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A341862
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a(n) is the even term in the linear recurrence signature for numerators and denominators of continued fraction convergents to sqrt(n), or 0 if n is a square.
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1
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0, 0, 2, 4, 0, 4, 10, 16, 6, 0, 6, 20, 14, 36, 30, 8, 0, 8, 34, 340, 18, 110, 394, 48, 10, 0, 10, 52, 254, 140, 22, 3040, 34, 46, 70, 12, 0, 12, 74, 50, 38, 64, 26, 6964, 398, 322, 48670, 96, 14, 0, 14, 100, 1298, 364, 970, 178, 30, 302, 198, 1060, 62, 59436
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OFFSET
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0,3
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COMMENTS
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The Everest et al. link states that "the continued fraction expansion of a quadratic irrational is eventually periodic, which implies that the numerators px and denominators qx of its convergents satisfy linear recurrence relations".
Let k be the period length minus one of the continued fraction of sqrt(n). Then the linear recurrence signatures with constant coefficients have the form (0, 0, ..., 0, a(n), 0, 0, ..., 0, (-1)^(n+1)), with k zeroes before and behind a(n).
a(n) is twice the numerator of the convergent to sqrt(n) with index k (starting with 0).
These properties result from the mirrored structure of the period of such continued fractions.
The sequence has remarkably many terms in common with A180495 and with 2*A033313.
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LINKS
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Graham Everest, Alf van der Poorten, Igor Shparlinski, and Thomas Ward, Recurrence Sequences, AMS Mathematical Surveys and Monographs, Volume 104 (2003) p. 8, 5th paragraph.
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FORMULA
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a(n) = 2*A006702(n) if n is not square, otherwise 0.
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EXAMPLE
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The numerators for sqrt(13) begin with 3, 4, 7, 11, 18, 119, ... (A041018) and have the signature (0,0,0,0,36,0,0,0,0,1). The continued fraction has period [1,1,1,1,6], so k=4 and a(13) = 2*A041018(4) = 2*18 = 36. The signature ends with (-1)^4.
The numerators for sqrt(19) begin with 4, 9, 13, 48, 61, 170, 1421, ... (A041028) and have the signature (0,0,0,0,0,340,0,0,0,0,0,-1). The continued fraction has period [2,1,3,1,2,8], so k=5 and a(19) = 2*A041028(5) = 2*170 = 340. The signature ends with (-1)^5.
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CROSSREFS
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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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