The giggol is a 1 followed by a 1 followed by a 1 etc etc and repeating this proces 99 times then the sentence ends by .... followed by a 1 with 10^10 zero's zero's zero's......
So therefore i would write it as 1>(1>(1>...99times the 1> term.....(1>10)..99braces..))
It's more easy to show it with the decker number that is in arrow notation (10,10,2)
is exactly
10^10^10^10^10^10^10^10^10^10=
10^......8 tens before.......^(1>10)=
10^......7 tens before.......^(1>(1>10))=
10^.....1 ten before........^(1>(1>(1>(1>(1>(1>(1>(1>10))))))))=
9 times the term "(1>" and then >10 and 9 braces.
So it also works out for the giggol in arrow notation defined as (10,100,2)
99 times the term "(1>" and then >10 and 99 braces
Now
E decker therefore is
E 1>1>1>1>1>1>1>1>1>10 =
1>(1>1>1>1>1>1>1>1>10) > (1>1>1>1>1>1>1>1>10)
To calculate this one we first look at the patterns of smaller similar E powers like
E 1>1>10 or E 1>1>1>10
E 1>10 = 1>1>11
E 1>1>10 = 1>1>((1>8)~10)
E 1>1>1>10 = 1>1000.....1>10.....>1000.....1>10.....=
1>1> (1000.....1>10.... + 1>10)=
1>1> ((1> ((1>10)-(10+1)))~(1>10))=
1>1> ((1> 9 999 999 989)~(1>10))=
1>(1>((1>(9re8~89))~(1>10)))
Now i use a new notation to indicate multiple "(>1" term with
(m)^>1
so for example 1>1>1>10 then is (3^>1)>10
so
E (4^>1)>10 = 1>(1000.....1>1>10.....>10000.....1>1>10......) =
1>1>1>(1>1>10 - 1>10)~1>1>10=
1>1>1>(9re(1>10-(10+1)))~8~(9re10)~(1>(1>10))=
1>(1>((1>((9re(9re8~89))~8~(9re10)))~(1>(1>10))))
What becomes clear already is that
E 1>(m^>1)>10 = 1>(1>(1>((1>((((m-3)^>1)>10) - (1+(1>(((m-4)^>1)>10))))) ~ (1>(((m-3)^>1)>10))))
as long as m=4 or more
so the only 'difficult' part in the number formation is the part of
(1>(((m-1)^>1) >10)) - (1+ (1>(((m-2)^>1)>10))))
We need to find it for m=9 and then for the giggol for m=99.
Lets try m=9;
As the pattern already shows
it becomes
1>1>1>(9re(1>((m-4)^>1>10)-(1+(1>m-5^>1>10))))~8~(9re1>(m-5)^>1>10)
so actually only
m=9 has to be placed in the equation here;
9re((1>((9-4)^>1)>10)-(1+((1>(9-5)^>1)>10)=
9re(1>1>1>1>1>10 - (1>1>1>1>10 + 1)) =
9re (9re (1>1>1>1>10 - (1>1>1>10 + 1)))~8~9re1>1>1>10 =
9re (9re 9re (1>1>1>10 - (1>1>10 + 1)) ~8~9re1>1>10)~8~9re1>1>1>10=
9re 9re 9re 9re 1>1>10 - 1>10 + 1))~8~9re1>10)~8~9re1>1>10.....=
9re (((9re (((9re (((9re ((9re ((9re8)~89))~(8~(9re10))))~(8~(9re(1>10)))))~(8~(9re(1>(1>10))))))~(8~(9re(1>(1>(1>10))))))
Thus on a list from E 1>1 to E 1>((m)^>10) ;
E 1>1 = 1>(1>1)
E 1>(1>1) =
E 1>(1^>1)>1 = 1>(1>11)
E 1>(1>(1>1) =
E 1>(2^>1)>1 = 1>(1>((1>8)~(1>1)))
E 1>(3^>1)>1 = 1>(1>((1>((9re8)~8~(9re1)))~(1>(1^>1)>1)))
E 1>(4^>1)>1 = 1>(1>(((1>((9re ((9re8)~8~(9re1))) ~8~(9re (1>1))))~(1>(2^>1)>1)))
E 1>(5^>1)>1 = 1>(1>(((1>((9re(9re ((9re8)~8~(9re1)))) ~8~(9re(1>(1^>1)>1))))~(1>(3^>1)>1)))
E 1>(6^>1)>1 = 1>(1>(((1>((9re(9re(9re ((9re8)~8~(9re1))))) ~8~(9re(1>(2^>1)>1))))~(1>(4^>1)>1)))
now i create for the ......9re(9re(9re(9..... part the notation ['(9re're(m)]~~ so for E 1>(6^>1)>1 i now write
1>(1>(((1>(['(9re're3]~~ ((9re8)~8~(9re1)))))..............
also with braces i can do that in the form of ['('re(m)] what can be handy if the number of braces becomes really big like over the 30.
in the case of E 1>(6^>1)>1 i write it as
1>(1>(((1>(['(9re're3]~~ ((9re8)~8~(9re1~~[')'re5] ~8~(.............
So the formula of E 1>((m)^>1)>1 can be formulated to be
1>(1>(((1>(['(9re're(m-3)]~~((9re8)~8~(9re1~~[')'re(m-1)]~8~(9re(1>((m-4)^>1)>1))))~(1>((m-2)^>1)>1)))
for a giggol is equal to
1>(100^>1)>1 m is equal to 100 so
E giggol =
E (10,100,2) = 1>(1>(((1>['(9re're97]~~((9re8)~8~(9re1~~[')'re99]~8~(9re(1>(96^>1)>1))))~(1>(98^>1)>1)))
and for the decker;
E decker =
E (10,10,2) = 1>(1>(((1>(['9re're 7]~~((9re8)~8~(9re1~~ [')'re 9]~8~(9re(1>( 6^>1)>1))))~(1>(( 8^>1)>1)))