User blog:Primussupremus/Examples of hyper notation values

1[0]1=2=1+1.

2[1]2=2+2=4

3[2]3=3*3=9

4[3]4=4^4=256

5[4]5=5↑↑5. 6[5]6=6↑↑↑6. 7[6]7=7↑↑↑↑7. 10[12]10=10↑(10)10 or ten followed by ten up arrows then ten. We could make a bigger number by making 10[12]10 the hyper operator we are going to use. 10[10[12]10]10 is quite large by layman standards but by Googologists standards it's puny. 10[10[10[12]10]10]10 would be an even larger number but it's still puny. 10[10[10[10[10[10[12]10]10]10]10]10]10 would be an even larger number but still it's puny. But what if the total number of tens expanding out in both directions was equal to 10[10[10[10[10[10[12]10]10]10]10]10]10 that would be humongous compared to the previous result. Now lets call this result "a" and then input it into the hyper operator slot to get 10[a]10. Now what if the number of tens expanding in each direction was equal to 10[a]10,essentially the number of things expanding in each direction is the previous result or the (n-1)th result. Let 10[a]10 be called b and using b produce out 10[b]10. Following this pattern we can deduce that 10[z]10 will be the highest value expressible with 1 variable in the hyper operator slot using the number 10 at either side. The next value would be 10[2a]10 after that we get 10[2b]10. After 10[2b]10 we get 10[2c]10 then 10[2d]10 and so on all the way up to 10[2z]10. Using this as a guideline the next limit would be 10[3z]10 after that it would be 10[4z]10 then 10[5z]10. We could continue this process on doing arbitrary nz's like 10[1000z]10 but this isn't going to be that powerful or interesting so instead we'll use an expression like 10[10z]10 as the basis for a large number where this will become the z. This would evaluate to 10[[10[10z]10]z]10 where 10[10z]10 go's into the z slot making and even larger number.