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

%I A051302 #48 Jun 10 2018 20:17:00
%S A051302 77976,223587,623808,894348,1788696,2105352,2989441,4298427,4672423,
%T A051302 4990464,5986575,6036849,7154784,8437832,9747000,14309568,16842816,
%U A051302 23915528,24147396,24770529,26745768,27948375,34387416,34634719,36570744,37379384,39923712,47892600
%N A051302 Numbers whose square can be expressed as the sum of two positive cubes in more than one way.
%C A051302 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
%C A051302 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
%C A051302 First differs from A145553 at A051302(172)=3343221000 where 3343221000^2 = 279300^3 + 2234400^3 = 790020^3 + 2202480^3 = 1256850^3 + 2094750^3.
%C A051302 This sequence is the union of A145553 and A155961.
%C A051302 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
%H A051302 Jon E. Schoenfield and Ray Chandler, <a href="/A051302/b051302.txt">Table of n, a(n) for n = 1..774</a>
%e A051302 2989441^2 = 1729^3+20748^3 = 15561^3+17290^3, so 2989441 is in the sequence.
%t A051302 (* 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. *)
%t A051302 max = 310000000; cubeFreeParts = {361, 8281, 33124, 159201, 169309, 221725, 565068, 628849, 917427, 1054729, 2370963, 2989441, 4672423, 8968323, 9402967, 9795747, 34634719};
%t A051302 r[x_] := Reduce[0 < y <= z && x^2 == y^3 + z^3, {y, z}, Integers];
%t A051302 okQ[primes_] := Intersection[{2, 3, 5, 7, 11, 13, 19, 31, 43, 61, 67, 79, 127, 139, 151, 181}, primes] == primes;
%t A051302 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]];
%t A051302 A051302 = crop /@ cubeFreeParts // Flatten // Sort (* _Jean-François Alcover_, Jul 02 2017 *)
%o A051302 (PARI) T=thueinit('x^3+1, 1);
%o A051302 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
%Y A051302 Cf. A050801, A001235, A011541, A145553, A155961.
%K A051302 nonn,nice
%O A051302 1,1
%A A051302 _Jud McCranie_
%E A051302 Definition corrected by _Jon E. Schoenfield_, Aug 27 2006
%E A051302 More terms from _Jon E. Schoenfield_, Oct 08 2006
%E A051302 Extended by _Ray Chandler_, Nov 22 2011