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A157119
Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+103)^2 = y^2.
5
0, 84, 105, 309, 765, 884, 2060, 4712, 5405, 12257, 27713, 31752, 71688, 161772, 185313, 418077, 943125, 1080332, 2436980, 5497184, 6296885, 14204009, 32040185, 36701184, 82787280, 186744132, 213910425, 482519877, 1088424813, 1246761572
OFFSET
1,2
COMMENTS
Corresponding values y of solutions (x, y) are in A157120.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (11+3*sqrt(2))/(11-3*sqrt(2)) for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2))*(11-3*sqrt(2))^2/(11+3*sqrt(2))^2 for n mod 3 = 0.
For the generic case x^2 + (x + p)^2 = y^2 with p = m^2 - 18 a (prime) number, m >= 11 (p >= 103), the first three consecutive solutions are (0, p), (6*m+18, m^2+6*m+18), (3*m^2-30*m+72, 5*m^2-42*m+90), and the other solutions are defined by (X(n), Y(n)) = (3*X(n-3) + 2*Y(n-3) + p, 4*X(n-3) + 3*Y(n-3) + 2*p), X(n) = 6*X(n-3) - X(n-6) + 2*p, and Y(n) = 6*Y(n-3) - Y(n-6) (can be easily proved using X(n) = 3*X(n-3) + 2*Y(n-3) + p, and Y(n) = 4*X(n-3) + 3*Y(n-3) + 2*p). - Mohamed Bouhamida, Jun 03 2026
FORMULA
a(n) = 6*a(n-3)-a(n-6)+206 for n > 6; a(1) = 0, a(2) = 84, a(3) = 105, a(4) = 309, a(5) = 765, a(6) = 884.
G.f.: x*(84+21*x+204*x^2-48*x^3-7*x^4-48*x^5)/((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 103*A001652(k) for k >= 0.
EXAMPLE
For p=103 (m=11) the first three consecutive solutions are (0, 103), (84, 205), (105, 233).
PROG
(PARI) {forstep(n=0, 1300000000, [1, 3], if(issquare(2*n^2+206*n+10609), print1(n, ", ")))}
CROSSREFS
Cf. A157120, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A157121 (decimal expansion of 11+3*sqrt(2)), A157122 (decimal expansion of 11-3*sqrt(2)), A157123 (decimal expansion of (11+3*sqrt(2))/(11-3*sqrt(2))).
Sequence in context: A214866 A383932 A111313 * A209204 A219801 A316833
KEYWORD
nonn,easy,changed
AUTHOR
Klaus Brockhaus, Feb 25 2009
STATUS
approved