About Triangular Square Numbers

        One might call these prime numbers the p parameter, such that either

t(N) = p2(N)                 or                t(N) = p2(N) - 1.

Then I wanted to find out how p(N) was increasing.  I hypothesized that p(N) increased geometrically, so I found the ratios between successive p's:

3/1 = 3

7/3 = 2.333…

17/7 = 2.4285714285…

41/17 = 2.41176470588…

Notice that these ratios successively alternate between being either larger than  or smaller than .

        Let N1, N2, N3, … be the sequence of triangular squares.  Let Ni and Ni+1 be a pair of successive triangular squares, such that

t(Ni) = p2(Ni) - 1,

t(Ni+1) = p2(Ni+1);

then the index (subscript) i will be even (whenever the two conditions are met), and it will be found that

.

For example, N2 = 36 and N3 = 1225, s(N2)=6, t(N2)=8,

and

The only other example possibly -- so far -- is with N4=41616 and N5=1413721.  Then s(N4)=204, t(N4)=288,

and

However, since the last triangular square found by the calculator has an odd index (N5 has index 5), the rule just discovered cannot be applied to find p(N6), which would yield t(N6), which would yield N6.

        So some other rule has to be found (induced) which should be applicable also to triangular squares with odd index.  I found (induced) such a rule.  Here it is:

Only there is a problem: it should not be possible to generate prime numbers by means of a finite, bounded, and fixed algebraic formula.  Actually it turns out that p(N) is not, generally, a prime: the first five might just be accidental.

        The last rule can be restated:

Then, since p(N5) = 41, p(N4) = 17, and p(N3) = 7, it follows that

99=9⋅11, it is not a prime number.

        Since the index, 6, is even, it follows that

t(N6) = p2(N6) - 1

         = 992 - 1

         = 9800.

So N6 is the 9800th triangular number, viz.

,

,

so N6 is the 6930th perfect square.

it is beginning to look like

1.4142… = .

        Now one can apply the first rule:

to find p(N7),

Now one can apply the second rule to see if it yields the same result:

Voila!  The same result.

        Since the index, 7, is odd, it follows that

t(N7) = p2(N7)

         = 2392

         = 57121.

So N7 is the 57121th triangular number, viz.

,

and

,

so N7 = 1631432881 is the 40391th perfect square, and it is the seventh triangular square.

        Repeating the paradigm shown above, it can be found that the eighth triangular square is:

N8 = 55420693056,

that it is the square of 235416, and that it is the 332928th triangular number.  Also,

which is closer to  = 1.41421356237…

        My conjecture is that this algorithm generates a countable infinity of triangular squares, and that any and all triangular squares which exist will be generated by this algorithm.

        But I have not proven that the algorithm works, nor do I know why it works, nor have I proven that there are no triangular squares lurking somewhere in the numerical space between, say, N7 and N8, or generally, between Ni and Ni+1.  I still have faith in this algorithm, though.

        NOVAM RATIŌNEM NUMERŌRUM HODIĒ DISCĒBAM.

January 20th

I applied the G-transformation to the finite difference formula

,

and letting x0 = 1, x1 = 1, x2 = 3, solved for xn, obtaining

But since p(N) has the property that

,

with p(N1) = 1, p(N2) = 3, p(N3) = 7, it follows that

.

Also, I found a general formula for t in terms of p; viz.

.

It follows that

Note that: i  is not -- here -- the square root of -1, but just an index variable (or a pointer) as used in computer science.

        Then one can derive the general formula for the ith triangular square:

.

This simplifies to

.

        I just induced another relation: notice that p(Ni) always divides into s(Ni).  E.g.

,

,

, etc.

Here is a list of the s/p ratios for the first eight triangular squares:

i                p(Ni)                s(Ni)                s/p

1                1                1                1

2                3                6                2

3                7                35                5

4                17                204                12

5                41                1189                29

6                99                6930                70

7                239                40391                169

8                577                235416        408

Looking at the  column, notice that

5 = 2(2) + 1,

12 = 2(5) + 2,

29 = 2(12) + 5,

70 = 2(29) + 12 = 58 + 12,

169 = 2(70) +29 = 140 + 29,

408 = 2(169) + 70 = 338 + 70.

Therefore one can induce the finite difference formula:

.

This is equivalent to the finite difference equation

.

