On this page we will try to estimate the Mega, the number equal to 2 in a pentagon in Steinhaus-Moser notation.
256 in one triangle is about 3.2317*10^616, and 256 in two triangles is greater than 10^(10^619), and 256 in three triangles is greater than 10^(10^(10^619)). Adding another triangle will raise 10 to the power of this lower-bound, so the Mega can be lower-bounded by:
10^(10^(10^(10^(...(10^(10^(10^619)))...)))) with 256 10s
We can similarly prove that replacing the 619 in the expression above with 620 will give an upper-bound (here the 3.2317*10^616 is replaced with 3.2318*10^616):
(3.2318*10^616)^(3.2318*10^616) < 10^(617*3.2318*10^616) < 10^(1000*3.2318*10^616) = 10^(3.2318*10^619) < 10^(10^620)
256 in 3 triangles < (10^(3.2318*10^619))^(10^(3.2318*10^619)) = 10^(3.2318*10^619*(10^(3.2318*10^619))) < 10^(10^(3.2318*10^619 + 620)) < 10^10^10^620
...
256 in 256 triangles < 10^10^10^10^...^10^10^10^10^620 with 256 10s
But we can still do better. First we will try to improve the lower bound, and then try to improve the upper bound. We will begin by using a more exact value for 256 in 2 triangles:
(256^256)^(256^256) > (3.2317*10^616)^(3.2317*10^616) > 10^(616*3.2317*10^616) = 10^(1.9907272*10^619)
256 in 3 triangles > 10^(1.9907272*10^619)^(10^(1.9907272*10^619)) = 10^(1.9907272*10^619*10^(1.9907272*10^619)) ~> 10^(10^(619 + 10^(1.9907272*10^619))
> 10^(10^(1.9907272*10^619))
It can be similarly shown that 256 in 4 triangles is greater than 10^(10^(10^(1.9907272*10^619))), and that 256 in 5 triangles is greater than 10^(10^(10^(10^(1.9907272*10^619)))). So the Mega is lower-bounded by 10^(10^(10^(10^(...(10^(10^(10^(1.9907272*10^619))))...)))) with 255 10s before the 1.9907272*10^619. This lower-bound is obtained by rounding the logarithm of 256^256 down to 616 and using an approximation of 256^256 to 5 digits.
Now, we can improve the upper-bound:
(256^256)^(256^256) < (3.2318*10^616)^(3.2318*10^616) < 10^(617*3.2318*10^616) = 10^(1.9940206*10^619)
256 in 3 triangles < (10^(1.9940206*10^619))^(10^(1.9940206*10^619)) < (10^10^620)^(10^(1.9940206*10^619)) = 10^(10^(1.9940206*10^619 + 620))
An additional triangle at this point would have barely any effect other than adding another 10 at the bottom of the tower. So we have the following bounds for the Mega so far:
10^(10^(10^(...(10^(10^(10^(1.9907272*10^619))))...))) with 255 10s < mega < 10^(10^(10^(...(10^(10^(10^(1.9940206*10^619 + 620))))...))) with 255 10s