What is the next triplet in the below series of triplets:
(2,3,4), (47,48,49), (674,675,676), (9407,9408,9409), ___?
Few points about the above series:
(i) the terms of the triplets are consecutive
(ii) there is at least one square among the three terms
(iii) the sum of the terms of each triplet is also a square
I tried to find an underlying series/formula for finding such triplets, but in vain.
So, can you please help me out!
We are basically searching all the triples $(x-2,x-1,x)$ such that:
$$(x-1)+(x-2)+x=3x-3=\beta^2 \text{ and } x=\alpha^2$$
1 Condition
$$3x-3=\beta^2 \Rightarrow 3(x-1)=\beta^2$$
If $3|\beta^2$ then $9|\beta^2$ so $x-1=3k^2$. So the triples we are searching are of this form $(3k^2-1,3k^2,3k^2+1)$
2 Condition
We must now impose that:
$$x=\alpha^2 \Rightarrow \alpha^2-3k^2=1 $$
Which is a standard Pell's equation. Solving it with Wolfram Alpha(I can do this alone but I haven't time ) we find that the first solution are:
$$k=1$$$$k=4$$$$k=15$$$$k=56$$$$k=209$$
The first 4 work and we find again the given triples. To find the next one we must consider $k=209$ that gives in output:
$$(131042,131043,131044)$$
:)
Moreover searching on OEIS it seems that the central number of each triples has a particular propriety connected with standard deviation: http://oeis.org/A007654