login
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
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^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
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
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 *)
PROG
(Haskell)
import Data.List (elemIndices)
a081324 n = a081324_list !! (n-1)
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
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