OFFSET
1,2
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).
LINKS
Colin Barker, Table of n, a(n) for n = 1..317
Index entries for linear recurrences with constant coefficients, signature (1443,-1443,1).
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.: -x*(61*x^2-242*x+1) / ((x-1)*(x^2-1442*x+1)). - corrected by Colin Barker, Jan 02 2015
a(n) = 1443*a(n-1)-1443*a(n-2)+a(n-3).
MATHEMATICA
LinearRecurrence[{1443, -1443, 1}, {1, 1201, 1731661}, 20] (* Harvey P. Dale, Feb 13 2022 *)
PROG
(PARI) Vec(-x*(61*x^2-242*x+1)/((x-1)*(x^2-1442*x+1)) + O(x^100)) \\ Colin Barker, Jan 02 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Richard Choulet, Sep 21 2007, corrected Sep 29 2007
EXTENSIONS
More terms from Paolo P. Lava, Sep 26 2008
STATUS
approved