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A086378
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Sum of successive remainders in computing Euclidean algorithm for (1,1/sqrt(n)) is rational.
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5
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1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 55, 56, 64, 70, 72, 81, 90, 100, 110, 121, 132, 144, 153, 155, 156, 169, 180, 182, 196, 210, 225, 240, 256, 272, 289, 305, 306, 324, 342, 361, 377, 380, 400, 420, 441, 462, 484, 504, 505, 506, 529, 546, 552, 576, 600
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OFFSET
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1,2
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COMMENTS
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For a real number x, take (a_0,b_0) = (1,x), then (a_(i+1),b_(i+1)) = (b_i,a_i-b_i*floor(a_i/b_i)), for i>=0 and call kappa(x) = b_1+b_2+b_3+... If kappa(1/sqrt(n)) is rational (which can be easily evaluated thanks to the periodicity of the process for a quadratic number), then n is in the sequence.
An infinity of 2nd degree polynomial functions take all their values over N in the sequence (such as x^2, x^2+x, 36*x^2+17*x+2, 100*x^2+150*x+55, 196*x^2+97*x+12, ...).
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LINKS
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EXAMPLE
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7137 is in the sequence because kappa(1/sqrt(7137)) = 7/1098 (in Q).
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MATHEMATICA
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kappa[n_] := Module[{a, b, i, p}, If[(a = Sqrt[n] - Floor[Sqrt[n]]) == 0, Return[0]]; i = a = Simplify[1/a]; p = 1; b = 0; Do[p = Simplify[a*p]; b = Simplify[a*b - Floor[a] + a]; If[(a = Simplify[1/(a - Floor[a])]) == i, Break[]], {Infinity}]; Return[Simplify[(1/a + b/(p-1))/Sqrt[n], Sqrt]]];
Reap[For[n = 1, n <= 600, n++, If[Element[kappa[n], Rationals], Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Apr 15 2017, translated from MuPAD *)
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PROG
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(MuPAD) kappa_1_over_sqrt := proc(n) local a, b, i, p; begin if (a := sqrt(n)-isqrt(n)) = 0 then return(0) end_if: i := a := simplify(1/a, sqrt); p := 1; b := 0; repeat p := simplify(p*a, sqrt); b := simplify(b*a+a-floor(a), sqrt); until (a := simplify(1/(a-floor(a)), sqrt)) = i end_repeat: return(simplify((b/(p-1) + 1/a)/sqrt(n), sqrt)); end_proc:
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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