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Numbers k for which there are numbers 0 < m <= k such that k^3 + m^3 is a square.
2

%I #36 Jan 22 2025 08:53:31

%S 2,8,18,21,26,32,37,46,50,65,70,72,84,88,91,98,104,105,112,128,148,

%T 162,184,189,190,200,234,242,249,260,273,280,288,312,330,333,336,338,

%U 345,352,354,364,371,392,407,414,416,420

%N Numbers k for which there are numbers 0 < m <= k such that k^3 + m^3 is a square.

%C The sequence is infinite since if u is in the sequence then so is u*t^2, t, u >= 1. - _Marius A. Burtea_ and _David A. Corneth_, Oct 23 2018

%C For the subsequence k= 8, 18, 32, 50,65, 72, 98, 104, 105,... two or more m exist satisfying the equation. - _R. J. Mathar_, Jan 22 2025

%e 8^3 + 4^3 = 512 + 64 = 576 = 24^2, so 8 is part of the sequence.

%e 18^3 + 9^3 = 5832 + 729 = 6561 = 81^2, so 18 is part of the sequence.

%e 91^3 + 65^3 = 753571 + 274625 = 1028196 = 1014^2, so 91 is part of the sequence.

%e 7^3 + 0^3 = 343 + 0 = 343, 7^3 + 1^3 = 343 + 1 = 344, 7^3 + 2^3 = 343 + 8 = 351,7^3 + 4^3 = 343 + 64 = 407, 7^3 + 5^3 = 343 + 125 = 468, 7^3 + 6^3 = 343 + 216 = 559 and 7^3 + 7^3 = 343 + 343 = 686. Numbers 343, 344, 351, 407, 468, 559 and 686 are not squares, so 7 is not part of the sequence.

%p A320662 := proc(n)

%p option remember ;

%p local m,k ;

%p if n =1 then

%p 2

%p else

%p for k from procname(n-1)+1 do

%p for m from 1 to k do

%p if issqr(k^3+m^3) then

%p return k ;

%p end if;

%p end do:

%p end do:

%p end if;

%p end proc:

%p seq(A320662(n),n=1..40) ; # _R. J. Mathar_, Jan 22 2025

%t Select[Range@ 420, AnyTrue[Range[#1]^3 + #2, IntegerQ@ Sqrt@ # &] & @@ {#, #^3} &] (* _Michael De Vlieger_, Nov 05 2018 *)

%o (PARI) is(n) = for(m=1, n, if(issquare(n^3+m^3), return(1))); 0 \\ _Felix Fröhlich_, Oct 22 2018

%Y Cf. A003325, A003997, A004999, A024670, A086119, A282639 (subsequence for coprime m,k), A050801 (bases of the squares).

%K nonn

%O 1,1

%A _Marius A. Burtea_, Oct 18 2018