Science Forums 2

[Turtle]

Originally Posted by Don Blazys

Interesting. Now we have a palindromic pattern of last digits that is eleven digits long and does not contain 2, 6, 7 or 8.

Don.

So it is. Well, the terminating 0 in the pattern makes the whole non-palindromic, but the the ten-element section has the goods.

Now from here it is apparent we might try this with all bases (or as many as we can carry), and as luck has it that is what I have done & recorded with my graphs.

Now I have to go aside because I was getting to the patterns of remainders by applying the Katabatak function (my name) to the powers written in different bases. This function has other names (can't find a Wolfram article just now) but it is simply the re-iterative adding of a numbers digits until arriving at a single digit.

So, taking the base eleven powers of two one at a time and operating on them with the K function gives us this:

Now what we have is the pattern we found in the ending digits of the base ten list of powers of two, namely {1 4 9 6 5 6 9 4 1 A}. Note the last digit/symbol differs in that the base ten ending-digit list/pattern terminates with 0 and the base eleven Katabatak transform pattern gives A, but since A is ten and ends in 0 in base ten then it is really the same result, i.e. a remainder of ten.

So I never needed to add the digits after changing base, rather only needed to take the last digits from the new base. Well, as I came at it from the adding digits algorithm in the first place I don't feel as bad as I could over so many years of extra work.

More to come - as if - and I am continuing to add graphs to the first post.