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Numbers k such that k^3 is the sum of two positive 7th powers.
0

%I #7 Apr 05 2021 20:43:01

%S 32,4096,69984,524288,2500000,8957952,26353376,67108864,153055008,

%T 320000000,623589472,1146617856,2007952544,3373232128,5467500000,

%U 8589934592,13130837536,19591041024,28603895648,35723051649,40960000000,57634833312,79819452416,108954414304

%N Numbers k such that k^3 is the sum of two positive 7th powers.

%C When a^7 + b^7 = m, (m^2*a)^7 + (m^2*b)^7 = m^15 is a cube.

%C When k in this sequence, k*(n^7) (n = 2, 3, ... ) is also in this sequence.

%e 32^3 = 4^7 + 4^7, so 32 is in the sequence.

%e 35723051649^3 = 16641^7 + 33282^7, so 35723051649 is in the sequence.

%t lst={};Do[If[IntegerQ[(n^3-a^7)^(1/7)],AppendTo[lst,n]],{n,1.467*10^11},{a,(n^3/2)^(1/7)}]; lst

%Y Cf. A000578, A001015, A000404, A009003, A050801.

%K nonn

%O 1,1

%A _XU Pingya_, Sep 03 2017