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The primus is composite all over, in particular:

U

_{2n-1}= (U

_{n}- U

_{n-1})(U

_{n}+U

_{n-1}) (primus theorem 2.2).

Since the tertius and quartus can be derived from the primus by taking subsequent differences and sums of its terms, the primus on odd indices equals

*tertius*quartus*.

This caught my attention.

__Sum-indexing__

Let's have a look at tertius- & quartus-factorization, but now with

*'sum-indexing'*derived from the parent series in the following way:

T_{2n-1} = U_{n} - U_{n-1} | Defines sum-indexing for the tertius. |

Q_{2n-1} = U_{n} + U_{n-1} | Defines sum-indexing for the quartus. |

The tertius and quartus, obtained by taking the differences and sums of subsequent terms of the primus, are now indexed by the sums of the indices of subsequent terms of that series. This has remarkable consequences:

- With sum-indexing factors of indices correspond one-to-one with factors of the corresponding terms.

This

*structural*factorization means that for every prime factor of the index one factor of the term can be predicted. It includes equal primes so the number of factors that can be predicted on grounds of the method involved, equals the sum of the exponents of the different primes of the index.

Of these factors some are again predictably composite on grounds of the tertius' & quartus' basic properties that are restated below.

With sum-indexing, irrespective of the factor 'F' of the series:

- If a term on a prime index is prime, it is of the 'near multiple' form:
*2k*index±1*.

If it is composite, its factors are either the index itself or a near-multiple.

There can be no structural factorization of terms with a prime index. To prove this, it's enough to prove it for one factor and for the quartus this is easily done if we take a look at the series U(0,1,0)2, otherwise known as the

*primus of 2*or the

*series of integers*.

The factor's tertius is characterized by U

_{n}=1 while its quartus is the series of odd integers. If we take a look at that one in sum-indexing, it's obvious that Q

_{2n+1}=U

_{n+1}+U

_{n}=2n+1 will always be prime if its index is prime :)

With sum-indexing, Un | Un+k(2n-1), the basic property of the tertius and the quartus for every integer k, looks like this:

T_{2n-1} | T_{(2k-1)(2n-1)} |
Basic property tertius |

Q_{2n-1} | Q_{(2k-1)(2n-1)} |
Basic property quartus |

U_{2n-1} = T_{(2n-1)} * Q_{(2n-1)} |
Primus theorem 6 |

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