A051302 Numbers whose square can be expressed as the sum of two positive cubes in more than one way.

77976, 223587, 623808, 894348, 1788696, 2105352, 2989441, 4298427, 4672423, 4990464, 5986575, 6036849, 7154784, 8437832, 9747000, 14309568, 16842816, 23915528, 24147396, 24770529, 26745768, 27948375, 34387416, 34634719, 36570744, 37379384, 39923712, 47892600

Offset: 1

Author

Jud McCranie

Keywords

• nonn
• nice

Comments

Observations regarding terms through a(64)=306761364: All are multiples of 7^2, 13^2, and/or 19^2. Other than 2, 3, 5 and 11, their only prime factors are 7, 13, 19, 31, 43, 61, 67, 79, 127, 151, and 181 (each of which exceeds a multiple of 6 by 1). None is a cube or higher power; the ones that are squares are a(7), a(12), a(17), a(19), a(20), a(32), a(33), a(41), a(49), a(55), and a(58). - Jon E. Schoenfield, Oct 08 2006

Many of the terms beyond a(64) have prime factors other than those found in a(1) through a(64); however, each term through a(774) has at most one distinct prime factor p > 5 that does not exceed a multiple of 6 by 1, and p, if such a prime factor exists, has a multiplicity m=3, with only a few exceptions: n=651 and n=713 (where p^m is 11^2), n=346 and n=770 (where p^m is 17^2), n=699 and n=740 (where p^m is 23^2), and n=741 (where p^m is 11^6). - Jon E. Schoenfield, Oct 20 2013

First differs from A145553 at A051302(172)=3343221000 where 3343221000^2 = 279300^3 + 2234400^3 = 790020^3 + 2202480^3 = 1256850^3 + 2094750^3.

This sequence is the union of A145553 and A155961.

This sequence is infinite. If n is a member of this sequence, then n^2 = a^3 + b^3 = c^3 + d^3 where (a, b) and (c, d) are distinct pairs. If n^2 = a^3 + b^3 = c^3 + d^3, then (n*k^3)^2 = n^2*k^6 = k^6*(a^3 + b^3) = k^6*(c^3 + d^3) = (a*k^2)^3 + (b*k^2)^3 = (c*k^2)^3 + (d*k^2)^3. It is obvious that if (a, b) and (c, d) are distinct, then (k^2*a, k^2*b), (k^2*c, k^2*d) are also distinct for all nonzero values of k. So if n is in this sequence, then n*k^3 is in this sequence for all k > 0. - Altug Alkan, May 10 2016

Examples

			2989441^2 = 1729^3+20748^3 = 15561^3+17290^3, so 2989441 is in the sequence.
		

Links

• Jon E. Schoenfield and Ray Chandler, Table of n, a(n) for n = 1..774

Crossrefs

Cf. A050801, A001235, A011541, A145553, A155961.

Programs

Mathematica

(* Warning: this script is only a recomputation of the original b-file of 64 terms from Jon E. Schoenfield, and should not be used to extend the data. *)
max = 310000000; cubeFreeParts = {361, 8281, 33124, 159201, 169309, 221725, 565068, 628849, 917427, 1054729, 2370963, 2989441, 4672423, 8968323, 9402967, 9795747, 34634719};
r[x_] := Reduce[0 < y <= z && x^2 == y^3 + z^3, {y, z}, Integers];
okQ[primes_] := Intersection[{2, 3, 5, 7, 11, 13, 19, 31, 43, 61, 67, 79, 127, 139, 151, 181}, primes] == primes;
crop[n_] := Reap[For[m = 1, True, m++, x = n*m^3; If[x > max, Break[]]; If[okQ[FactorInteger[x][[All, 1]]], If[Head[rx = r[x]] === Or, Print["x = ", x, " ", rx]; Sow[x]];]]][[2, 1]];
A051302 = crop /@ cubeFreeParts // Flatten // Sort (* Jean-François Alcover, Jul 02 2017 *)

PARI

T=thueinit('x^3+1, 1);
is(n)=my(v=thue(T, n^2)); sum(i=1, #v, v[i][1]>=0 && v[i][2]>=v[i][1])>1 \\ Charles R Greathouse IV, May 10 2016

Extensions

Definition corrected by Jon E. Schoenfield, Aug 27 2006

More terms from Jon E. Schoenfield, Oct 08 2006

Extended by Ray Chandler, Nov 22 2011