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A186885
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Numbers whose squares are the average of two distinct positive cubes.
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3
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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
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OFFSET
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1,1
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COMMENTS
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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.
The table gives all 396 triples (n, a, b) such that n^2 = (a^3 + b^3)/2 and n < 5*10^5.
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LINKS
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FORMULA
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n^2 is average of two cubes: n^2 = (a^3 + b^3)/2, 0 < a < b.
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EXAMPLE
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6^2 = (2^3 + 4^3)/2;
42^2 = (11^3 + 13^3)/2;
147^2 = (7^3 + 35^3)/2.
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MATHEMATICA
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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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PROG
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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),
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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