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Can you prove why the recurrence $3P_{n} - 2P_{n + 1} = P_{n - 1}$ holds for some (a subset of the) prime numbers?

Mathematics Asked by Joebloggs on December 25, 2020

Can you prove why this recurrence relation holds for some (a subset of the) prime numbers? And does it go on infinitely? (I think it does)

$$3P_{n} – 2P_{n + 1} = P_{n – 1}$$

First few values for which it applies:

$5,
7,
13,
20,
25,
26,
28,
45,
49,
60,
64,
78,
88,
89,
93,
95,
113,
136,
144,
148,
152,
159,
182,
212,
225,
229,
230,
236,
243,
247,
249,
262,
265,
277,
286,
288,
291,
302,
315,
323$

E.G: For $n = 5, P_{n} = 11, P_{n + 1} = 13, P_{n – 1} = 7:$

$3(11) – 2(13) = 7$

$3P_{5} – 2P_{6} = P_{4}$

…&c for the others

One Answer

Using the standard prime gap notation gives that

$$g_n = p_{n+1} - p_{n} implies p_{n+1} = p_{n} + g_{n} tag{1}label{eq1A}$$

$$p_{n} = p_{n-1} + g_{n-1} tag{2}label{eq2A}$$

Next, eqref{eq2A} in eqref{eq1A} gives

$$p_{n+1} = p_{n-1} + g_{n-1} + g_{n} tag{3}label{eq3A}$$

Substituting eqref{eq2A} and eqref{eq3A} into your relation results in

$$begin{equation}begin{aligned} & 3p_{n} - 2p_{n + 1} = p_{n - 1} \ & 3(p_{n-1} + g_{n-1}) - 2(p_{n-1} + g_{n-1} + g_{n}) = p_{n - 1} \ & 3p_{n-1} + 3g_{n-1} - 2p_{n-1} - 2g_{n-1} - 2g_{n} = p_{n - 1} \ & p_{n-1} + g_{n-1} - 2g_{n} = p_{n - 1} \ & g_{n-1} = 2g_{n} end{aligned}end{equation}tag{4}label{eq4A}$$

Thus, your relation holds whenever any prime gap is double that of the next prime gap. With your example of $n = 5$, we have $g_{n-1} = 11 - 7 = 4$ and $g_{n} = 13 - 11 = 2$.

As for whether or not there's an infinite number of primes where eqref{eq4A} holds, I'm quite certain that is currently unknown, but it's conjectured to be the case. Note there are infinitely many prime $k$-tuples which are admissible and, in the cases where they also represent a prime constellation, then eqref{eq4A} is true, e.g., your example of $(0, 4, 6)$ (which is called a prime triplet), $(0, 8, 12)$, $(0, 12, 18)$, etc. The "Admissibility" section also states

It is conjectured that every admissible $k$-tuple matches infinitely many positions in the sequence of prime numbers. However, there is no admissible tuple for which this has been proven except the $1$-tuple ($0$). Nevertheless, by Yitang Zhang's famous proof of $2013$ it follows that there exists at least one $2$-tuple which matches infinitely many positions; subsequent work showed that some $2$-tuple exists with values differing by $246$ or less that matches infinitely many positions.[2]

In particular, the prime $k$-tuple conjecture states

... that every admissible pattern for a prime constellation occurs infinitely often ...

This conjecture implies there are an infinite number of primes which satisfy eqref{eq4A}.

Answered by John Omielan on December 25, 2020

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