see my "proof" of the hypothesis?
Message: 828 - Contents - Hide Contents Date: Thu, 23 Aug 2001 18:22:15 Subject: Re: Chromatic = commatic? From: Paul Erlich --- In tuning-math@y..., graham@m... wrote:> In-Reply-To: <9m1338+rgjv@e...> > Paul wrote: >>>> This is because you're expecting to get an octave-specific result. >> The>>> octave-equivalent equivalent of an equal temperament is an MOS. >>>> In what way? What do you mean by that? That certainly doesn't seem to >> be true. >> There both one dimensional sets of points.Yes, but it seemed you were talking about a more "specific" type of "equivalence".> >>> (Or>>> family of MOS scales, there's nothing special about the particular >> numbers>>> of generators that give two interval sizes.) >>>> As opposed to the ones that don't?? >> Not for these purposes.Well I disagree. The hypothesis, in fact, is all about showing that there _is_ something special about the particular numbers of generators that give two interval sizes, as opposed to those that don't.
Message: 829 - Contents - Hide Contents Date: Thu, 23 Aug 2001 18:28:24 Subject: Re: Chromatic = commatic? From: Paul Erlich --- In tuning-math@y..., graham@m... wrote:> In-Reply-To: <9m12rc+9cqs@e...> > Paul wrote: >>>> The octave-equivalent algebra doesn't distinguish the different >> repeating >>> blocks. >>>> Sure it does. The symmetrical decatonic scale (LssssLssss) has 10 >> notes -- two repeating blocks per octave. >> But how can you say "per octave" in a system that doesn't recognize > octaves?What do you mean, doesn't recognize octaves? Octave-equivalence is assumed right from the beginning, in the construction of the 7-limit periodicity block.>>>> The linear temperament result tells you the number of >>> times the period goes into the interval of equivalence (octave) >>>> Exactly! So the WF definition, where the period _is_ the interval of >> equivalence, won't do. >> Only if their definition of "interval of equivalence" is the same as > yours. Whatever the definition, I don't expect they intended to exclude > this case.Then why on earth didn't they list the Messaien scale and the diminished (octatonic) scale and the augmented (hexatonic) scale, etc. in their list?? Certainly the list was geared toward those thinking about a tuning of 12 notes per _octave_.> > >>> and the>>> mapping in terms of generators. Again, you need the octave- >> specific>>> algebra, or the metric, to get the mapping by generators *and* >> periods.>>> Effectively, the octave-equivalent algebra collapses so that the >> interval>>> of equivalence becomes the period in any given context. >>>> What do you mean? For the symmetrical decatonic, the interval of >> equivalence remains _two_ periods whether you're looking octave- >> equivalently or not. >> It doesn't look that way to me. That scale can be defined as a > periodicity block by > > [-1 2 0] > [ 0 2 -2] > [-2 0 -1] > > The adjoint is > > [-2 2 -4] > [ 4 1 -2] > [ 4 -4 -2] > > The left hand column defines the generator mapping, which we can write as > [ 1] > -2*[-2] > [-2] > > The minus sign isn't important. The 2 tells us the new interval of > equivalence is half the old one.That's your interpretation. But it isn't correct! We have 10 equivalence classes in this group, and this half-octave takes you from one equivalence class to another -- not to the same one.> The > > [ 1] > [-2] > [-2] > > gives us the mapping by the generator modulo this new interval of > equivalence. 16:15 is [-1 -1 0] times > > [ 1] > [-2] > [-2] > > or 1 generator. 3:2 is [1 0 0] or 1 generator. Where does the algebra > tell us these two intervals are not equivalent?Where does it tell us they are? Yes, they're _both_ generators. Because of the symmetry, there is more than one possible generator. Think of a prime-numbered ET -- how many generators does that have?
