This has to do with the 2nd order recursive sequence [tex]\{...a, b,(adsbygoogle = window.adsbygoogle || []).push({});

c ...\}[/tex] where a,b,c are any three sucessive terms and [tex]c = 6b-a + 2k[/tex].

I found that it has the following property.

[tex]8ab - (a+b-k)^2 = 8bc - (b+c-k)[/tex] That is eight times the product of two adjacent terms always equals the square of the difference between [tex]k[/tex] and the sum of the adjacent terms plus a number that is independent of three sucessive terms of the series are chosen.

A related property is the following. Again it is independent of which three sucessive terms of this sequence is chosen.

[tex]8a(a+k) - (b-3a-k)^2 = 8b(b+k) - (c-3b-k)^2[/tex]

The proof of both is by induction

The proof of the first relation follows:

[tex]8ab - (a+b-k)^2 = 8bc - (b+c-k)^2[/tex]

[tex]\quad = 8b(6b - a +2k) - (7b - a + k)^2[/tex]

[tex]\quad = (48-49)b^2 -a^2 -k^2 +(-8 +14)ab +(16-14)bk +2a[/tex]

[tex]\quad = 8ab -(b^2 +a^2 +k^2 +2ab -2bk -2ak)[/tex]

[tex]\quad = 8ab -(b+a-k)^2[/tex]

Thus the above property is proven.

Now if [tex]ab[/tex] is a triangular number for the first two terms of a

sequence, to make the product [tex]ab[/tex] to be a triangular number regardless of which pair of adjacent terms [tex]a \text{and} b[/tex] are in the sequence, we need to select a "k" such that [tex]8ab+1=(a+b-k)^2[/tex]

I will leave the proof of the second part and how it relates to square triangular numbers up to the reader.

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# Composite numbers and squares from recursive series

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