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Number of solutions to 5+B^2=p^2+q^2 with B=2n, p,q>0 and 2p^2<5+B^2.
0

%I #3 Feb 27 2009 03:00:00

%S 1,1,0,1,0,0,1,0,1,0,1,0,0,0,0,2,0,0,1,0,0,2,0,0,1,0,0,0,1,0,0,0,0,1,

%T 0,2,1,0,0,1,0,0,0,0,0,2,1,0,1,0,0,0,0,1,0,2,0,1,0,0,0,0,1,1,0,0,2,0,

%U 0,0,0,0,0,0,0,0,0,0,2,0,2,1,1,0,1,0,0,0,0,0,2,1,0,0,0,0,0,0,0,1,0,0,1,0,0

%N Number of solutions to 5+B^2=p^2+q^2 with B=2n, p,q>0 and 2p^2<5+B^2.

%C The number of matrices with entries in Z to G^2=G+1 not of the form gI or g'I (g, the golden number and g'=1-g are the solutions to x^2=x+1), hence of the form (1+p q-B | q+B 1-p) with p^2+q^2=5+B^2 is given by 8a(n) for n!=1 and by 4a(1)=4 for n=1.

%e a(0)=1 because 5+0^2=5=1^2+2^2. a(15)=2 because 5+30^2=905=8^2+29^2=11^2+28^2.

%Y Cf. A104768.

%K easy,nonn

%O 0,16

%A Michele Dondi (blazar(AT)lcm.mi.infn.it), Mar 24 2005