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COMMENTS
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The problem is to find p and r such that 6*(2*p-1)^2=5*(2*r+1)^2+1 equivalent to 3*p^2-3*p+1=0.5*(5*r^2+5*r+2). The Diophantine equation (6*X)^2=30*Y^2+6 is such that
X is given by 1, 21, 461, 10121,... with a(n+2)=22*a(n+1)-a(n) and also a(n+1)=11*a(n)+(120*a(n)^2-20)^0.5
Y is given by 1, 23, 805, 11087,... with a(n+2)=22*a(n+1)-a(n) and also a(n+1)=11*a(n)+(120*a(n)^2+24)^0.5
r is given by 0,11, 252, 5543, 121704,... with a(n+2)=22*a(n+1)-a(n)+10 and also a(n+1)=11*a(n)+5+(120*a(n)^2+120*a(n)+36)^0.5
p is given by 1, 11, 231, 5061,... with a(n+2)=22*a(n+1)-a(n)-10 and also a(n+1)=11*a(n)-5+(120*a(n)^2-120*a(n)+25)^0.5
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FORMULA
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a(n+2)=482*a(n+1)-a(n)-150 a(n+1)=241*a(n)-75+11*(480*a(n)^2-300*a(n)+45)^0.5
G.f.: z*(1-152*z+z^2)/((1-z)*(1-482*z+z^2))
a(n)=(5/16)+(11/32)*[241+44*sqrt(30)]^n-(1/16)*sqrt(30)*[241-44*sqrt(30)]^n+(11/32)*[241-44 *sqrt(30)]^n+(1/16)*[241+44*sqrt(30)]^n*sqrt(30), with n>=0 [From Paolo P. Lava, Sep 26 2008]
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