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# Moser's number

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### Moser's number

In mathematics, Steinhaus–Moser notation is a notation for expressing certain extremely large numbers. It is an extension of Steinhaus's polygon notation.

## Definitions

a number n in a triangle means nn.
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 with "the number n inside n squares, which are all nested."

etc.: n written in an (m + 1)-sided polygon is equivalent with "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 nn inside one triangle, which is equivalent to nn raised to the power of nn.

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

## Special values

Steinhaus defined:

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

Moser's number is the number represented by "2 in a megagon", where a megagon is a polygon with "mega" 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:
• $M\left(n,1,3\right) = n^n$
• $M\left(n,1,p+1\right) = M\left(n,n,p\right)$
• $M\left(n,m+1,p\right) = M\left(M\left(n,1,p\right),m,p\right)$
• and
• mega = $M\left(2,1,5\right)$
• megiston = $M\left(10,1,5\right)$
• moser = $M\left(2,1,M\left(2,1,5\right)\right)$

## 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(3.2 × 10616)...))) [254 triangles] = ...

Using the other notation:

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

With the function $f\left(x\right)=x^x$ we have mega = $f^\left\{256\right\}\left(256\right) = f^\left\{258\right\}\left(2\right)$ 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) = $\left(256^\left\{\,\!256\right\}\right)^\left\{256^\left\{256\right\}\right\}=256^\left\{256^\left\{257\right\}\right\}$
• M(256,3,3) = $\left(256^\left\{\,\!256^\left\{257\right\}\right\}\right)^\left\{256^\left\{256^\left\{257\right\}\right\}\right\}=256^\left\{256^\left\{257\right\}\times 256^\left\{256^\left\{257\right\}\right\}\right\}=256^\left\{256^\left\{257+256^\left\{257\right\}\right\}\right\}$$256^\left\{\,\!256^\left\{256^\left\{257\right\}\right\}\right\}$

Similarly:

• M(256,4,3) ≈ $\left\{\,\!256^\left\{256^\left\{256^\left\{256^\left\{257\right\}\right\}\right\}\right\}\right\}$
• M(256,5,3) ≈ $\left\{\,\!256^\left\{256^\left\{256^\left\{256^\left\{256^\left\{257\right\}\right\}\right\}\right\}\right\}\right\}$

etc.

Thus:

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

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:

• $M\left(256,1,3\right)\approx 3.23\times 10^\left\{616\right\}$
• $M\left(256,2,3\right)\approx10^\left\{\,\!1.99\times 10^\left\{619\right\}\right\}$ ($\log _\left\{10\right\} 616$ is added to the 616)
• $M\left(256,3,3\right)\approx10^\left\{\,\!10^\left\{1.99\times 10^\left\{619\right\}\right\}\right\}$ ($619$ is added to the $1.99\times 10^\left\{619\right\}$, which is negligible; therefore just a 10 is added at the bottom)
• $M\left(256,4,3\right)\approx10^\left\{\,\!10^\left\{10^\left\{1.99\times 10^\left\{619\right\}\right\}\right\}\right\}$

...

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

## Moser's number

It has been proven that in Conway chained arrow notation,

$\mathrm\left\{moser\right\} < 3\rightarrow 3\rightarrow 4\rightarrow 2,$

and, in Knuth's up-arrow notation,

$\mathrm\left\{moser\right\} < f\left(f\left(f\left(4\right)\right)\right), \text\left\{ where \right\} f\left(n\right) = 3 \uparrow^n 3.$

Therefore Moser's number, although incomprehensibly large, is vanishingly small compared to Graham's number:

$\mathrm\left\{moser\right\} \ll 3\rightarrow 3\rightarrow 64\rightarrow 2 < f^\left\{64\right\}\left(4\right) = \text\left\{Graham\text{'}s number\right\}.$