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)?

Define the three numbers involved by:                 (a + 1)³ - a³ = b²

This simplifies to  3a² + 3a + 1 = b²  and then, after multiplying by 4 and rearranging, becomes   12a² + 12a + 3 + 1 = 4b²  which then gives:

            (2b)² - 3(2a + 1)² = 1                                         (1)

This is Pell’s equation (usually written as p² - Nq² = 1), so the possible values of p and q (i.e. 2b and 2a + 1) are given by the numerators and denominators of particular convergents of sqrt(3), found by writing sqrt(3) as a continued fraction. These convergents are:
1/1, 2/1, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571,....

The p/q values that satisfy p² - 3q² = 1 are:   2/1, 7/4, 26/15, 97/56,...
but we need the p value to be even and the q value to be odd, so our solutions are given by 2/1, 26/15, 362/209,...... then every 4th convergent thereafter.

So   2a + 1 = 1, 15, 209,.....  and, correspondingly,  2b = 2, 26, 362,....    
 
giving the solutions for (a, b) as:             (0, 1), (7, 13), (104, 181),...

Denoting these pairs by (a[1], b[1]), (a[2], b[2]), etc., the recurrence relations which derive from Pell’s results are:

     a[n] = 14a[n-1] - a[n-2] + 6  and   b[n] = 14b[n-1] - b[n-2]        (2)

and can be used to calculate further pairs: ...(1455, 2521), (20272, 35113)..etc

Assuming a solution of the form b[n] proportional to mⁿ, equation (2) then gives
mⁿ = 14m^{n - 1} - m^{n - 2} , which simplifies to m² - 14m + 1 = 0 with the solutions:
m = 7 + 4*sqrt(3) or  7 - 4*sqrt(3). The general solution is therefore
b[n] = A(7 + 4sqrt(3))ⁿ + B(7 - 4sqrt(3))ⁿ where A and B are constants.
Using the initial conditions b[1] = 1 and b[2] = 13 gives (after some algebra!):
A = (2 - sqrt(3))/4     and B = (2 + sqrt(3))/4 yielding the following formula:

    b[n] = 0.25(2 - sqrt(3))(7 + 4sqrt(3))ⁿ + 0.25(2 + sqrt(3))(7 - 4sqrt(3))ⁿ
  
A formula for a[n] can then be found using equation (1) or a similar approach.

As n tends to infinity, (7 + 4sqrt(3))ⁿ predominates over (7 - 4sqrt(3))ⁿ so that
b[n] tends to 0.25(2 - sqrt(3))(7 + 4sqrt(3))ⁿ, with the ratio of successive values of b[n] therefore tending to 7 + 4sqrt(3), which is approximately 13.92820 as pointed out by Jim Keneipp.

Posted by Harry on 2010-02-27 21:14:12