Changes

==Idea==

==Idea==

Let <math>\text{p}_i</math> be the <math>i^\text{th}</math> prime, where the positive integer <math>i</math> is called the ''index'' of the prime  <math>\text{p}_i</math> and the indices are taken in such a way that <math>\text{p}_1 = 2.</math>  Thus the sequence of primes begins as follows:

Let <math>\text{p}_i\!</math> be the <math>i^\text{th}\!</math> prime, where the positive integer <math>i\!</math> is called the ''index'' of the prime  <math>\text{p}_i\!</math> and the indices are taken in such a way that <math>\text{p}_1 = 2.\!</math>  Thus the sequence of primes begins as follows:

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The prime factorization of a positive integer <math>n</math> can be written in the following form:

The prime factorization of a positive integer <math>n\!</math> can be written in the following form:

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| <math>n ~=~ \prod_{k = 1}^{\ell} \text{p}_{i(k)}^{j(k)},</math>

| <math>n ~=~ \prod_{k = 1}^{\ell} \text{p}_{i(k)}^{j(k)},\!</math>

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where <math>\text{p}_{i(k)}^{j(k)}</math> is the <math>k^\text{th}</math> prime power in the factorization and <math>\ell</math> is the number of distinct prime factors dividing <math>n.</math>  The factorization of <math>1</math> is defined as <math>1</math> in accord with the convention that an empty product is equal to <math>1.</math>

where <math>\text{p}_{i(k)}^{j(k)}\!</math> is the <math>k^\text{th}\!</math> prime power in the factorization and <math>\ell\!</math> is the number of distinct prime factors dividing <math>n.\!</math>  The factorization of <math>1\!</math> is defined as <math>1\!</math> in accord with the convention that an empty product is equal to <math>1.\!</math>

Let <math>I(n)</math> be the set of indices of primes that divide  <math>n</math> and let <math>j(i, n)</math> be the number of times that <math>\text{p}_i</math> divides <math>n.</math>  Then the prime factorization of <math>n</math> can be written in the following alternative form:

Let <math>I(n)\!</math> be the set of indices of primes that divide  <math>n\!</math> and let <math>j(i, n)\!</math> be the number of times that <math>\text{p}_i\!</math> divides <math>n.\!</math>  Then the prime factorization of <math>n\!</math> can be written in the following alternative form:

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| <math>n ~=~ \prod_{i \in I(n)} \text{p}_{i}^{j(i, n)}.</math>

| <math>n ~=~ \prod_{i \in I(n)} \text{p}_{i}^{j(i, n)}.\!</math>

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Each index <math>i</math> and exponent <math>j</math> appearing in the prime factorization of a positive integer <math>n</math> is itself a positive integer, and thus has a prime factorization of its own.

Each index <math>i\!</math> and exponent <math>j\!</math> appearing in the prime factorization of a positive integer <math>n\!</math> is itself a positive integer, and thus has a prime factorization of its own.

Continuing with the same example, the index <math>504</math> has the factorization <math>2^3 \cdot 3^2 \cdot 7 = \text{p}_1^3 \text{p}_2^2 \text{p}_4^1</math> and the index <math>529</math> has the factorization <math>{23}^2 = \text{p}_9^2.</math>  Taking this information together with previously known factorizations allows the following replacements to be made in the above expression:

Continuing with the same example, the index <math>504\!</math> has the factorization <math>2^3 \cdot 3^2 \cdot 7 = \text{p}_1^3 \text{p}_2^2 \text{p}_4^1\!</math> and the index <math>529\!</math> has the factorization <math>{23}^2 = \text{p}_9^2.\!</math>  Taking this information together with previously known factorizations allows the following replacements to be made in the above expression:

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The <math>1</math>'s that appear as indices and exponents are formally redundant, conveying no information apart from the places they occupy in the resulting syntactic structure.  Leaving them tacit produces the following expression:

The <math>1\!</math>'s that appear as indices and exponents are formally redundant, conveying no information apart from the places they occupy in the resulting syntactic structure.  Leaving them tacit produces the following expression:

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Applying the same procedure to any positive integer <math>n</math> produces an expression called the ''doubly recursive factorization'' of <math>n</math> and notated as <math>\operatorname{drf}(n).</math>

Applying the same procedure to any positive integer <math>n\!</math> produces an expression called the ''doubly recursive factorization'' of <math>n\!</math> and notated as <math>\operatorname{drf}(n).\!</math>

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