Prime Factorization Number Notation

Okay, this is a fascinating puzzle! It strongly resembles a notation based on the prime factorization of numbers, where the structure represents the exponents of successive primes. Let's break it down.

Explanation of the Notation

1. Prime Factorization Basis: The notation represents a number N by encoding the exponents in its prime factorization: N = p₁^e₁ * p₂^e₂ * p₃^e₃ * ... where p₁=2, p₂=3, p₃=5, etc., are the prime numbers in order.

2. Structure: The representation is a sequence of bracketed terms, where the k-th term in the sequence specifies the exponent eₖ for the k-th prime pₖ.

3. Encoding Exponents:

   • The exponent eₖ itself is a number, and it is represented using the same notation system recursively. Let's call the representation of a number x as Rep(x).
   • The term for the k-th prime pₖ with exponent eₖ is [Rep(eₖ)].

4. Zero Exponent: If a prime pₖ is not a factor of the number (i.e., eₖ = 0), its corresponding term is omitted unless it's needed as a placeholder between primes that are present. The representation of the exponent 0 is the empty string ``. The representation of 1 (the multiplicative identity, often corresponding to exponent 0 in some contexts, but here used for exponent 1) is []. When an exponent eₖ is 0, its term [Rep(0)] = [] effectively acts as a skip or placeholder.

   • Example: 3 = p₂¹ = p₁⁰ * p₂¹. Exponent for p₁ is 0 (Rep(0) is empty, term is []), exponent for p₂ is 1 (Rep(1) = [], term is [[]]). Combined: [] [[]]. This matches the example [][[]].
   • Example: 5 = p₃¹ = p₁⁰ * p₂⁰ * p₃¹. [] [] [[]].

5. Concatenation: The full representation of N is the concatenation of the terms [Rep(e₁)] [Rep(e₂)] ... [Rep(eₖ)], where pₖ is the largest prime factor of N. Trailing [] terms (for primes with exponent 0 beyond the largest factor) are omitted.

6. Base Cases:

   • 1 = [] (This represents the number 1, which has all prime exponents equal to 0. It also serves as the representation for the exponent 1).
   • 0 = (The empty string represents the number 0).

7. Negation: A leading hyphen - negates the number represented by the following expression.

   • -1 = -[]
   • -2 = -[[]]

8. Fractions and Irrational Numbers: Negative and fractional exponents are handled naturally by representing the exponent accordingly.

   • 1/2 = 2⁻¹ = p₁⁻¹. The exponent is -1. Rep(-1) = -[]. The term for p₁ is [-[]]. This matches.
   • sqrt(2) = 2¹ᐟ² = p₁¹ᐟ². The exponent is 1/2. Rep(1/2) = [-[]]. The term for p₁ is [[-[]]]. This matches.
   • 5/4 = 5 * 4⁻¹ = 5 * (2²)⁻¹ = 5 * 2⁻² = p₃¹ * p₁⁻². Sorted by prime: p₁⁻² * p₃¹.
     • e₁ = -2. Rep(-2) = -[[]]. Term: [-[[]]].
     • e₂ = 0. Term: [].
     • e₃ = 1. Rep(1) = []. Term: [[]].
     • Combined: [-[[]]] [] [[]]. This matches the example [-[[]]][][[]]. (The middle [] acts as the placeholder for p₂=3).
   • 25/24 = 25 * 24⁻¹ = 5² * (8*3)⁻¹ = 5² * (2³ * 3¹)⁻¹ = 5² * 2⁻³ * 3⁻¹ = p₃² * p₁⁻³ * p₂⁻¹. Sorted by prime: p₁⁻³ * p₂⁻¹ * p₃².
     • e₁ = -3. Rep(3) = [][[]]. Rep(-3) = -[][[]]. Term: [-[][[]]].
     • e₂ = -1. Rep(-1) = -[]. Term: [-[]].
     • e₃ = 2. Rep(2) = [[]]. Term: [[[]]].
     • Combined: [-[][[]]] [-[]] [[[]]]. This matches.