Applying the G-transformation to this equation yields the following G-transform:

.

Now let x0 = 0, x1 = 1 (x2 = 2⋅x1 + x0 = 2, x3 = 2⋅x2+x1 = 5, etc.) and solve for G(xn) to obtain:

which is a proper rational fraction.  Convert it to a sum of rational fractions:

.

Then one can apply the rule of the inverse G-transform, G-1:

,

therefore

.

(Both G and G-1 are linear transformations.)  I will put down for the record that the rule for the forward G-transform is:

.

This I obtained from the book Advanced Engineering Mathematics by Louis C. Barrett (Professor Emeritus, Montana State University).  It is not the only rule for G-transforms, but it is the only one which I needed to apply in this case.

        Anyway, ending the digression, the point is that there is a fixed-form formula (non-recursive) for obtaining the s/p ratio:

.

Then, since we also know that

one can derive s(Ni):

and since Ni = s2(Ni), this gives another formula for Ni.  Here it is:

Expand the square,

,

and simplify the last term;

,

so that

.

        Now I will prove that the two formulas for Ni are equivalent.  Starting from the first formula for Ni, which is

Expand both terms;

then simplify noticing any case where

Multiply the second term by 4/4 and combine into a single term:

.

(-1)2i equals 1, so the last term becomes -8.  Then, by cancelling out, one obtains

which is the second formula; the two formulas for Ni have just been shown to be equivalent.

        If p(Ni) is a natural number whenever i  is a natural number, then t(Ni) must also be an integer (not a non-integral rational nor an irrational number).  And if t(Ni) is a natural number, then the first formula for Ni must be a natural.  Then the second formula for Ni must yield a natural, which is the square of s(Ni).

        What I am trying to say is that if the formula for p(Ni) is known to yield natural numbers, and if the formula for s(Ni) is known to yield natural numbers, then the formula for Ni must yield triangular squares.  But it is not obvious that the formula for p(Ni) yields natural numbers, nor is it obvious by inspection that the formula for s(Ni) yields natural numbers.  Both formulas contain powers of irrational numbers.

        Consider this: even powers of are natural, odd powers of  are irrational.  Then think of  as a power series whose terms have the form

and think of  as a power series whose terms have the form

whose sign alternates.  When the two series are added together, the terms with odd powers of n cancel out, and the terms which are left have even powers of n, and so they contain only natural powers of .  The coefficients must be whole numbers because

is the number of combinations (n-tuples) of n objects possible out of a domain of i  objects; if i  is a whole number then must be a whole number.

        Then, since  is a power series containing only even powers of , and since even powers of are natural powers of 2, it follows that this power series produces an even number, and a half of an even number is a whole number, therefore

is a formula which produces whole numbers.

        If instead, one were to find the difference between and , one should find that even powers of n cancel out; only odd powers are left, and all of these contain powers of ; they are all irrational.  But all their irrationalities have something in common: odd powers of are actually equivalent to the product of a natural power of 2 and , so all these terms contain only once.  So dividing by produces a natural number.

        The terms in  which did not cancel out are the result of "constructive interference", i.e. they show up twice, once in and once again in ; so all the terms of  will have even coefficients.  When this difference is divided by , the terms of will all be even, and so

will produce only natural numbers.  Since one now knows that p(Ni) and  are both natural (for natural i), it follows that

is also natural for any i.  Therefore the formula for Ni yields only triangular squares.

        This means that there is an algorithm which I have now proven to produce an infinity of triangular squares.  The only thing that would remain to be proved is that this algorithm produces all the possible triangular squares.

        As far as I know, this is an original problem in number theory: I invented it (ILLUD INVENIĒBAM), I induced a solution, and then I deduced it.  I never did that before.

        I imagine that somebody, somewhere, some time ago, already contemplated the infinitude of triangular squares, and wrote about it; I have read many times about triangular numbers and about perfect squares, but never about triangular squares.  This is the best I have ever done in number theory.

        If I were to show this to Pythagoras, he would admit me into his secret society: tetraktys all the way!

January 28th, 2003

I looked up "triangular square numbers" with Google and found at least one website which not only mentioned triangular squares, but also analyzed them, in a fashion similar to mine, so my discovery is sublunary.

NIHIL SUB SŌLE NOVUM  (Ecclesiastes.)