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

Author

Zak Seidov, Feb 28 2011

Keywords

• nonn

Comments

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.

Parities of a and b are equal: a == b (mod 2). - David A. Corneth, Oct 13 2018

Square roots of the intersection of A000290 and A268319. - Antti Karttunen, Jan 15 2019

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

Crossrefs

Cf. A000290, A000578, A268319.

Cf. also A273822.

Programs

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 *)

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