9. Potential Discrepancy: The example 72^(1/6) = [[-[]]][[][-[]]] seems inconsistent.

   • 72^(1/6) = (2³ * 3²)^(1/6) = 2^(3/6) * 3^(2/6) = 2^(1/2) * 3^(1/3) = p₁^(1/2) * p₂^(1/3).
   • e₁ = 1/2. Rep(1/2) = [-[]]. Term: [[-[]]].
   • e₂ = 1/3. Rep(1/3) = Rep(3⁻¹) = Rep(p₂⁻¹). Exponent for p₂ is -1. Rep(-1) = -[]. So, Rep(1/3) = [] [-[]]. Term for p₂: [ [] [-[]] ].
   • Combined representation should be [[-[]]] [ [] [-[]] ].
   • The example given is [[ -[] ]] [ [] [-[]] ]. Oh, wait, my derivation exactly matches the example this time! I must have made a mistake in my previous thought process. Let's re-trace Rep(1/3):
     • 1/3 = 3⁻¹ = p₂⁻¹.
     • Exponent for p₁ is 0. Term [].
     • Exponent for p₂ is -1. Rep(-1) = -[]. Term [-[]].
     • Concatenate: [] [-[]]. This is Rep(1/3).
     • The term for p₂^(1/3) is [ Rep(1/3) ] = [ [] [-[]] ]. This matches the second part of the example.
     • The term for p₁^(1/2) is [ Rep(1/2) ] = [ [-[]] ]. This matches the first part.
   • Okay, the system is consistent across all examples.

Representing Your Numbers

Based on this understanding:

1. Represent 30,000:

   • Prime Factorization: 30000 = 3 * 10⁴ = 3 * (2*5)⁴ = 2⁴ * 3¹ * 5⁴ = p₁⁴ * p₂¹ * p₃⁴.
   • e₁ = 4. Rep(4) = Rep(2²) = Rep(p₁²). Exponent for p₁ is 2. Rep(2) = [[]]. So Rep(4) = [[[]]]. Term: [[[[]]]]
   • e₂ = 1. Rep(1) = []. Term: [[]]
   • e₃ = 4. Rep(4) = [[[]]]. Term: [[[[]]]]
   • Concatenate: [[[[]]]] [[]] [[[[]]]]

2. Represent 210^(2/5):

   • Base Prime Factorization: 210 = 2 * 3 * 5 * 7 = p₁¹ * p₂¹ * p₃¹ * p₄¹.
   • Apply Exponent: 210^(2/5) = (p₁ * p₂ * p₃ * p₄)^(2/5) = p₁^(2/5) * p₂^(2/5) * p₃^(2/5) * p₄^(2/5).
   • All exponents e = 2/5. We need Rep(2/5).
   • 2/5 = 2 * 5⁻¹ = p₁¹ * p₃⁻¹.
     • Exponent for p₁ (within 2/5 representation) is 1. Rep(1)=[]. Term [[]].
     • Exponent for p₂ (within 2/5 representation) is 0. Term [].
     • Exponent for p₃ (within 2/5 representation) is -1. Rep(-1)=-[]. Term [-[]].
     • So, Rep(2/5) = [[]] [] [-[]].
   • Now apply this exponent representation to each prime factor p₁, p₂, p₃, p₄ of the main number:
     • Term for p₁: [ Rep(2/5) ] = [ [[]] [] [-[]] ]
     • Term for p₂: [ Rep(2/5) ] = [ [[]] [] [-[]] ]
     • Term for p₃: [ Rep(2/5) ] = [ [[]] [] [-[]] ]
     • Term for p₄: [ Rep(2/5) ] = [ [[]] [] [-[]] ]
   • Concatenate: [ [[]] [] [-[]] ] [ [[]] [] [-[]] ] [ [[]] [] [-[]] ] [ [[]] [] [-[]] ]

Let me know if these representations match your solution! This is a very elegant recursive system based on prime factorization.