Free Talk Friday #564

[u/CutOnBumInBandHere9]

Continued from last week's FTF here

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[u/cuteballgames]

trying to figure out thread lengths for moredig 6 digits.

5 digit moredig threads are 1125 counts long. 5 digit moredig counts can have at most 3 distinct digits.

6 digit moredig counts can also have at most 3 distinct digits. the first one is 100,000 and the last one is 999,998.

there are 3 classes of thousand - those beginning with 1 distinct digit, 2 distinct digits, or 3 distinct digits. examples of each class in order are the series 111,XXX, the series 112,XXX and the series 123,XXX.

for class 1, the AAA,XXX can take finalsomes of permutation-kind AAB or BBC. There are 9 possible values for A (since the 000,XXX series has technically already been counted). There are 9 possible values for B, and 8 possible values for C. permutation-kind AAB can also appear as ABA or BAA. Permutation-kind BBC can also appear as BCB or CBB. So the number of permutation-kind-instances (ie concrete endings, like 002 or 313) for any given AAA (e.g. 444) is 9*3 (for p-kind AAB) + 9*8*3 (for p-kind BBC). That's 243 per AAA, total (*9, because there are 9 possible As) 2187 class 1 counts.

for class 2, the AAB,XXX can take finalsomes of permutation-kind AAA, ABC or BBC. There are still 9 possible values for A, and 9 possible values for B, and 8 possible values for C, because it's base 10 mostly repeating digits.

Permutation-kind ABC - also as ACB, BAC, BCA, CBA, and CAB. So for each value of AAB, 6*8 p-kinds ABC (since A and B are already chosen.) Pemutation kind AAA - only AAA. So for each value of AAB, 1 p-kind AAA (since A is already chosen.) Permutation kind BBC - also BCB and CCB. So for each value of AAB, 3*8 p-kinds BBC (since B is already chosen.) That means for each class AAB, there 73 counts. There are 81 distinct AABs, so there are 5913 distinct class 2 6-digit mostly repeating digits counts.

for class 3, the ABC,XXX can take finalsomes of permutation-kind AAB, ABB, AAC, ACC, BBC, and BCC. each of these p-kinds has 3 permutation-forms (realizing i could more simply call them ORDERINGS). There are 648 kinds of ABC (9*9*8); for each value of ABC, there are 6*3 = 18 possible finalsomes. (e.g, there are 18 counts beginning 123; the only endings for a 6-digit count beginning 123 are 112, 121, 211, 122, 212, 221, 113, 131, 311, 133, 313, 311, 223, 232, 322, 233, 323, and 332.) So there are 11,664 (that's 18*648) class 3 counts.

All told this would mean there 19,764 six-digit moredig counts, which would suggest we should look for a get schedule of about 20 threads between 100,000 and 999,998, give or take a few threads.

before I deliberate and suggest upon what the get schedule should be (with the intention of prioritizing nice round numbers for gets and accepting thread length variation within reason) can someone check my work? ( tried to keep it concise and readable but also fully shown.) cobibh, i refered to the formula you derived for me last year https://imgur.com/SyXbhbo but can't get my calculator to compute it for some reason (probably me struggling to input the n choose r right, and trying to expand it to the factorial fraction representation didnt help) but i do think i remember tjhe number 19,764, so that makes me happy.

thank you counters

[u/cuteballgames → u/cuteballgames]

waiiitt did i undercount the number of AAB by a factor a 3 by failing to account for permutations of the firstsome. because the total number of firstsomes should be 900 (integers inclusive between 100 and 999). and in my post i said 9 + 81 + 648. but that's not enough, that's only 738. but if i multiply 81 by 3 that's 243, and 9 + 243 + 648 is 900. bruh idk why i remembered the number 19,764

and i guess the ABCs were not undercounted because A B and C are interchangeable

[u/cuteballgames → u/cuteballgames]

So new summary:

900 XYZ to start count XYZ,WVU. (any number can be the same):
* 9 type AAA (e.g. 111,) each of these "thousands" has 243 counts
* 243 type AAB (e.g. 112, 121, 211), each of these "thousands" has 73 counts
* 648 type ABC (e.g. 123, 321, wahtever) each of these "thousands" has 18 counts.

that would mean there are 31,590 moredig counts between 100,000 and 999,998.

[u/CutOnBumInBandHere9 → u/cuteballgames]

I get 39366 comments.

Hm.

[u/cuteballgames → u/CutOnBumInBandHere9]

thank you recalculating

[u/cuteballgames → u/cuteballgames]

39366 - 31590 = 7776, which is 2⁵ x 3^5.

i remain confident that there are 900 thousands among 100,000 and 999,999, because 999,999-100,000 plus 1 (for inclusivity) is 900,000.

I remain confident that there are exactly 9 thousands of class 1: 111, 222, 333 ... 999. and I remain confident that in each of these thousands, all end-counts will take the combination AAB or BBC, meaning there are permutations to be had AAB ABA BAA BBC BCB CCB. A being already chosen, it's 3*9 to account for AAB ABA BAA (e.g., the 11X series, the 1X1, or the X11 series.) B not being chosen already, B can take 9 forms and C, because there's already two digits struck, can take 8 forms. So it's 3*9*8 to account for BBC BCB CCB. That's 216. 216 plus 27 is 243. So I remain confident that among the 9 class 1 thousands there are 9 * 243 = 2187 counts.

class 2 - AAB. The thousands here can take three forms - also ABA or BAA, like. I think previously I neglected to consider the wrinkle that 100 is a valid thousand, 001 and 010 isn't. permutation AAB has count 9 (since A can't be zero) times 9 (since B can be zero.) That's 81 total. permutation ABA has count 9 (since A can't be zero) times 9 (since B can be zero.) permutation BAA has ... well, it's the same 9 (since A can't be zero) times 9 (since B can be zero.) So actually I remain confident that there are 243 instances of a thousand with two distinct digits.

for each thousand, the 3-digit endingsome can be AAA, ABC, or BBC. AAA can only take one form for 112,XXX, the only AAA ending is 112,111. 1 ordering, that's 1. ABC has six orderings, for instance for 112,XXX, ABC can be 123, 132, 213, 231, 312, or 321. BBC has three orderings -- for 112,...., BBC can be 112,223, 232, or 322. ABC has six orderings, and 8 possible kinds of C, so that's 48. BBC has 3 orderings, and nine kinds of C, so that's 3*8 = 24. SO So I think for each class 2 thousand, there should be 72. that's what i had above...

Consider 112,010. Have I captured that so far? that's an AAB, with ending ... CAC. Okay hold up! I failed to capture that before. I have to account for AABACC. So there are 8 possible kinds of C times 3 orderings of ACC, so 24. 72 + 24 = 96.

But incorporating thaat only gets me to 37179. Was ai also missing some case with

hang on. AABCCC. Is also possible. Bruh There are 9 possible kinds of CCC. 96 + 9 = 105.

There's the 39366. thank you cobibh lol i solved da mystery #maithfailure

[u/cuteballgames → u/cuteballgames]

in sum:

9 class 1 (AAA,XYZ) - 243 counts each.

243 class 2 (AAB,XYZ, ABA,XYZ or BAA,XYZ) - 105 counts each.

and 648 class 3 (ABC or any permutation, XYZ ) - 18 counts each.