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A009759
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Expansion of (3 - 21*x + 4*x^2)/((x-1)*(x^2 - 6*x + 1)).
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1
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-3, 0, 17, 116, 693, 4056, 23657, 137900, 803757, 4684656, 27304193, 159140516, 927538917, 5406093000, 31509019097, 183648021596, 1070379110493, 6238626641376, 36361380737777, 211929657785300, 1235216565974037
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OFFSET
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0,1
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COMMENTS
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If k satisfies the previous condition, i.e., P(k) = 2*k^2 + 14*k + 25 is a square, then there exists an integer m such that (k+3)^2 + (k+4)^2 = m^2; finding such numbers k is equivalent to finding Pythagorean triples (X,X+1,Z) with X in A001652 (1st formula). For k >= 0, P(k) = A001844(k+3), and the successive values of P(k) that are squares are in A008844. - Bernard Schott, Mar 24 2019
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REFERENCES
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Albert H. Beiler, Recreations in the Theory of Numbers, Dover, New-York, 1964, pp. 122-124.
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LINKS
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FORMULA
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a(n) = ( (1+sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 14 )/4.
a(n) = (Q(2*n+1) -14)/4 = (4*P(n)*P(n+1) + (-1)^n - 7)/2, where P(n) = A000129(n) (Pell) and Q(n) = A002203(n) (Pell-Lucas). - G. C. Greubel, Apr 04 2019
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MATHEMATICA
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CoefficientList[Series[(3-21x+4x^2)/((x-1)(x^2-6x+1)), {x, 0, 30}], x] (* or *) LinearRecurrence[{7, -7, 1}, {-3, 0, 17}, 30] (* Harvey P. Dale, Dec 12 2016 *)
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PROG
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(PARI) my(x='x+O('x^30)); Vec((3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))) \\ G. C. Greubel, Feb 12 2018
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))) ) // G. C. Greubel, Feb 12 2018
(Sage) ((3-21*x+4*x^2)/((x-1)*(x^2-6*x+1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 04 2019
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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G.f. and Binet formula corrected by R. J. Mathar, Aug 24 2016
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STATUS
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approved
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