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COMMENTS
| We solve r^2+(r+1)^2=5*p^2-5*p+1 equivalent to 2*(2*r+1)^2=5*(2*p-1)^2-3. the Diophantine equation (2*X)^2=10*Y^2-6 is such that
X is given by 1, 49,1861,70669,... with a(n+2)=38*a(n+1)-a(n) and also a(n+1)=19*a(n)+(360*a(n)^2+540)^0.5
Y is given by 1, 31,1177,44695,... with a(n+2)=38*a(n+1)-a(n) and also a(n+1)=19*a(n)+(360*a(n)^2-216)^0.5
r is given by 0, 24,930,35334,... with a(n+2)=38*a(n+1)-a(n)+18 and also a(n+1)=19*a(n)+9+(360*a(n)^2+360*a(n)+225)^0.5 (new sequence it seems)
p is given by 1, 16,589, 22345,... with a(n+2)=38*a(n+1)-a(n)-18 and also a(n+1)=19*a(n)-9+(360*a(n)^2-360*a(n)+36)^0.5 (new sequence it seems)
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FORMULA
| a(n+2)=1442*a(n+1)-a(n)-180,
a(n+1)=721*a(n)-90+38*(360*a(n)^2-90*a(n)-45)^0.5.
G.f.: z*(1-242*z+z^2)/((1-z)*(1-442*z+z^2))
a(n)=(1/8)+(7/16)*[721-228*sqrt(10)]^n-(1/8)*[721-228*sqrt(10)]^n*sqrt(10)+(1/8)*[721+228 *sqrt(10)]^n*sqrt(10)+(7/16)*[721+228*sqrt(10)]^n, with n>=0 [From Paolo P. Lava (paoloplava(AT)gmail.com), Sep 26 2008]
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