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A049685
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a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).
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35
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1, 6, 41, 281, 1926, 13201, 90481, 620166, 4250681, 29134601, 199691526, 1368706081, 9381251041, 64300051206, 440719107401, 3020733700601, 20704416796806, 141910183877041, 972666870342481
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OFFSET
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0,2
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COMMENTS
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In general, Sum_{k=0..n} binomial(2*n-k,k)j^(n-k) = (-1)^n*U(2n, I*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
Take 7 numbers consisting of 5 ones together with any two successive terms from this sequence. This set has the property that the sum of their squares is 7 times their product. (R. K. Guy, Oct 12 2005.) See also A111216.
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
All positive solutions of the Diophantine equation x^2 + y^2 - 7*x*y = -5 are given by [x(n) = S(n, 7) - S(n-1, 7), y(n) = x(n-1)], for all integer numbers n, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-n, x) = -S(n-2, x), for n >= 2. x(n) = a(n), for n >= 0.
This indefinite binary quadratic form has discriminant D = +45. There is only this family representing -5 properly with x and y positive, and there are no improper solutions.
All proper and improper solutions of the generalized Pell equation X^2 - 45*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n) from the preceding comment, by X(n) = x(n) + x(n-1) = S(n-1, 7) - S(n-2, 7) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 7), for all integer numbers n. For positive integers X(n) = A056854(n) and Y(n) = A004187(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper family of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer numbers n.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
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LINKS
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FORMULA
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Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i); then q(n, 5)=a(n); a(n) = 7a(n-1) - a(n-2). - Benoit Cloitre, Nov 10 2002
a(n+2) = 7a(n+1) - a(n).
G.f.: (1-x)/(1-7x+x^2).
a(n)*a(n+3) = 35 + a(n+1)*a(n+2). (End)
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*5^k. - Paul Barry, Aug 30 2004
If another "1" is inserted at the beginning of the sequence, then A002310, A002320 and A049685 begin with 1, 2; 1, 3; and 1, 1; respectively and satisfy a(n+1) = (a(n)^2+5)/a(n-1). - Graeme McRae, Jan 30 2005
a(n) = (-1)^n*U(2n, i*sqrt(5)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = S(n, 7) - S(n-1, 7) with Chebyshev S polynomials S(n-1, 7) = A004187(n), for n >= 0. - Wolfdieter Lang, Feb 09 2021
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EXAMPLE
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MATHEMATICA
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Table[LucasL[4*n+2]/3, {n, 0, 50}] (* or *) LinearRecurrence[{7, -1}, {1, 6}, 50] (* G. C. Greubel, Dec 17 2017 *)
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PROG
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(Sage) [lucas_number1(n, 7, 1)-lucas_number1(n-1, 7, 1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009
(Magma) [Lucas(4*n+2)/3: n in [0..30]]; // G. C. Greubel, Dec 17 2017
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CROSSREFS
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Cf. similar sequences listed in A238379.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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