A081324 Twice a square but not the sum of 2 distinct squares.

0, 2, 8, 18, 32, 72, 98, 128, 162, 242, 288, 392, 512, 648, 722, 882, 968, 1058, 1152, 1458, 1568, 1922, 2048, 2178, 2592, 2888, 3528, 3698, 3872, 4232, 4418, 4608, 4802, 5832, 6272, 6498, 6962, 7688, 7938, 8192, 8712, 8978, 9522, 10082, 10368, 11552

Offset: 1

Author

Benoit Cloitre, Apr 20 2003

Keywords

• nonn

Comments

Conjecture: for n>1 this is A050804.

From Altug Alkan, Apr 12 2016: (Start)

Conjecture is true. Proof :

If n = a^2 + b^2, where a and b are nonzero integers, then n^3 = (a^2 + b^2)^3 = A^2 + B^2 = C^2 + D^2 where;

A = 2*a^2*b + (a^2-b^2)*b = 3*a^2*b - b^3,

B = 2*a*b^2 - (a^2-b^2)*a = 3*a*b^2 - a^3,

C = 2*a*b^2 + (a^2-b^2)*a = 1*a*b^2 + a^3,

D = 2*a^2*b - (a^2-b^2)*b = 1*a^2*b + b^3.

Obviously, A, B, C, D are always nonzero because a and b are nonzero integers. Additionally, if a^2 is not equal to b^2, then (A, B) and (C, D) are distinct pairs, that is, n^3 can be expressible as a sum of two nonzero squares more than one way. Since we know that n is a sum of two nonzero squares if and only if n^3 is a sum of two nonzero squares (see comment section of A000404); if n^3 is the sum of two nonzero squares in exactly one way, n must be a^2 + b^2 with a^2 = b^2 and n is the sum of two nonzero squares in exactly one way. That is the definition of this sequence, so this sequence is exactly A050804 except "0" that is the first term of this sequence. (End) [Edited by Altug Alkan, May 14 2016]

Conjecture: sequence consists of numbers of form 2*k^2 such that sigma(2*k^2)==3 (mod 4) and k is not divisible by 5.

The reason of related observation is that 5 is the least prime of the form 4*m+1. However, counterexamples can be produced. For example 57122 = 2*169^2 and sigma(57122) == 3 (mod 4) and it is not divisible by 5. - Altug Alkan, Jun 10 2016

For n > 0, this sequence lists numbers n such that n is the sum of two nonzero squares while n^2 is not. - Altug Alkan, Apr 11 2016

2*k^2 where k has no prime factor == 1 (mod 4). - Robert Israel, Jun 10 2016

Links

• Evan M. Bailey, Table of n, a(n) for n = 1..20000 (first 115 terms from Reinhard Zumkeller, terms 116-1000 from Zak Seidov)
• Evan M. Bailey, a081324.cpp

Crossrefs

Cf. A050804, A025284, A063725, A125022, A018825, A000404, A004431.

Programs

Haskell

import Data.List (elemIndices)
a081324 n = a081324_list !! (n-1)
a081324_list = 0 : elemIndices 1 a063725_list
-- Reinhard Zumkeller, Aug 17 2011

Maple

map(k -> 2*k^2, select(k -> andmap(t -> t[1] mod 4 <> 1, ifactors(k)[2]), [$0..100])); # Robert Israel, Jun 10 2016

Mathematica

Select[ Range[0, 12000], MatchQ[ PowersRepresentations[#, 2, 2], {{n_, n_}}] &] (* Jean-François Alcover, Jun 18 2013 *)

PARI

concat([0,2],apply(n->2*n^2, select(n->vecmin(factor(n)[, 1]%4)>1, vector(100,n,n+1)))) \\ Charles R Greathouse IV, Jun 18 2013

Formula

A063725(a(n)) = 1. [Reinhard Zumkeller, Aug 17 2011]

a(n) = 2*A004144(n-1)^2 for n > 1. - Charles R Greathouse IV, Jun 18 2013

Extensions

a(19)-a(45) from Donovan Johnson, Nov 15 2009

Offset corrected by Reinhard Zumkeller, Aug 17 2011