2023-12-18 Q1(m=6)

Rearrange the digits in ⟨1263045⟩ to meet the rules below.

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

4th → a, 2nd → b, a+b=3

⟦1,5⟧ ∋ 2,3

⟨       ³ʳᵈa     ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 51

6th → a, 1st → b, a+b=1+4n

⛔Avoid

1st → 4

----- Information -----

🔲 「4th → a, 2nd → b, a+b=3」

480 permutations match this pattern.

Examples: ⟨0426135⟩, ⟨2135046⟩, ⟨6204351⟩.

🔲 「⟦1,5⟧ ∋ 2,3」

The closed interval given by 1 and 5 contains 2, 3.

840 permutations match this pattern.

Examples: ⟨0413256⟩, ⟨0532461⟩, ⟨1403625⟩.

🔲 「⟨       ³ʳᵈa     ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 51」

(ab)₁₀ ∶= 10a + b.

1320 permutations match this pattern.

Examples: ⟨3605241⟩, ⟨4016253⟩, ⟨0216354⟩.

🔲 「6th → a, 1st → b, a+b=1+4n」

[6th] + [1st] = 1 | 5 | 9 | ⋯ .

1440 permutations match this pattern.

Examples: ⟨5021643⟩, ⟨5601243⟩, ⟨6054231⟩.

🔳 「1st → 4」

720 permutations match this pattern.

Examples: ⟨1206345⟩, ⟨2306541⟩, ⟨6132045⟩.

#125034_v2.2

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │   │   │   │ 6 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │   │   │ 2 │ 6 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │   │   │ 2 │ 6 │ 1 │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │   │ 3 │ 2 │ 6 │ 1 │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 3 │ 2 │ 6 │ 1 │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 3 │ 2 │ 6 │ 1 │   │ 4 │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 5 │ 3 │ 2 │ 6 │ 1 │ 0 │ 4 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2023-12-18 Q1(m=6)
═══════════════════════════

Notation: if nth -> a, then we write [nth] = a.

By ✅「4th → a, 2nd → b, a+b=3」, we have

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│*4 │3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │   │ X │   │ Y │   │   │
└───┴───┴───┴───┴───┴───┴───┘

(1) where {X,Y} = {0,3} or {1,2}.

(2) On the other hand, by ✅「⟨       ³ʳᵈa     ⁰ᵗʰb ⟩, (ab)₁₀ ≥ 51」, we have [3rd] = 5 or 6. We claim that [3rd] = 6.

------------------------------

It is proved by contradiction. Suppose on the contrary [3rd] = 5.

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │   │ X │ 5 │ Y │   │   │
└───┴───┴───┴───┴───┴───┴───┘

Note that ✅「⟦1,5⟧ ∋ 2,3」 implies 1,5 are not adjacent. Therefore, it follows from (1) that {X,Y} = {0,3}. Combining this with the preceding pattern, there are two possibilities:

    ┌───┬───┬───┬───┬───┬───┬───┐
    │6th│5th│ 4▲│3rd│ 2▲│1st│0th│
    ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(3) │   │   │ 0 │ 5 │ 3 │ 2 │ 1 │
    ├───┼───┼───┼───┼───┼───┼───┤
(4) │ 1 │ 2 │ 3 │ 5 │ 0 │   │   │
    └───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │   │   │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

We cannot match ✅「6th → a, 1st → b, a+b=1+4n」 if it is case (3). In case (4), we need [1st] = 4 to match the preceding pattern, but then we match ⛔「1st → 4」 too, which is a contradiction.

------------------------------

Our claim in (2) is verified. Accordingly

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │   │ X │ 6 │ Y │   │   │
└───┴───┴───┴───┴───┴───┴───┘

(5) Now, we consider whether {X,Y} = {0,3} or {1,2}. We claim that the latter holds actually.

------------------------------

It is again proved by contradiction. If on the contrary {X,Y} = {0,3}, then the idle numbers become

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │   │   │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Observe that to match ✅「6th → a, 1st → b, a+b=1+4n」, namely to satisfy

A + B = 1|5|9, A := [6th], B := [1st],

we need {A,B} = {1,4} or {4,5}. A fortiori, 4 has to be the value of [6th] or [1st]. Combining this with ⛔「1st → 4」, we find that

┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │   │ X │ 6 │ Y │ B │   │
└───┴───┴───┴───┴───┴───┴───┘

where B = 1 or 5.

Recall that {X,Y} = {0,3} by assumption. So, regardless of B = 1 or 5, we cannot match ✅「⟦1,5⟧ ∋ 2,3」. This shows a contradiction.

------------------------------

Our claim in (5) is thus verified. It means that we have

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│   │   │ X │ 6 │ Y │   │   │
└───┴───┴───┴───┴───┴───┴───┘

where {X,Y} = {1,2}. Combining this with ✅「⟦1,5⟧ ∋ 2,3」, there are two possibilities:

    ┌───┬───┬───┬───┬───┬───┬───┐
    │6th│5th│*4 │3rd│*2 │1st│0th│
    ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(6) │   │   │ 1 │ 6 │ 2 │ 3 │ 5 │
    ├───┼───┼───┼───┼───┼───┼───┤
(7) │ 5 │ 3 │ 2 │ 6 │ 1 │   │   │
    └───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │   │   │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

We cannot match ✅「6th → a, 1st → b, a+b=1+4n」 if case (6) holds. So case (7) holds instead. In view of ⛔「1st → 4」, we finish by ⟨5326104⟩.

Q.E.D.

#125034_v2.2