Message: 830 - Contents - Hide Contents Date: Thu, 23 Aug 2001 18:30:16 Subject: Re: Generators and unison vectors From: Paul Erlich --- In tuning-math@y..., graham@m... wrote:> In-Reply-To: <9m12f6+qhkq@e...> > Paul wrote: >>>> You can't define the 5:4 approximation by adding octaves and >> twelfths. >>>> Why not? We've been doing so all along. The 5:4 approximation is four >> twelfths minus six octaves. >> Do you remember, back when you were at school, a distinction being made > between "adding" and "subtracting"?Ha ha. I thought we weren't scared of negative numbers anymore.>>>> Probably it'd be better written as >>> >>> v v v v v v v v v v v v p >>> >>> where v is a fifth and p is a Pythagorean comma. That means p is >> the>>> (chromatic) unison vector. That leaves v as "the thing you have >> left when>>> you take out all the unison vectors". That's what we need a name >> for. >> >> The generator? >> Yes, isn't that where I started? In octave-specific terms, when you take > out 25:24 and 81:80 you're left with a generator equivalent to 16:15. In > octave-invariant terms, when you take out 81:80 you're left with a > generator of a fifth. In octave-specific terms, when you take out only > 81:80 you have a choice of generators. They could be 2:1 and 3:2 or 16:15 > and 25:24. Taking 2:1 and 25:24 as a pair is different, because the > generators and unison vectors don't form a unitary matrix. So perhaps > that's why we call one an "interval of equivalence" and the other a > "chromatic unison vector".You lost me. Gene, can you shed any light?
Message: 832 - Contents - Hide Contents Date: Thu, 23 Aug 2001 18:38:01 Subject: Re: EDO consistency and accuracy tables (was: A little research...) From: Paul Erlich --- In tuning-math@y..., BobWendell@t... wrote:> Never mind, Dave! Just got there through the menu (duh!). there are > no column labels. I inferred from the structure that the consisitency > tables showed levels of consistency through successive odd limits? Right. > Apparently the accuracy table only shows the accuracy for the > intervals that are consistent in that temperament? Right. > > Also what does the number for consistency levels mean?A consistency "level" of N, for N > 1.5, simply means that the largest error is 1/2N steps of the ET. Paul Hahn had something else in mind but this is the way I think of it, since it's mathematically equivalent.
Message: 833 - Contents - Hide Contents Date: Thu, 23 Aug 2001 19:34:17 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@j... --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> Lost me there. Did you see my "proof" of the hypothesis?No--where is it hiding?
Message: 834 - Contents - Hide Contents Date: Thu, 23 Aug 2001 20:04:31 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: >>> Lost me there. Did you see my "proof" of the hypothesis? >> No--where is it hiding?Gene, I'm getting the feeling that somehow you're missing a good number of my posts. Like the chromatic vs. commatic unison vector deal . . . I explained that quite a number of times before you seemed to acknowledge it. Anyway, as I tried to tell you before, my sketch of a proof is in message #591.
Message: 835 - Contents - Hide Contents Date: Thu, 23 Aug 2001 20:06:01 Subject: Re: EDO consistency and accuracy tables (was: A little research...) From: Paul Erlich I wrote,> A consistency "level" of N, for N > 1.5, simply means that the > largest error is 1/2N steps of the ET.Should be "the largest error is _less than_ 1/2N steps of the ET".
Message: 837 - Contents - Hide Contents Date: Thu, 23 Aug 2001 21:18:22 Subject: Re: EDO consistency and accuracy tables (was: A little research...) From: Paul Erlich --- In tuning-math@y..., BobWendell@t... wrote:> So 111-EDO (3*37 divisions) looks like it wins hands down on both > consistency and accuracy?Well sort of . . . but consider that 120-tET _can't_ have more than a 5 cent error for _anything_ (since it's increments of 10 cents). So one would have hoped for a "special" tuning with much fewer than 120 notes to satisfy your criteria. Unfortunately there isn't one.> > By the way, 72-EDO does indeed look compelling, but 13/8 (50 steps) > is flat by 7.2 cents, so it doesn't meet the criteria I specified. > Considering its other merits and the high order prime 13, I suppose > 7.2 cents ain't half bad!Right . . . but I think the whole idea of 72 or 111 or 121 as a least- common-denominator way of describing ideal musical practice kind of falters when _adaptive JI_ comes into the picture . . . doesn't it?
