## Hypertetrahedral Polytopic Roots

The $n^{\rm th}$ triangular number is given by

$t = \bigtriangleup_2(n) = \bigtriangleup(n) = \dfrac{n(n+1)}{2} = \dbinom{n+1}{2}$.

For example, $\bigtriangleup(4)=10$.

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It turns out the inverse formula is (inventing some notation):

$n = \sqrt[\bigtriangleup]{t} = \dfrac{\sqrt{8t+1}-1}{2}$

Thus, for example, $\sqrt[\bigtriangleup]{10} = \dfrac{\sqrt{80+1}-1}{2} = 4$.

Can this be generalised?

### Triangular Roots (2-Dimensional)

Given a $t \text{ where } \bigtriangleup(n) = t$, it would be good to have a formula to find $n$. This $n$ could be called the triangular root, and denoted by $\sqrt[\bigtriangleup]{t}$.

There is an approximate formula $\bigtriangleup(n) \approx \dfrac{(n+\frac{1}{2})^2}{2}$ for large $n$. Therefore we have $\sqrt[\bigtriangleup]{t} \approx \sqrt{2t}-\frac{1}{2}$.

If we know that our $t$ is a triangular number, we also have that $\sqrt[\bigtriangleup]{t} = \lfloor \sqrt{2t} \rfloor$.

But what about an exact formula?

If we note that $\bigtriangleup(n) = t$ leads to the polynomial $0 = \bigtriangleup(n) - t$, then we can use the usual formula so solve this. If fact, it’s easier if we take the equation as $0 = 2 (\bigtriangleup(n) - t) = n^2+n-2t$. Thus

$n = \dfrac{\pm \sqrt{8t+1}-1}{2}$ and we choose +, as we’re interested only in positive $n$:

$\sqrt[\bigtriangleup]{t} = \dfrac{\sqrt{8t+1}-1}{2}$

### Tetrahedral Roots (3-Dimensional)

Similarly, the $n^{\rm th}$ tetrahedral number (the 3-dimensional equivalent of triangular number) is given by

$t = \bigtriangleup_3(n) = \dfrac{n(n+1)(n+2)}{3!} = \dbinom{n+2}{3}$.

For example, $\bigtriangleup_3(6)=56$.

[The letter $t$ is a convenient choice, as it stands for triangular, tetrahedral and (poly)tope.]

Can the tetrahedral root be found? Can this be generalised?

Yes, to at least some extent it can.

We have the formulae for large $n$: $\bigtriangleup_3(n) \approx \dfrac{(n+1)^3}{6}$ and
$\sqrt[\bigtriangleup_3]{t} \approx \sqrt[3]{6t} - 1$.

To attempt to find and exact formula, we can use the techniques of the algebraic solution of polynomials.

The corresponding polynomial for the tetrahedral case is $0 = 6 (\bigtriangleup_3(n) - t) = n^3+3n^2+2n-6t$.

Depress this with $n = m - 1$ to $0 = m^3 - m - 6t$. Substitute $m = r+\dfrac{1}{3r}$, and multiply through by $r^3$ to obtain $0 = r^6 - 6t r^3 + \frac{1}{27}$, which is essentially quadratic (in $r^3$). Thus $r = \sqrt[3]{3t \pm \sqrt{9t^2-\frac{1}{27}}}$. Back-substituting after choosing + and the real cube (or cubic) root, we obtain the formula for the tetrahedral root:

$\sqrt[\bigtriangleup_3]{t} = \dfrac{3r^2-3r+1}{3r}$

where $r = \sqrt[3]{3t + \sqrt{9t^2-\frac{1}{27}}}$.

[We leave this in this form, as all the $r$s need to be the same value, and mathematical notation doesn't easily cater for formulae that are DAGs (directed acyclic graphs) rather than trees.]

