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bonanova

Next numbers in this series

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4, 6, 12, 18, 30, 42, 60, 72, 102, 108, ...
 
4+2(n-1)+2(n-1)(n-2)-(4/6)(n-1)(n-2)(n-3)...
 
Continue in this fashion  to build all the terms then slap on any scalar along with (n-1)...(n-10) to generate infinite endings
 
Note:
n=1 gives 4
 
 

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I see what you mean.

Kind of like being able to draw an nth-order polynomial through any set of n points.

 

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I d not see that coming.

 

The classic solution is to find numbers that sandwiched between 2 prime numbers.

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OK well since I will define this puzzle not to be an algebra class assignment ...

hhh3 has the correct answer. Anyone else see why?

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