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A236506
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Conjectured least index k > 2 of Fibonacci-like sequence f(i+2) = f(i+1) + f(i), with f(1)=1 and f(2)=n, such that f(k) is a square, or k=0 if squares do not exist in the corresponding sequence.
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3
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12, 11, 3, 4, 0, 10, 0, 3, 11, 10, 0, 4, 0, 0, 3, 0, 0, 11, 15, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 3, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 3, 0, 0, 12, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 11, 10, 0, 4
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OFFSET
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1,1
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COMMENTS
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n=1 corresponds to the usual Fibonacci sequence, for which f(12) = 144 = 12^2, so a(1)=12.
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REFERENCES
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J. H. E. Cohn, On square Fibonacci numbers, J. London Math. Soc. 39 (1964), 537-540.
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LINKS
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EXAMPLE
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a(6)=10 since f(10) = 361 = 19^2 for sequence starting f(1)=1, f(2)=6.
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MATHEMATICA
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squareQ[n_] := IntegerQ[Sqrt[n]]; nn = 100; Table[f = {1, n}; Do[AppendTo[f, f[[-1]] + f[[-2]]], {i, 3, nn}]; k = 2; While[k++; k <= nn && ! squareQ[f[[k]]]]; If[k > nn, k = 0]; k, {n, 100}] (* T. D. Noe, Jan 28 2014 *)
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CROSSREFS
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Cf. A233513 (triangular array for f(1)=m and f(2)=n).
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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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