Thus, for example: $n = \sqrt[\bigtriangleup_3]{56} = \dfrac{3r^2-3r+1}{3r}$ where $r = \sqrt[3]{3t + \sqrt{9t^2-\frac{1}{27}}}$ with $t=56$. So, $r = \sqrt[3]{3\times 56 + \sqrt{9\times 56^2-\frac{1}{27}}} \approx 6.952$, so $n = 6$.

### Pentatope and other Hypertetrahedral Polytope Numbers

We define

$t = \bigtriangleup_d(n) = \dbinom{n+d-1}{d}$.

The corresponding approximate formulae, for large $n$ are:

$\bigtriangleup_d(n) \approx \dfrac{\left( n+\dfrac{d-1}{2} \right)^d}{d!}$;

$\sqrt[\bigtriangleup_d]{t} \approx \sqrt[d]{d! t} - \dfrac{d-1}{2}$.

### Pentatopic Roots (4-Dimensional)

$t = \bigtriangleup_4(n) = \dbinom{n+3}{4} = \frac{1}{24}n^4 + \frac{1}{4}n^3 + \frac{11}{24}n^2 + \frac{1}{4}n$
$0 = n^4 + 6n^3 + 11n^2 + 6n - 24t$

Depress this with $n=m-\frac{3}{2}$:
$0 = m^4 - \frac{5}{2}m^2 + (\frac{9}{16} - 24t)$
which is just a quadratic in $m^2$, because is just so happens that the depression has caused the $m$ term in addition to the $m^3$ term to vanish; thus this case is simpler than the 3-dimensional, tetrahedral, case.

(The result of this depression should not really surprise us, since the result is
$0 = (n - \frac{3}{2})(n - \frac{1}{2})(n + \frac{1}{2})(n + \frac{3}{2})$
and we know that
$(n-c)(n+c) = n^2 - c^2$.)

Thus we have
$m^2 = \dfrac{(\frac{5}{2} \pm \sqrt{(\frac{5}{2})^2 - 4(\frac{9}{16} - 24t)})}{2} = \frac{5}{4} \pm \sqrt{24t + 1}$.

So, choosing +, and back-substituting, we get the pentatopic root formula:

$\sqrt[\bigtriangleup_4]{t} = n = \sqrt{\frac{5}{4} + \sqrt{24t + 1}} - \frac{3}{2}$.

Example:
$\bigtriangleup_4(5) = 70$
$\sqrt[\bigtriangleup_4]{70} = \sqrt{\frac{5}{4} + \sqrt{24 \times 70 + 1}} - \frac{3}{2} = \sqrt{\frac{5}{4} + \sqrt{1681}} - \frac{3}{2} = \sqrt{\frac{169}{4}} - \frac{3}{2} = \frac{13}{2} - \frac{3}{2} = 5$.

### 5 Dimensions?

Unfortunately, when the formula that arises in the 5-dimensional case $0 = n^5 + 10n^4 + 35n^3 + 50n^2 + 24n - 120t$ is depressed by $n=m-2$, we obtain $0 = m^5 - 5m^3 + 4m - 120t$. Although both the $m^4$ and $m^2$ terms have vanished, there is an inconvenient contant term. This may be soluble; I don’t know.

However, the 4-dimensional case makes me quite optimistic about the 6- and 8-dimensional cases…

### 6-Dimensional Hypertetrahedral Roots

$t = \bigtriangleup_6(n) = \dbinom{n+5}{6}$
$0 = n^6 + 15n^5 + 85n^4 + 225n^3 + 274n^2 + 120n - 720t$

Depress this with $n=m-\frac{5}{2}$:
$0 = m^6 - \frac{35}{4}m^4 + \frac{259}{16}m^2 - (\frac{225}{64} + 720t)$
which is just a cubic in $m^2$, so depress again with $m^2 = k + \frac{35}{12}$:
$0 = k^3 - \frac{28}{3}k - (\frac{160}{27} + 720t)$
Substitute $k = h + \frac{28}{9h}$ and multiply by $h^3$:
$0 = h^6 - (720t+\frac{160}{27})h^3 + \frac{21952}{739}$
which is now just a quadratic in $h^3$, so
$h^3 = \dfrac{720t + \frac{160}{27} \pm \sqrt{(720t + \frac{160}{27})^2 - 4 \times \frac{21952}{729}}}{2}$
$= 360t + \frac{80}{27} \pm \sqrt{\frac{64}{3}(6075t^2 + 100t - 1)}$.

