Still remember that 2007 Halloween 5.6 earthquake.

Yeah mid-4.0 earthquakes are pretty fun.

What a shame that it had to happen almost exactly on the 10000th topic.

Also, pseudocode example for really fast generation:

initarray := [a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z]
mod := initarray
while 1:{
for loopvar from 1 to dim(mod):{
output mod[loopvar]}
newmod := []
for loopvar from 1 to 26:{
temp := initarray[loopvar] + mod
newmod := newmod + temp
}
mod := newmod
}

ok this really kills the memory, whoops
and you'll have really long pauses for when the computer is calculating the next string length

EDIT: figured out a constant-time algorithm, same memory problems though
pseudocoded for viewing pleasure:
list := [a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z]
queue := list
while forever{
pop queue => string
print string
concatenate string, list => list2
push list2 into queue
}


I can't vote either.

By the way, you can also make unclimbable walls by creating ledges.

division by 2 semihack

128391284905932852 / 2:
lump up even blocks:
12 8 3912 8 4 90 5932 8 52 /2
divide by two without carry on each digit
01 4 1401 4 2 40 2411 4 21
add a bunch of 5's
+
05 0 0555 0 0 05 0555 0 05
06 4 1956 4 2 45 2966 4 26
= 64195642452966426.

@Xeda: Any polynomial interpolation will be bound to lose accuracy pretty quickly, considering that f(x)>2^(x-1) which is exponential.

Progress on creating an explicit formula, I believe, can be done the following way:
1) Create a recursive CA110 step function. For example CAstep110(1) = 3, CAstep110(3)=7, etc.
2) Find an explicit formula for this, using characteristic polynomials.
3) Recursively apply the explicit formula to 1, to generate this.

So far, for step 1, I have the following:
CAstep110(2n) = 2CAstep110(n)
CAstep110(4n+1) = CAstep110(4n)+3 = 4CAstep110(n)+3
CAstep110(8n+3) = CAstep110(8n+2)+1 = 2CAstep110(4n+1)+1 = 8CAstep110(n)+7
CAstep110(8n+7) = CAstep110(8n+6)-1 = 2CAstep110(4n+3)-1

The fourth equation is the problematic one.