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%I #36 Jan 19 2019 03:37:56
%S 6,42,48,78,147,162,196,336,384,456,624,722,750,1050,1134,1176,1296,
%T 1342,1568,1573,1674,1694,2028,2058,2106,2366,2387,2450,2522,2646,
%U 2688,2899,3072,3087,3211,3648,3698,3969,4374,4992,5250,5292,5550,5776,5915,6000
%N Numbers whose squares are the average of two distinct positive cubes.
%C If m is in this sequence, then so is m*k^3 for all k >= 1: e.g., both m = 6 and 6000 = m*10^3 are in this sequence. Also, there are no primes in this sequence.
%C The table gives all 396 triples (n, a, b) such that n^2 = (a^3 + b^3)/2 and n < 5*10^5.
%C Parities of a and b are equal: a == b (mod 2). - _David A. Corneth_, Oct 13 2018
%C Square roots of the intersection of A000290 and A268319. - _Antti Karttunen_, Jan 15 2019
%H David A. Corneth, <a href="/A186885/b186885.txt">Table of n, a(n) for n = 1..3948</a>
%H Zak Seidov, <a href="/A186885/a186885.txt">Triples {n,a,b} for n's up to 5*10^5</a>
%F n^2 is average of two cubes: n^2 = (a^3 + b^3)/2, 0 < a < b.
%e 6^2 = (2^3 + 4^3)/2;
%e 42^2 = (11^3 + 13^3)/2;
%e 147^2 = (7^3 + 35^3)/2.
%t nn = 13552; lim = Floor[(2 nn^2)^(1/3)]; Sort[Reap[Do[num = (a^3 + b^3)/2; If[IntegerQ[num] && num <= nn^2 && IntegerQ[Sqrt[num]], Sow[Sqrt[num]]], {a, lim}, {b, a - 1}]][[2, 1]]]
%t (* Second program: *)
%t Sqrt[#]&/@Select[Mean/@Subsets[Range[500]^3,{2}],IntegerQ[Sqrt[ #]]&]// Union (* _Harvey P. Dale_, Oct 13 2018 *)
%t upto[m_] := Module[{res = {}, n = m*m, i, j, k}, For[i = 1, i <= Floor[ Quotient[n, 2]^(1/3)], i++, For[j = i+2, j <= Floor[(n-i^3)^(1/3)], j += 2, If[IntegerQ[k = Sqrt[(i^3 + j^3)/2]], AppendTo[res, k]]]]; Sort[res]]; upto[20000] (* _Jean-François Alcover_, Jan 17 2019, after _David A. Corneth_ *)
%o (PARI) upto(n) = {my(res = List(), k); n*=n; for(i = 1, sqrtnint(n \ 2, 3), forstep(j = i + 2, sqrtnint(n - i^3, 3), 2, if(issquare((i^3 + j^3) / 2, &k),
%o listput(res, k)))); listsort(res); res} \\ _David A. Corneth_, Nov 25 2018
%Y Cf. A000290, A000578, A268319.
%Y Cf. also A273822.
%K nonn
%O 1,1
%A _Zak Seidov_, Feb 28 2011
%E Edited by _M. F. Hasler_, Dec 10 2018
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