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A007655
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Standard deviation of A007654.
(Formerly M4948)
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22
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0, 1, 14, 195, 2716, 37829, 526890, 7338631, 102213944, 1423656585, 19828978246, 276182038859, 3846719565780, 53577891882061, 746243766783074, 10393834843080975, 144767444036350576
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| a(n)=A001353(2n)/4. a(n) corresponds also to one-sixth the area of Fleenor-Heronian triangle with middle side A003500(n). - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 15 2002
a(n) give all (nontrivial, integer) solutions of Pell equation b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)=A011943(n), n>=0.
Number of units of a(n) belongs to a periodic sequence: 0, 1, 4, 5, 6, 9.We conclude that a(n) and a(n+6) have the same number of units. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 05 2009]
For n>=3, a(n) equals the permanent of the (n-2)X(n-2) tridiagonal matrix with 14's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). [From John M. Campbell, Jul 08 2011]
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REFERENCES
| D. A. Benaron, personal communication.
E. K. Lloyd (E.K.Lloyd(AT)maths.soton.ac.uk), "The standard deviation of 1, 2, .., n, Pell's equation and rational triangles", preprint.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..100
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to linear recurrences with constant coefficients
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
| a(n) = 14*a(n-1) - a(n-2). G.f.: (x^2)/(1-14*x+x^2).
a(n+1) ~ 1/24*sqrt(3)*(2 + sqrt(3))^(2*n) - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n+1) = S(n-1, 14), n>=0, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n+1) = ((7+4*sqrt(3))^n - (7-4*sqrt(3))^n)/(8*sqrt(3)).
a(n+1) = sqrt((A011943(n)^2 - 1)/48), n>=0.
Chebyshev's polynomials U(n-2, x) evaluated at x=7.
4*a(n+1) + A046184(n) = A055793(n+2) + A098301(n+1) 4*a(n+1) + A098301(n+1) + A055793(n+2) = A046184(n+1) (4*a(n+1))^2 = A098301(2n+1) (conjectures) - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Nov 02 2004
(4*a(n))^2 = A103974(n)^2 - A011922(n-1)^2. - Paul D. Hanna (pauldhanna(AT)juno.com), Mar 06 2005
a(n) = 13*(a(n-1)+a(n-2))-a(n-3), a(n) = 15*(a(n-1)-a(n-2))+a(n-3). a(n)=14*a(n-1)-a(n-2). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), May 26 2007
a(n)=b such that (-1)^n/4*Integral_{x=-Pi/2..Pi/2} (sin((2*n-2)*x))/(2-sin(x)) dx = c+b*ln(3). [From Francesco Daddi, Aug 02 2011]
a(n+2) = Sum_{k, 0<=k<=n} A101950(n,k)*13^k. - DELEHAM Philippe, Feb 10 2012
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MATHEMATICA
| lst={}; Do[AppendTo[lst, GegenbauerC[n, 1, 7]], {n, 0, 8^2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 11 2008]
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PROG
| sage: [lucas_number1(n, 14, 1) for n in xrange(0, 20)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008
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CROSSREFS
| Cf. A001353, A003500.
Cf. A011945, A067900.
Cf. A103974, A011922.
Sequence in context: A086946 A158530 A171319 * A207030 A207065 A207074
Adjacent sequences: A007652 A007653 A007654 * A007656 A007657 A007658
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KEYWORD
| nonn,easy,changed
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Chebyshev comments from W. Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Nov 08 2002
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