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)?
Each line of the output from the program shows, in order, n, n^3, (n+1)^3, (n+1)^3-n^3 and sqrt((n+1)^3-n^3):
run
0 0 1 1 1
7 343 512 169 13
104 1124864 1157625 32761 181
1455 3080271375 3086626816 6355441 2521
20272 8330859163648 8332092086417 1232922769 35113
282359 22511524427594279 22511763608256000 239180661721 489061
3932760 60826430604528576000 60826477004344027081 46399815451081 6811741
OK
list
10 for I=1 to 10000000
20 Cube=I*I*I
30 D=Cube-Pcube:Sr=int(sqrt(D)+0.5)
40 if Sr*Sr=D then
50 :print I-1;Pcube;Cube;D;Sr
90 Pcube=Cube
100 next
OK
Unfortunately Sloane's OLEIS seems to be unavailable this morning. Otherwise it might have a formula for this.
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Posted by Charlie
on 2010-02-27 14:29:40 |
computer exploration
