

A081324


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


9



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
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OFFSET

1,2


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

Reinhard Zumkeller and Zak Seidov, Table of n, a(n) for n = 1..1000 (first 115 terms from Reinhard Zumkeller)


FORMULA

A063725(a(n)) = 1. [Reinhard Zumkeller, Aug 17 2011]
a(n) = 2*A004144(n1)^2 for n > 1.  Charles R Greathouse IV, Jun 18 2013


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_}}] &] (* JeanFrançois Alcover, Jun 18 2013 *)


PROG

(Haskell)
import Data.List (elemIndices)
a081324 n = a081324_list !! (n1)
a081324_list = 0 : elemIndices 1 a063725_list
 Reinhard Zumkeller, Aug 17 2011
(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


CROSSREFS

Cf. A050804, A025284, A063725, A125022, A018825, A000404, A004431.
Sequence in context: A001105 A051787 A050804 * A190787 A018229 A166830
Adjacent sequences: A081321 A081322 A081323 * A081325 A081326 A081327


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Apr 20 2003


EXTENSIONS

a(19)a(45) from Donovan Johnson, Nov 15 2009
Offset corrected by Reinhard Zumkeller, Aug 17 2011


STATUS

approved



