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Proof: (related to "Sieve for Twin Primes")

By hypothesis, $n\ge 1$ is an integer and let $f_s:=\left\lfloor\frac {s+1}6\right\rfloor$ using the floor function.

$”\Rightarrow”$

$”\Leftarrow”$

1 $s$ has either the form $6\nu-1$ or $6\nu+1$ for some $\nu\ge 1.$ If $s=6\nu-1$ then $6f_s-1=6\left\lfloor\frac{6\nu-1+1}{6}\right\rfloor-1=6\nu-1=s$. If $s=6\nu+1$ then $6f_s-1=6\left\lfloor\frac{6\nu+1+1}{6}\right\rfloor-1=6\lfloor\nu+\frac 13\rfloor-1=6\nu-1=s.$ In both cases, $s=6f_s-1.$

q.e.d

| | | | created: 2020-03-14 10:29:29 | modified: 2020-03-14 14:17:25 | by: bookofproofs | references: [6913]

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Bibliography (further reading)

[6913] Piotrowski, Andreas: Anmerkungen zur Verteilung der Primzahlzwillinge, Master’s thesis, Frankfurt am Main, 1999