A186885 Numbers whose squares are the average of two distinct positive cubes.
6, 42, 48, 78, 147, 162, 196, 336, 384, 456, 624, 722, 750, 1050, 1134, 1176, 1296, 1342, 1568, 1573, 1674, 1694, 2028, 2058, 2106, 2366, 2387, 2450, 2522, 2646, 2688, 2899, 3072, 3087, 3211, 3648, 3698, 3969, 4374, 4992, 5250, 5292, 5550, 5776, 5915, 6000
Offset: 1
Keywords
Examples
6^2 = (2^3 + 4^3)/2; 42^2 = (11^3 + 13^3)/2; 147^2 = (7^3 + 35^3)/2.
Links
- David A. Corneth, Table of n, a(n) for n = 1..3948
- Zak Seidov, Triples {n,a,b} for n's up to 5*10^5
Programs
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Mathematica
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]]] (* Second program: *) Sqrt[#]&/@Select[Mean/@Subsets[Range[500]^3,{2}],IntegerQ[Sqrt[ #]]&]// Union (* Harvey P. Dale, Oct 13 2018 *) 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 *)
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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), listput(res, k)))); listsort(res); res} \\ David A. Corneth, Nov 25 2018
Formula
n^2 is average of two cubes: n^2 = (a^3 + b^3)/2, 0 < a < b.
Extensions
Edited by M. F. Hasler, Dec 10 2018
Comments