Message: 838 - Contents - Hide Contents Date: Thu, 23 Aug 2001 21:36:56 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@j... --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> Gene, I'm getting the feeling that somehow you're missing a good > number of my posts. Like the chromatic vs. commatic unison vector > deal . . . I explained that quite a number of times before you seemed > to acknowledge it.Sorry about that. I did read that message before, but that was in an attempt to understand definitions--what is a MOS, hyper-MOS, PB, unison vector and so forth. The definitions are gradually becoming clear, so these things began to make sense to me. You were discussing the two-scale-step condition, which made it sound to me as if the whole business was defined in terms of large and small steps, without reference to tuning. I like the semiconvergents at this point!
Message: 839 - Contents - Hide Contents Date: Thu, 23 Aug 2001 21:50:03 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> You were discussing the two-scale-step condition, which made it sound > to me as if the whole business was defined in terms of large and > small steps, without reference to tuning. I like the semiconvergents > at this point!In my proof, I actually get at _two different_ (but mathematically equivalent) definitions of MOS. One concerns not a "two-scale-step" condition, as you say, but rather a "Myhill" condition, which says that every generic interval, not just steps but _any_ interval, aside from the interval of repetition, will come in exactly two step sizes. But the other definition of MOS I was trying to get at in my proof is, I believe, the "semiconvergent" one. I'm talking about vectors which have length M/N in the direction of the period boundaries. This is clearly very closely related (much more so than the Myhill stuff) to the continued fraction approximation process. Do you see that? Anyway, if you're beginning to think the hypothesis is true, then for any set of n-1 (or n-2, if we're counting your way) commatic unison vectors, you should be able to find the generator of the resulting linear temperament. Do you see a direct mathematical way of determining what this generator is?
Message: 840 - Contents - Hide Contents Date: Thu, 23 Aug 2001 21:56:51 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich I wrote,> > In my proof, I actually get at _two different_ (but mathematically > equivalent) definitions of MOS. One concerns not a "two-scale-step" > condition, as you say, but rather a "Myhill" condition, which says > that every generic interval, not just steps but _any_ interval, aside > from the interval of repetition, will come in exactly two step sizes.Ack! You got me doing it! I meant _two sizes_ at the end there, scratch the word "step"!
Message: 842 - Contents - Hide Contents Date: Thu, 23 Aug 2001 22:17:25 Subject: Re: Now I think "the hypothesis" is true :) From: genewardsmith@j... --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: One concerns not a "two-scale-step">> condition, as you say, but rather a "Myhill" condition, which says >> that every generic interval, not just steps but _any_ interval, > aside>> from the interval of repetition, will come in exactly two step > sizes.> Ack! You got me doing it! I meant _two sizes_ at the end there, > scratch the word "step"!Let's see if I can phrase this purely in terms of steps, without reference to sizes: Let S be a sequence consisting of steps of sizes L and s, which is periodic with period P. Every set of N contiguous steps will consist of p L's and q s's, such that p+q = N. Then the Myhill' condition says that exactly two counts of L's, p1 and p2, occur for every N. (Hence of course only two counts of s's, so that we have p1+q1 = N and p2+q2 = N.) Is the Myhill' condition the same as the Myhill condition?
Message: 843 - Contents - Hide Contents Date: Thu, 23 Aug 2001 23:36:04 Subject: Re: EDO consistency and accuracy tables (was: A little research...) From: Paul Erlich --- In tuning-math@y..., BobWendell@t... wrote:> > I don't have the experience with these tunings in practice to address > that, Paul. Not by a long shot (how about zero?) As far as a daptive > JI, you may be right. (How would I know?)Well clearly you understand the Vicentino adaptive JI scheme, based on two 1/4-comma meantone chains. Since 53-tET is a "scale of commas", you would need 53*4=212-tET to implement this scheme. See?> But I was looking at it as > more of a potential compositional tool for microtonal polyphony.If you don't care about melody or "necessarily tempered chords" (such as CEGAD) and only care about just-style harmony, then sure, that's great thinking. Looking forward to hearing some harmonically-oriented 111-tET music!
