I've been told that prime-number factoring is based on a "roman-doll" sequence of matrices, where a seed matrix of height Y and width X exists consisting of all zeros except for a single 1 at row Y column X. For every x columns added, the matrix grows by y, and two closely related formulas sum the column vectors and the row vectors.

An additional constraint that results in the "roman doll" effect are: $$ \forall I \in \mathcal{N} \cdot I\ge X\implies \sum_{j\in \mathcal{N}} h_{I,j}=\sum_{j=Y+(y/x-m)(I-X)}^{Y+y(I-X)/x} h_{I,j} $$

I've been told that prime-number factoring is based on a "russian-doll" sequence of matrices, where a seed matrix of height Y and width X exists consisting of all zeros except for a single 1 at row Y column X. For every x columns added, the matrix grows by y, and two closely related formulas sum the column vectors and the row vectors.

An additional constraint that results in the "russian doll" effect are: $$ \forall I \in \mathcal{N} \cdot I\ge X\implies \sum_{j\in \mathcal{N}} h_{I,j}=\sum_{j=Y+(y/x-m)(I-X)}^{Y+y(I-X)/x} h_{I,j} $$

$$ \forall I \in \mathcal{N} \cdot \sum_{j\in \mathcal{N}}\sum_{i=1}^{I} h_{i,j}=\sum_{n=1}^{\lfloor\sqrt{I}\rfloor} (\lfloor\frac{I}{n}\rfloor-n) $$

$$ \forall I \in \mathcal{N} \cdot \sum_{j\in \mathcal{N}}\sum_{i=1}^{I} h_{i,j}=\sum_{n=1}^{\lfloor\sqrt{I}\rfloor} (\lfloor\frac{I}{n}\rfloor-n)=(D(x)-u)/2 $$

where D(x) is the Divisor Summatory Function

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Reference

I don't have a good online reference for this, but the methods are readily understandable by anyone with a junior high education level. A similar approach is described at http://www.gbbservices.com/math/squarediff.html#Cracking_RSA

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Examples