OFFSET
1,1
COMMENTS
n such that n^4 = r^3 + s^3 has a solution with r>0, s>0.
By multiplying n^4 = r^3 + s^3 by n^3, also numbers whose 7th power is expressible as the sum of positive cubes.
When n is the sum of 2 positive cubes (A003325) there is a trivial solution: e.g., 133 is a term in A003325, 133=2^3+5^3 and 133^4=(2*133)^3+(5*133)^3. - Zak Seidov, Oct 17 2011
From Robert Israel, Jun 01 2015: (Start)
Slightly more generally, if x^3 + y^3 = u*v^4, then (u*v*w^3)^4 = (u*w^4*x)^3 + (u*w^4*y)^3, so u*v*w^3 is in the sequence for any w >= 1.
There are at least five pairs of adjacent numbers in the sequence: (133,134),(182,183), (854,855), (1842,1843), (3473,3474). Are there infinitely many?
(End)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
EXAMPLE
134^4 = 469^3 + 603^3.
MAPLE
N:= 1000: # to get all terms <= N
Cubes:= {seq(x^3, x=1..floor(N^(4/3)))}:
select(n -> nops(map(t -> n^4-t, Cubes) intersect Cubes)>0, [$1..N]); # Robert Israel, Jun 01 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved