OFFSET
1,1
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
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
FORMULA
n^2 is average of two cubes: n^2 = (a^3 + b^3)/2, 0 < a < b.
EXAMPLE
6^2 = (2^3 + 4^3)/2;
42^2 = (11^3 + 13^3)/2;
147^2 = (7^3 + 35^3)/2.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, Feb 28 2011
EXTENSIONS
Edited by M. F. Hasler, Dec 10 2018
STATUS
approved