Message: 844 - Contents - Hide Contents Date: Thu, 23 Aug 2001 23:39:46 Subject: Re: Now I think "the hypothesis" is true :) From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote: > > One concerns not a "two-scale-step">>> condition, as you say, but rather a "Myhill" condition, which > says>>> that every generic interval, not just steps but _any_ interval, >> aside>>> from the interval of repetition, will come in exactly two step >> sizes. >>> Ack! You got me doing it! I meant _two sizes_ at the end there, >> scratch the word "step"! >> Let's see if I can phrase this purely in terms of steps, without > reference to sizes: > > Let S be a sequence consisting of steps of sizes L and s, which is > periodic with period P. Every set of N contiguous steps will consist > of p L's and q s's, such that p+q = N. Then the Myhill' condition > says that exactly two counts of L's, p1 and p2, occur for every N. > (Hence of course only two counts of s's, so that we have p1+q1 = N > and p2+q2 = N.) Seems correct. > > Is the Myhill' condition the same as the Myhill condition?Almost. The difference is that there may be an interval of repetition, which is exactly a factor of the number of steps per interval of equivalence, for which there may be only one count of L's. For example, for the scale LssssLssss, if N=5, then p1 always equals 1 and p2 always equals 4.
Message: 845 - Contents - Hide Contents Date: Fri, 24 Aug 2001 01:31:25 Subject: Re: Now I think "the hypothesis" is true :) From: Dave Keenan --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> --- In tuning-math@y..., genewardsmith@j... wrote:>> --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote: >>>>> What makes the apparent relationship between MOS and >>> tempered periodicity blocks hard to fathom ... >>>> Well, tastes differ. I certainly hope your definition is correct, >> It's almost correct. There can be an interval of repetition which is > 1/P of the interval of equivalence, with P an integer.Huh? I included that. I used "i" where you've used "P" and be careful because I used "P" to mean something else, i.e. the interval of repetition (for which "period" is a shorter word) as a frequency ratio p = log(P). And i is not just any integer, it's a whole-number (not zero, not negative). -- Dave Keenan
Message: 846 - Contents - Hide Contents Date: Fri, 24 Aug 2001 01:51:45 Subject: Re: Tetrachordal alterations From: Dave Keenan --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:>> Hmm. So in the case where the disjunction consists of a single > step,>> you insist that that step (always an approx 8:9) appears in the >> tetrachords, before calling a scale tetrachordal? > > Right.Does John Chalmers agree with this?
Message: 847 - Contents - Hide Contents Date: Fri, 24 Aug 2001 02:30:41 Subject: Re: Now I think "the hypothesis" is true :) From: Dave Keenan --- In tuning-math@y..., genewardsmith@j... wrote:> --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote: >>> What makes the apparent relationship between MOS and >> tempered periodicity blocks hard to fathom ...It's not so hard as I first thought. In MOS we are dealing with the log of the generator over the log of the period. In PB's we are dealing with vectors of logs (except they are all logs to different prime bases).> Well, tastes differ. I certainly hope your definition is correct, > because: ... > (3) It explains why Paul's hypothesis is true. > > For the last, ...Gene, you seem to be only proving (or making plausible) that cardinality of MOS = denominator of convergent or semiconvergent. While it's nice to have you confirm it, I believe Carey and Clampitt proved this in 1989 and we have accepted it. This is not Paul's hypothesis (or should it be called conjecture, I thought hypotheses were for science, not math?). -- Dave Keenan
Message: 848 - Contents - Hide Contents Date: Fri, 24 Aug 2001 04:38:40 Subject: Re: Tetrachordal alterations From: Paul Erlich --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:>> --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:>>> Hmm. So in the case where the disjunction consists of a single >> step,>>> you insist that that step (always an approx 8:9) appears in the >>> tetrachords, before calling a scale tetrachordal? >> >> Right. >> Does John Chalmers agree with this?Oops! OK, let's call his "tetrachordal" and mine "strongly tetrachordal" or something.
Message: 849 - Contents - Hide Contents
Date: Fri, 24 Aug 2001 04:40:51
Subject: Jacks
From: genewardsmith@j...
Let me start by reviewing Farey sequences. The first row of the Farey
sequence is [0/1, 1/1] and the nth row is obtained by inserting the
fraction (p1+q1)/(p2+q2) between p1/q1 and p2/q2 if it is in reduced
form (that is, if gcd(p1+q1, p2+q2)=1) and if q1+q2 <= n. Hence the
second row is [0/1, 1/2, 1/1] and so forth.
