1. A @ B = A*B^2 (B^2-1)/8 and
2. A Ó B = A*B (B-1)/2

Proof
In each case,
(A # Y) # Y` = (A*Y (Y-1)/K) *Y` (Y`-1)/K
= A*Y*Y` Y`*(Y-1)/K (Y`-1)/K
= A*(Y*Y`) (Y*Y` - Y` Y` -1)/K
= A*(Y*Y`) (Y*Y` - 1)/K
= A # (Y*Y`).

Now with regard to a triangular number minus a square, let the difference be A, S be the number that is squared and Q be the argument of T such that T(Q) - S^2 = A, then T(Q Ó B) - (S*B)^2 = A @ B.

Proof:
(S*B)^2 = T(Q Ó B) – A @ B
= - <(T(Q) – S^2) @ B> T(Q*B {B-1}/2)
= (S^2 – T(Q)) * B^2 - (B^2-1)/8 T(Q*B B/2 - 1/2)
= (S*B)^2 - (B^2)*Q(Q 1)/2 - (B^2-1)/8 <(Q*B B/2) –1/2><(Q*B B/2) 1/2>/2
= (S*B)^2 - {(Q*B)^2}/2 - (B^2)*Q/2 - (B^2-1)/8 <((Q*B B/2)^2)/2 –1/8>
= (S*B)^2 - <(Q*B)^2 (Q*B)*B (B^2)/4>/2 < (Q*B B/2)^2>/2
= (S*B)^2

The form you have given is a # b = a * b^p + (b^p - 1)/k for some k.

That is, a # b = (a + 1/k)*(b^p) - 1/k

The generalisation which this suggests is: Let f be a function with a right inverse g. Define a$b to be g(f(a) * b).
Then (a$b)$c = g(f(g(f(a) * b)) * c)
=g(f(a) * b * c) since f(g(x)) = x for all x
= a$(b*c)

(Note that for # we have f(x) = (x + 1/k)^(1/p))

Thankyou

sfwc
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