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%I #24 Oct 30 2018 10:31:02
%S 2,9,16,28,35,54,65,72,91,126,128,133,134,152,182,183,189,201,217,219,
%T 224,243,250,273,278,280,309,341,344,351,370,399,407,422,432,453,468,
%U 497,513,520,539,559,576,579,637,651,658,686,728,730,737,756,793,854
%N Numbers whose 4th power is the sum of two positive cubes.
%C n such that n^4 = r^3 + s^3 has a solution with r>0, s>0.
%C By multiplying n^4 = r^3 + s^3 by n^3, also numbers whose 7th power is expressible as the sum of positive cubes.
%C 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
%C From _Robert Israel_, Jun 01 2015: (Start)
%C 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.
%C 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?
%C (End)
%H Chai Wah Wu, <a href="/A051386/b051386.txt">Table of n, a(n) for n = 1..10000</a>
%e 134^4 = 469^3 + 603^3.
%p N:= 1000: # to get all terms <= N
%p Cubes:= {seq(x^3,x=1..floor(N^(4/3)))}:
%p select(n -> nops(map(t -> n^4-t, Cubes) intersect Cubes)>0, [$1..N]); # _Robert Israel_, Jun 01 2015
%Y Cf. A003325, A051387.
%K nonn
%O 1,1
%A _Jud McCranie_
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