Now suppose r = u/v is any rational number 0<r<=1, where u/v is in
reduced form. We can define a sequence s_i by setting s_0 = a_0/b_0
to be the Farey fraction adjacent to and below r in the vth row Farey
sequence and setting
s_{i+1} = a_{i+1}/b_{i+1} = (u+a_i)/(v+b_i),
so that s_i is the sequence of Farey fractions approaching r from
below.
Now let us define a function B for any positive integer n and any
rational number 0<r<1, and for any for any integer n>1 when r=1, by
setting
B(r, n) = s_n/s_{n-1}.
If p1/q1 and p2/q2 are two successive Farey fractions, then p2q1 -
p1q2 = 1, which implies p2q1/p1q2 is a superparticular ratio; hence B
takes values in superparticular ratios--that is in numbers of the
form n/(n-1). In exactly the same way, if r is a rational number
0<r<1 and n is a positive integer, we can define a function A(r, n)
of superparticular ratios obtained from Farey fractions approaching r
from above. Since the numerators and denominators of the sequences
s_i used to define A and B grow linearly, the numerators and
denominators of the values of A and B grow quadratically as a
function of n.
Some particular examples of this are
B(1, n) = n^2/(n^2-1) = n^2/(n+1)(n-1)
A(1/2, n) = (2n^2+n)/(2n^2+n-1) = n(2n+1)/(n+1)(2n-1)
B(1/2, n) = (2n^2+3n+1)/(2n^2+3n) = (n+1)(2n+1)/n(2n+3)
A(2/3, n) = (6n^2-n-1)/(6n^2-n-2) = (2n-1)(3n+1)/(2n+1)(3n-2)
B(2/3, n) = (6n^2+n-1)/(6n^2+n-2) = (2n+1)(3n-1)/(2n-1)(3n+2)
The numerators of B(1,n) are squares, and we get the sequence 4/3,
9/8, 16/15, 25/24, 36/35, ... . from it. The numerators of A(1/2, n),
when reduced to lowest terms, are triangular numbers for figurate
triangles with even sides; that is numbers 1/2 x (x+1) where x is
even. It gives us the sequence 3/2, 10/9, 21/20, 36/35, ... where we
might note that 36, which is both a square and a triangular number,
appears on both lists. The numerators of B(1/2, n) are triangular
numbers for triangles with odd size, giving us the sequence 6/5,
15/14, 28/27 and so forth.
These functions lead a multitude of multiplicative relationships,
including
B(1, n) = A(1/2, n)B(1/2, n-1)
A(1/2, n) = B(1, 2n)B(1, 2n+1)
B(1/2, n) = B(1, 2n+1)B(1, 2n+2)
Since this posting was inspired by one of Paul's which mentioned
Blackjack scales, I am going to propose the following definition:
A *jack* is a number which is the value of A(r, n) for n positive and
1/2 <= r < 1, or B(r, n) for n positive and 1/2 <= r <= 1.
(Since this is not being submitted to a peer-reviewed journal, no one
can stop me! Aint the Internet wonderful?)
It can happen that two jacks (n+1)/n and n/(n-1) are adjacent Farey
fractions. When this happens, the ratio of these two jacks is also a
jack, which "jumps" from one jack to another. This leads to the
following definition:
A *jumping jack* is the ratio of two jacks (n+1)/n and n/(n-1) which
are adjacent Farey fractions, so that the jumping jack is B(1, n) =
n^2/(n^2-1). Examples are 81/80 and 2401/2400 where the numerator is
the square of a square--ie, a fourth power--and 225/224, where the
numerator is the square of a triangular number.
Finally, we can have a jack which is defined as a jack in more than
one way. To weed out uninteresting cases, let us consider only A(p/q,
n) and B(p/q, n) where n>=q. We then may define a *high jack* as a
jack which is a jack in two different ways, with the condition on n
above for both of the ways. Since 36 is both the 8th triangular and
6th square, it defines the high jack 36/35.
When we focus on one particular type of jumping jack or high jack, we
can determine all jacks of that type by solving a Pell's equation,
which is a Diophantine equation (an equation for which we seek
solutions in the integers) of the form x^2 - dy^2 = N, where d and N
are fixed positive integers and d is square-free. We then seek all
solutions in pairs x, y of integers. In this way we can determine a
formula (and a recurrence relationship) for all square numbers which
are one less than a triangular number, and so forth.
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