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A084109 n is congruent to 1 (mod 4) and is not the sum of two squares. 8

%I #25 Jul 01 2016 22:46:24

%S 21,33,57,69,77,93,105,129,133,141,161,165,177,189,201,209,213,217,

%T 237,249,253,273,285,297,301,309,321,329,341,345,357,381,385,393,413,

%U 417,429,437,453,465,469,473,489,497

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

%C 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

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

%C 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.

%C 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.

%C 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

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

%H Jean-Christophe Hervé, <a href="/A084109/b084109.txt">Table of n, a(n) for n = 1..1000</a>

%H H. Ehlich, <a href="http://dx.doi.org/10.1007/BF01111249">Determinantenabschätzungen für binäre Matrizen</a>, Math. Z. 83 (1964) 123-132.

%H D. Raghavarao, <a href="http://dx.doi.org/10.1214/aoms/1177705664">Some aspects of weighing designs</a>, Ann. Math. Stat. 31 (1960) 878-884.

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

%p N:= 1000: # to get all entries <= N

%p S:= {seq(i,i=1..N,4)} minus

%p {seq(seq(i^2+j^2, j=1..floor(sqrt(N-i^2)),2),i=0..floor(sqrt(N)),2)}:

%p sort(convert(S,list)); # _Robert Israel_, Oct 25 2015

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

%o (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

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

%K easy,nonn

%O 1,1

%A _William P. Orrick_, Jun 18 2003

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