A145553 Numbers n such that n^2 can be expressed as the sum of 2 positive cubes in exactly 2 different ways.

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

Iain Renfrew (iain.renfrew(AT)btinternet.com), Oct 13 2008

Keywords

• nonn

Comments

This is conjectured to be an infinite sequence.

Subsequence of A051302. [R. J. Mathar, Oct 14 2008]

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

If n is a term 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 and n*k^3 is not in A155961, then n*k^3 is in this sequence for all k > 0. If this sequence is not infinite, then there are infinitely many consecutive k values for any term n such that n*k^3 is in A155961. Is it possible? - Altug Alkan, May 10 2016

Examples

			a(1): 77976^2 = 6080256576 = 1824^3 + 228^3 = 1710^3 + 1026^3;
a(2): 223587^2 = 49991146569 = 3666^3 + 897^3 = 3276^3 + 2457^3;
a(3): 623808^2 = 389136420864 = 7296^3 + 912^3 = 6840^3 + 4104^3;
a(4): 894348^2 = 799858345104 = 9282^3 + 546^3 = 9009^3 + 4095^3.
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..771

Crossrefs

Cf. A145552, A051302.

Extensions

a(5)-a(15) from Zak Seidov, Oct 15 2008

Extended by Ray Chandler, Nov 22 2011