Novel Generalization Of Square Triangular Nos?

[ramsey2879]

Triangular numbers T(n) = n(n+1)/2. Certain triangular numbers are also square numbers: T(0) = 0 = 0^2; T(1)= 1 = 1^2; T(8) = 36 = 6^2, T(49) = 1225 = 35^2; T(288) = 41616 = 204^2; ...). It is well known that the arguments of the square triangular numbers {0,1,8,49,288 ...} have the recursive formula S(n) = 6*S(n-1) - S(n-2) + 2 while the square roots of the square triangular numbers {0,1,6,35,204,...} have the recursive formula Q(n) = 6*Q(n-1) - Q(n-2). As far as I know, no one has observed that the first two numbers of the S(n) series can be put into the generalized form 0 = 0 and 1 = 4a+1 while the first two numbers of the Q(n) series can be put into the form 0 = a and 1 = 3a + 1 where a can be any integer and that the general relationship is now T(S(n)) = (Q(n)+a)*(Q(n)-a).
Where "a" = 0, the infinite set of square triangular numbers are generated.
But where a is other than 0, an infinite set of triangular numbers that can be factored into a product of two integers differing by 2a is formed.
For instance let a = 1
S(n) = {0,5,32,189,...}
Q(n) = {1,4,23,134,...}

T(0) = 0*2=0
T(5) = 3*5=15
T(32) =22*24=528
T(189) =133*135=17955
...
Because the recursive formula of S(n) is S(n) = 6*S(n-1) - S(n-2) + 2 there is no common method to determine a closed form formula (non-recursive) for the nth term but there is a simple way around this. To find the close form equation for the nth term of the S(n) series for various values of "a", we consider the modified series S<n> formed by adding 1/2 to each term of S(n) so that the recursive formula is now S<n> = 6S<n-1> - S<n-2>. To determine the close form formula for such a series is straight forward. Then we can use the relation S(n) = S<n> - 1/2.

Can anyone suggest where I might find disclosures of my finding prior to mine. If there are any, I would like to know.

 

[chaoseverlasting]

Your post is not very accessable. Could you expand upon the last paragraph? Also what are you trying to prove/state?