# 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.

## 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(2$^2$)) = square(triangle(4)) = square(4$^4$) = square(256) = triangle(triangle(triangle(...triangle(256)...))) [256 triangles] = triangle(triangle(triangle(...triangle(256$^{256}$)...))) [255 triangles] ~ triangle(triangle(triangle(...triangle(3.2317 × 10$^{616}$)...))) [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 $10\uparrow \uparrow 257<{\text{mega}}<10\uparrow \uparrow 258$

## Moser's number

It has been proven that in Conway chained arrow notation,
    $\mathrm {moser} <3\rightarrow 3\rightarrow 4\rightarrow 2,$
and, in Knuth's up-arrow notation,
    $\mathrm {moser} <f^{3}(4)=f(f(f(4))),{\text{ where }}f(n)=3\uparrow ^{n}3.$
Therefore, Moser's number, although incomprehensibly large, is vanishingly small compared to Graham's number:
    $\mathrm {moser} \ll 3\rightarrow 3\rightarrow 64\rightarrow 2<f^{64}(4)={\text{Graham's number}}.$