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A041007
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Denominators of continued fraction convergents to sqrt(6).
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10
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1, 2, 9, 20, 89, 198, 881, 1960, 8721, 19402, 86329, 192060, 854569, 1901198, 8459361, 18819920, 83739041, 186298002, 828931049, 1844160100, 8205571449, 18255302998, 81226783441, 180708869880
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OFFSET
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0,2
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COMMENTS
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sqrt(6) = 4/2 + 4/9 + 4/(9*89) + 4/(89*881) + 4/(881*8721), ...; where sqrt(6) = 2.4494897427... and the sum of the first 5 terms of this series = 2.449489737... - Gary W. Adamson, Dec 21 2007
sqrt(6) = 2 + continued fraction [2, 4, 2, 4, 2, 4, ...] = 4/2 + 4/9 + 4/(9*89) + 4/(89*881) + 4/(881*8721) + ... - Gary W. Adamson, Dec 21 2007
For n > 0, a(n) equals the permanent of the n X n tridiagonal matrix with the main diagonal alternating sequence [2, 4, 2, 4, ...] and 1's along the superdiagonal and the subdiagonal. - Rogério Serôdio, Apr 01 2018
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LINKS
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FORMULA
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Empirical g.f.: (1+2*x-x^2)/(1-10*x^2+x^4). - Colin Barker, Dec 31 2011
Recurrence formula: a(n) = (3 + (-1)^n)*a(n-1) + a(n-2), a(0) = 1, a(1) = 2.
Some properties:
(1) a(n)^2 - a(n-2)^2 = (3+(-1)^n)*a(2*n-1), for n > 1;
(2) a(2*n+1) = a(n)*(a(n+1) + a(n-1)), for n > 0;
(3) a(2*n) = A142239(2*n), for n >= 0;
(4) a(2*n+1) = A041007(2*n+1)/2, for n >= 0;
(5) a(2*n-1)*A142239(2*n+1) = a(n)^2 - 1, for n > 0;
(7) Sum_{k=0..n} a(2*k+1)*(A142239(2*k) + A142239(2*(k+1))) = Sum_{k=0..n} a(3+4*k);
(8) Sum_{k=0..n} (a(2*k-1) + a(2*k+1))*A142239(2*k) = Sum_{k=0..n} A142239(3+4*k). (End)
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MATHEMATICA
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CROSSREFS
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KEYWORD
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nonn,cofr,frac,easy
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AUTHOR
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STATUS
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approved
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