User blog:Allam948736/Reptend hierarchy

I invented a hierarchy similar to the fast-growing hierarchy, except the functions have two arguments instead of one, and f0(a, b) is equal to the reptend in the reciprocal of a^b, or ((10^(p*(a^(b-1)))) - 1)/a^b (where p is the period of 1/a). (Except if a is 3, when it is the reptend in the reciprocal of 3^(b+2), because the reptend in 1/3 is 3 and the reptend in 1/9 is 1). If a has no factors that are coprime to 10, then the result is just zero.

r0(3, 1) = 37

r0(3, 2) = 12345679

r0(3, 3) = 4115226337448559670781893

r0(3, 4) = 1371742112482853223593964334705075445816186556927297668038408779149519890260631

r0(3, 5) ~ 4.5724737082761774*10^239

r0(7, 1) = 142857

r0(7, 2) = 20408163265306122448979591836734693877551

r0(11, 1) = 9

r0(11, 2) = 8264462809173553719

r1(a, b) is equal to r0 applied b times to b with a as the first argument, just as f1 in the FGH is f0 applied to n n times.

r1(3, 1) = r0(3, 1) = 37

r1(3, 2) = r0(3, r0(3, 2)) = r0(3, 12345679) = ((10^(3^12345679)) - 1)/3^12345681 = 1552069995.........4587456533 (7.16*10^5890385 digits)

r1(3, 3) = r0(3, r0(3, r0(3, 3))) = r0(3, r0(3, 4115226337448559670781893))

r1(7, 1) = r0(7, 1) = 142857

r1(7, 2) = r0(7, r0(7, 2)) = r0(7, 20408163265306122448979591836734693877551) = 3990206661............0856598393 (>10^(10^40) digits)

Similarly, r2(a, b) is equal to r1 applied to b b times with a as the first argument, and r3(a, b) is equal to r2 applied b times to b with a as the first argument. r3(3, 3) is actually greater than 3^^^^3 (the first term in the sequence defining Graham's number). I tried to come up with a name for this number, but couldn't come up with any ideas. What would be a good name for this number?