If you add 2,4,6..... To these four prime numbers then you will get a series of primes
5 5,7,11,17,25
1111,13,17,23,31,41,53,67,83,101,121
17 17,19,23,29,37,47,59,73,89,107,127,149,173,199,227,257,289
41 41,43,47,53,61,71,83,97,113,131,151,173,197,223,251,281,313,347,383,421,461,503,547,593,641,691,743,797,853,911,971,1033,1097,1163,1231,1301,1373,1447,1523,1601,1681
Notice how every series ends at the square of first number and also the total number of terms in series is equal to first number and also every first number is first number of twin prime pair. I conjecture that there is no other number producing such a series. Can you tell why?
Prime generation
Prime generation
Last edited by S_r on Fri Dec 29, 2017 2:38 am, edited 2 times in total.
Re: Prime generation
Anyone?
Last edited by S_r on Fri Dec 29, 2017 2:48 am, edited 1 time in total.
 sjhillier
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Re: Prime generation
What you have got there are 4 of the lucky numbers of Euler. The others are the rather more trivial smaller values. It has indeed been proved that no larger numbers of this type exist, but I suspect the proof is not so easy. A simpler insight into why this should be would be interesting.

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Re: Prime generation
I think this is a relatively known concept... Framing it in terms of equations, we have
$n(n+1)+p$ where $p \in{5, 7, 11, 41}$ and $n \geq 0$
The last part of everything ending in a square is trivial by using $n = p1$ in which case it quadratic becomes $p(p1)+p=p^2$
More on Prime Generating Polynomial where it is stated that the only lucky numbers of Euler are 2, 3, 5, 11, 17 and 41. Whether the reasoning is simple or not lies in the eye of the beholder
$n(n+1)+p$ where $p \in{5, 7, 11, 41}$ and $n \geq 0$
The last part of everything ending in a square is trivial by using $n = p1$ in which case it quadratic becomes $p(p1)+p=p^2$
More on Prime Generating Polynomial where it is stated that the only lucky numbers of Euler are 2, 3, 5, 11, 17 and 41. Whether the reasoning is simple or not lies in the eye of the beholder
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