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A157120
Positive numbers y such that y^2 is of the form x^2+(x+103)^2 with integer x.
4
73, 103, 205, 233, 515, 1157, 1325, 2987, 6737, 7717, 17407, 39265, 44977, 101455, 228853, 262145, 591323, 1333853, 1527893, 3446483, 7774265, 8905213, 20087575, 45311737, 51903385, 117078967, 264096157, 302515097, 682386227, 1539265205
OFFSET
1,1
COMMENTS
(-48, a(1)) and (A157119(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+103)^2 = y^2.
FORMULA
a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1) = 73, a(2) = 103, a(3) = 205, a(4) = 233, a(5) = 515, a(6) = 1157.
G.f.: x*(1-x)*(73+176*x+381*x^2+176*x^3+73*x^4) / (1-6*x^3+x^6).
a(3*k-1) = 103*A001653(k) for k >= 1.
Limit_{n -> oo} a(n)/a(n-3) = 3+2*sqrt(2).
Limit_{n -> oo} a(n)/a(n-1) = (3+2*sqrt(2))*(11-3*sqrt(2))^2/(11+3*sqrt(2))^2 for n mod 3 = 1.
Limit_{n -> oo} a(n)/a(n-1) = (11+3*sqrt(2))/(11-3*sqrt(2)) for n mod 3 = {0, 2}.
EXAMPLE
(-48, a(1)) = (-48, 73) is a solution: (-48)^2+(-48+103)^2 = 2304+3025 = 5329 = 73^2.
(A157119(1), a(2)) = (0, 103) is a solution: 0^2+(0+103)^2 = 10609 = 103^2,
(A157119(3), a(4)) = (105, 233) is a solution: 105^2+(105+103)^2 = 11025+43264 = 54289 = 233^2.
MATHEMATICA
Select[Table[Sqrt[x^2+(x+103)^2], {x, -50, 3*10^6}], IntegerQ] (* THe program generates the first 20 terms of the sequence. *) (* or *) LinearRecurrence[ {0, 0, 6, 0, 0, -1}, {73, 103, 205, 233, 515, 1157}, 50](* Harvey P. Dale, Aug 19 2020 *)
PROG
(PARI) {forstep(n=-48, 1100000000, [1, 3], if(issquare(2*n^2+206*n+10609, &k), print1(k, ", ")))}
CROSSREFS
Cf. A157119, A001653, 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: A116210 A180522 A238679 * A118220 A144050 A087878
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Feb 25 2009
EXTENSIONS
Typo corrected by Klaus Brockhaus, Mar 01 2009
STATUS
approved