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A051386 Numbers whose 4th power is the sum of two positive cubes. 3
2, 9, 16, 28, 35, 54, 65, 72, 91, 126, 128, 133, 134, 152, 182, 183, 189, 201, 217, 219, 224, 243, 250, 273, 278, 280, 309, 341, 344, 351, 370, 399, 407, 422, 432, 453, 468, 497, 513, 520, 539, 559, 576, 579, 637, 651, 658, 686, 728, 730, 737, 756, 793, 854 (list; graph; refs; listen; history; text; internal format)
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

Cf. A003325, A051387.

Sequence in context: A288484 A011193 A085960 * A003325 A101420 A275498

Adjacent sequences:  A051383 A051384 A051385 * A051387 A051388 A051389

KEYWORD

nonn

AUTHOR

Jud McCranie

STATUS

approved

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Last modified December 19 11:23 EST 2018. Contains 318246 sequences. (Running on oeis4.)