Steinhaus–Moser notation

In mathematics, Steinhaus–Moser notation is a notation for expressing certain large numbers. It is an extension (devised by Leo Moser) of Hugo Steinhaus's polygon notation.^[1]

Definitions

a number

n

in a triangle means

n n

.

a number

n

in a square is equivalent to "the number

n

inside

n

triangles, which are all nested."

a number

n

in a pentagon is equivalent to "the number

n

inside

n

squares, which are all nested."

etc.: $n$ written in an ( $m + 1$ )-sided polygon is equivalent to "the number $n$ inside $n$ nested $m$ -sided polygons". In a series of nested polygons, they are associated inward. The number $n$ inside two triangles is equivalent to $n n$ inside one triangle, which is equivalent to $n n$ raised to the power of $n n$ .

Steinhaus defined only the triangle, the square, and the circle , which is equivalent to the pentagon defined above.

Special values

Steinhaus defined:

mega is the number equivalent to 2 in a circle: ②
megiston is the number equivalent to 10 in a circle: ⑩

Moser's number is the number represented by "2 in a megagon". Megagon is here the name of a polygon with "mega" sides (not to be confused with the polygon with one million sides).

Alternative notations:

use the functions square(x) and triangle(x)
let M(n, m, p) be the number represented by the number n in m nested p-sided polygons; then the rules are:
- $M(n,1,3)=n^{n}$
- $M(n,1,p+1)=M(n,n,p)$
- $M(n,m+1,p)=M(M(n,1,p),m,p)$
and
- mega = $M(2,1,5)$
- megiston = $M(10,1,5)$
- moser = $M(2,1,M(2,1,5))$

Mega

A mega, ②, is already a very large number, since ② = square(square(2)) = square(triangle(triangle(2))) = square(triangle(2²)) = square(triangle(4)) = square(4⁴) = square(256) = triangle(triangle(triangle(...triangle(256)...))) [256 triangles] = triangle(triangle(triangle(...triangle(256²⁵⁶)...))) [255 triangles] ~ triangle(triangle(triangle(...triangle(3.2317 × 10⁶¹⁶)...))) [255 triangles] ...

Using the other notation:

mega = $M(2,1,5)=M(256,256,3)$

With the function $f(x)=x^{x}$ we have mega = $f^{256}(256)=f^{258}(2)$ where the superscript denotes a functional power, not a numerical power.

We have (note the convention that powers are evaluated from right to left):

$M(256,2,3)=$ $(256^{\,\!256})^{256^{256}}=256^{256^{257}}$
$M(256,3,3)=$ $(256^{\,\!256^{257}})^{256^{256^{257}}}=256^{256^{257}\times 256^{256^{257}}}=256^{256^{257+256^{257}}}$ ≈ $256^{\,\!256^{256^{257}}}$

Similarly:

$M(256,4,3)\approx$ ${\,\!256^{256^{256^{256^{257}}}}}$
$M(256,5,3)\approx$ ${\,\!256^{256^{256^{256^{256^{257}}}}}}$
$M(256,6,3)\approx$ ${\,\!256^{256^{256^{256^{256^{256^{257}}}}}}}$

etc.

Thus:

mega = $M(256,256,3)\approx (256\uparrow )^{256}257$ , where $(256\uparrow )^{256}$ denotes a functional power of the function $f(n)=256^{n}$ .

Rounding more crudely (replacing the 257 at the end by 256), we get mega ≈ $256\uparrow \uparrow 257$ , using Knuth's up-arrow notation.

After the first few steps the value of $n^{n}$ is each time approximately equal to $256^{n}$ . In fact, it is even approximately equal to $10^{n}$ (see also approximate arithmetic for very large numbers). Using base 10 powers we get:

$M(256,1,3)\approx 3.23\times 10^{616}$
$M(256,2,3)\approx 10^{\,\!1.99\times 10^{619}}$ ( $\log _{10}616$ is added to the 616)
$M(256,3,3)\approx 10^{\,\!10^{1.99\times 10^{619}}}$ ( $619$ is added to the $1.99\times 10^{619}$ , which is negligible; therefore just a 10 is added at the bottom)
$M(256,4,3)\approx 10^{\,\!10^{10^{1.99\times 10^{619}}}}$

...

mega = $M(256,256,3)\approx (10\uparrow )^{255}1.99\times 10^{619}$ , where $(10\uparrow )^{255}$ denotes a functional power of the function $f(n)=10^{n}$ . Hence $10\uparrow \uparrow 257<{\text{mega}}<10\uparrow \uparrow 258$