Editing Steinhaus–Moser notation (section)

==Mega==
A mega, ②, is already a very large number, since ② =
square(square(2)) = square(triangle(triangle(2))) =
square(triangle(2<sup>2</sup>)) = 
square(triangle(4)) =
square(4<sup>4</sup>) =
square(256) =
triangle(triangle(triangle(...triangle(256)...)))  [256 triangles] =
triangle(triangle(triangle(...triangle(256<sup>256</sup>)...)))  [255 triangles] ~
triangle(triangle(triangle(...triangle(3.2317 &times; 10<sup>616</sup>)...)))  [255 triangles]
...

Using the other notation:

mega = <math>M(2,1,5) = M(256,256,3)</math>

With the function <math>f(x)=x^x</math> we have mega = <math>f^{256}(256)  = f^{258}(2)</math> where the superscript denotes a [[Iterated function|functional power]], not a numerical power.

We have (note the convention that powers are evaluated from right to left):
*<math>M(256,2,3) =</math> <math>(256^{\,\!256})^{256^{256}}=256^{256^{257}}</math>
*<math>M(256,3,3) =</math> <math>(256^{\,\!256^{257}})^{256^{256^{257}}}=256^{256^{257}\times 256^{256^{257}}}=256^{256^{257+256^{257}}}</math>≈<math>256^{\,\!256^{256^{257}}}</math>
Similarly:
*<math>M(256,4,3) \approx</math> <math>{\,\!256^{256^{256^{256^{257}}}}}</math>
*<math>M(256,5,3) \approx</math> <math>{\,\!256^{256^{256^{256^{256^{257}}}}}}</math>
*<math>M(256,6,3) \approx</math> <math>{\,\!256^{256^{256^{256^{256^{256^{257}}}}}}}</math>
etc.

Thus:
*mega = <math>M(256,256,3)\approx(256\uparrow)^{256}257</math>, where <math>(256\uparrow)^{256}</math> denotes a functional power of the function <math>f(n)=256^n</math>.

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

After the first few steps the value of <math>n^n</math> is each time approximately equal to <math>256^n</math>. In fact, it is even approximately equal to <math>10^n</math> (see also [[Large numbers#Approximate arithmetic|approximate arithmetic for very large numbers]]). Using base 10 powers we get:
*<math>M(256,1,3)\approx 3.23\times 10^{616}</math>
*<math>M(256,2,3)\approx10^{\,\!1.99\times 10^{619}}</math> (<math>\log _{10} 616</math> is added to the 616)
*<math>M(256,3,3)\approx10^{\,\!10^{1.99\times 10^{619}}}</math> (<math>619</math> is added to the <math>1.99\times 10^{619}</math>, which is negligible; therefore just a 10 is added at the bottom)
*<math>M(256,4,3)\approx10^{\,\!10^{10^{1.99\times 10^{619}}}}</math>
...
*mega = <math>M(256,256,3)\approx(10\uparrow)^{255}1.99\times 10^{619}</math>, where <math>(10\uparrow)^{255}</math> denotes a functional power of the function <math>f(n)=10^n</math>. Hence <math>10\uparrow\uparrow 257 < \text{mega} < 10\uparrow\uparrow 258</math>