A084109 n is congruent to 1 (mod 4) and is not the sum of two squares.

21, 33, 57, 69, 77, 93, 105, 129, 133, 141, 161, 165, 177, 189, 201, 209, 213, 217, 237, 249, 253, 273, 285, 297, 301, 309, 321, 329, 341, 345, 357, 381, 385, 393, 413, 417, 429, 437, 453, 465, 469, 473, 489, 497

Offset: 1

Author

William P. Orrick, Jun 18 2003

Keywords

• easy
• nonn

Comments

Alternatively, n is congruent to 1 (mod 4) with at least 2 distinct prime factors congruent to 3 (mod 4) in the squarefree part of n. - Comment corrected by Jean-Christophe Hervé, Oct 25 2015

Applications to the theory of optimal weighing designs and maximal determinants: An (n+1) X (n+1) conference matrix is impossible.

The upper bound of Ehlich/Wojtas on the determinant of a (0,1) matrix of order congruent to 1 (mod 4) cannot be achieved for n X n matrices.

The bound of Ehlich/Wojtas on the determinant of a (-1,1) matrix of order congruent to 2 (mod 4) cannot be achieved for (n+1) X (n+1) matrices.

Numbers with only odd prime factors, of which a strictly positive even number are raised to an odd power and congruent to 3 (mod 4). - Jean-Christophe Hervé, Oct 24 2015

Examples

			a(1) = 3*7 = 21, a(2) = 3*11 = 33, a(3) = 3*19 = 57, a(14) = 3^3*7 = 189.
		

References

• F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 56.

Links

• Jean-Christophe Hervé, Table of n, a(n) for n = 1..1000
• H. Ehlich, Determinantenabschätzungen für binäre Matrizen, Math. Z. 83 (1964) 123-132.
• D. Raghavarao, Some aspects of weighing designs, Ann. Math. Stat. 31 (1960) 878-884.

Crossrefs

Cf. A000952, A003432, A003433, A022544, A001481.

Programs

Maple

N:= 1000: # to get all entries <= N
S:= {seq(i,i=1..N,4)} minus
   {seq(seq(i^2+j^2, j=1..floor(sqrt(N-i^2)),2),i=0..floor(sqrt(N)),2)}:
sort(convert(S,list)); # Robert Israel, Oct 25 2015

Mathematica

a[m_] := Complement[Range[1, m, 4], Union[Flatten[Table[j^2+k^2, {j, 1, Sqrt[m], 2}, {k, 0, Sqrt[m], 2}]]]]

PARI

is(n)=if(n%4!=1, return(0)); my(f=factor(n)); for(i=1, #f~, if(f[i, 1]%4==3 && f[i, 2]%2, return(1))); 0 \\ Charles R Greathouse IV, Jul 01 2016