Thus, if we choose +, and back-substitute:

$\sqrt[\bigtriangleup_6]{t} = n = \sqrt{(h+\frac{28}{9h} + \frac{35}{12}} - \frac{5}{2}$ where

$h = \sqrt[3]{360t + \frac{80}{27} + \sqrt{\frac{64}{3}(6075t^2 + 100t - 1)}}$

### 8-Dimensional Hypertetrahedral Roots

$t = \bigtriangleup_8(n) = \dbinom{n+7}{8}$
$0 = n^8 + 28 n^7 + 322 n^6 + 1960 n^5 + 6769 n^4 + 13132 n^3 + 13068 n^2 + 5040 n - 40320 t$.

Depress with $n=m-\frac{7}{2}$:

$0 = m^8 - 21m^6 + \frac{987}{8}m^4 + \frac{3229}{16}m^2 + (\frac{11025}{256} - 40320t)$
which is just a quartic in $m^2$, so depress again with $m^2 = k + \frac{21}{4}$:
$0 = k^4 - 42 k^2 - 64 k + (105 -40320t)$.

This is
$0 = k^4 + c k^2 + b k + a$
with
$c = -42$, $b = -64$ and $a = 105 - 40320 t$.

Note that we know $\langle c, b, a \rangle$.

We try to factor this as a biquadratic:
$0 = k^4 + c k^2 +b k + a$
$= (k^2 + p k + s)(k^2 + q k + r)$
$0 = k^4 + (p+q) k^3 + (pq+r+s) k^2 + (pr+qs) k + rs$.

Comparing coefficients, we thus have:
$0 = p+q$;
$c = pq+r+s$;
$b = pr+qs$; and
$a = rs$.

Therefore
$q = -p$;
$c = r+s-p^2$, so $r+s = c+p^2$;
$b = p(r-s)$, so $r-s = b/p$.

We also have
$s = \frac{1}{2}(c+p^2-b/p)$;
$r = \frac{1}{2}(c+p^2+b/p)$;
so if we find $p$, we will have $\langle p, q, r, s \rangle$,
and we can then obtain k from $0 = (k^2 + p k + s)$ and $0 = (k^2 + q k + r)$:

$k = \dfrac{-p \pm \sqrt{p^2 - 4s}}{2}$ or $\dfrac{-q \pm \sqrt{q^2 - 4r}}{2}$.

Now,
$(c+p^2)^2 = (r+s)^2 = r^2 + s^2 + 2rs$, and
$(b/p)^2 = (r-s)^2 = r^2 + s^2 - 2rs$; so
$4a = 4rs = (c+p^2)^2 - (b/p)^2$.

Multiplying that by $p^2$ and rearranging, we have

$0 = p^6 + 2c p^4 + (c^2 - 4a) p^2 - b^2$
$= p^6 - 84p^4 + 1344(120t+1)p^2 - 4096$.

This is just a cubic in $p^2$, so we can depress this with $p^2 = h+28$:

$0 = h^3 + 1008(160t-1)h + 1152(3920t-9)$.

Substitute $h = g - \dfrac{336(160t-1)}{g}$, then multiply by $g^3$:

$0 = g^6 + v g^3 + w$ where
$v = 1152(3290t - 9)$, and
$w = - (336(160t - 1))^3$.

And, that is just a quadratic in $g^3$.

So: $g = \sqrt[3]{\dfrac{-v \pm \sqrt{v^2 - 4w}}{2}}$.

Now, we just back-substitute to $p$, solve the two quadratics, and back-substitute to $n$. At various stages, there are multiple, complex roots, thus each must be tried.

The overall solution in this case is fairly complicated, but it does exist.