The difference of the cubes of two non-negative successive integers is a square of an integer.
Find few possible pairs.
How about a general formula or evaluation algorithm
for a(n)?

Some couples (a,b) for (a+1)^3-a^3=b^2 :
(0, 1), (7, 13), (104, 181),...
To get more, use the recurrence relations :
a[n] = 14a[n-1] - a[n-2] + 6 and
b[n] = 14b[n-1] - b[n-2]

Comments: (
You must be logged in to post comments.)