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)?
(In reply to
computer exploration by Charlie)
Sloane's OLEIS is back on line now.
Sequence A001921 is the sequence of the first number in each pair of successive integers whose cubes differ by a square. It's recursive formula is a(n) = 14a(n-1) - a(n-2) + 6 after a(1) = 0 and a(2) = 7. Sloan gives a direct formula:
a(n)=-1/2-(1/6)*sqrt(3)*[7-4*sqrt(3)]^n+(1/6)*sqrt(3)*[7+4*sqrt(3)]^n+(1/4)*[7+4*sqrt(3)]^n +(1/4)*[7-4*sqrt(3)]^n, with n>=0
though here it seems a(0) = 0 and a(1) = 7, etc.
The sequence of the square roots of the differences is given in A001570: Numbers n such that n^2 is simultaneously square and centered hexagonal, whose formula is a(n) = (1/4)*((2+sqrt(3))^(2*n+1)+(2-sqrt(3))^(2*n+1)).
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Posted by Charlie
on 2010-03-01 03:02:49 |
re: computer exploration -- Sloane back on-line
