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A160577
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Positive numbers y such that y^2 is of the form x^2+(x+409)^2 with integer x.
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3
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305, 409, 641, 1189, 2045, 3541, 6829, 11861, 20605, 39785, 69121, 120089, 231881, 402865, 699929, 1351501, 2348069, 4079485, 7877125, 13685549, 23776981, 45911249, 79765225, 138582401, 267590369, 464905801, 807717425, 1559630965
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OFFSET
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1,1
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COMMENTS
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(-136, a(1)) and (A129641(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+409)^2 = y^2.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (473+168*sqrt(2))/409 for n mod 3 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (204819+83570*sqrt(2))/409^2 for n mod 3 = 1.
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LINKS
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FORMULA
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a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=305, a(2)=409, a(3)=641, a(4)=1189, a(5)=2045, a(6)=3541.
G.f.: (1-x)*(305+714*x+1355*x^2+714*x^3+305*x^4) / (1-6*x^3+x^6).
a(3*k-1) = 409*A001653(k) for k >= 1.
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EXAMPLE
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(-136, a(1)) = (-136, 305) is a solution: (-136)^2+(-136+409)^2 = 18496+74529 = 93025 = 305^2.
(A129641(1), a(2)) = (0, 409) is a solution: 0^2+(0+409)^2 = 167281 = 409^2.
(A129641(3), a(4)) = (611, 1189) is a solution: 611^2+(611+409)^2 = 373321+1040400 = 1413721 = 1189^2.
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MATHEMATICA
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LinearRecurrence[{0, 0, 6, 0, 0, -1}, {305, 409, 641, 1189, 2045, 3541}, 50] (* or *) Select[Table[Sqrt[x^2+(x+409)^2], {x, -140, 10^6}], IntegerQ] (* The second program generates the first 16 terms of the sequence. To generate more, increase the x constant but the program may take a long time to run. *) (* Harvey P. Dale, Mar 14 2022 *)
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PROG
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(PARI) {forstep(n=-136, 10000000, [3, 1], if(issquare(2*n^2+818*n+167281, &k), print1(k, ", ")))}
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CROSSREFS
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Cf. A129641, A001653, A156035 (decimal expansion of 3+2*sqrt(2)), A160578 (decimal expansion of (473+168*sqrt(2))/409), A160579 (decimal expansion of (204819+83570*sqrt(2))/409^2).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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