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%I #73 Sep 08 2022 08:44:58
%S 1,6,41,281,1926,13201,90481,620166,4250681,29134601,199691526,
%T 1368706081,9381251041,64300051206,440719107401,3020733700601,
%U 20704416796806,141910183877041,972666870342481
%N a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).
%C 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
%C a(n) = L(n,7), where L is defined as in A108299; see also A033890 for L(n,-7). - _Reinhard Zumkeller_, Jun 01 2005
%C 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.
%C 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
%C 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
%C From _Wolfdieter Lang_, Feb 09 2021: (Start)
%C 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.
%C 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.
%C 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.
%C 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.
%C 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)
%H Indranil Ghosh, <a href="/A049685/b049685.txt">Table of n, a(n) for n = 0..1193</a>
%H Alex Fink, Richard K. Guy, and Mark Krusemeyer, <a href="https://doi.org/10.11575/cdm.v3i2.61940">Partitions with parts occurring at most thrice</a>, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H J.-C. Novelli and J.-Y. Thibon, <a href="https://arxiv.org/abs/1403.5962">Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions</a>, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
%H John Riordan, <a href="/A002720/a002720_3.pdf">Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences</a>. Note that the sequences are identified by their N-numbers, not their A-numbers.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (7,-1).
%F 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
%F From _Ralf Stephan_, May 29 2004: (Start)
%F a(n+2) = 7a(n+1) - a(n).
%F G.f.: (1-x)/(1-7x+x^2).
%F a(n)*a(n+3) = 35 + a(n+1)*a(n+2). (End)
%F a(n) = Sum_{k=0..n} binomial(n+k, 2k)*5^k. - _Paul Barry_, Aug 30 2004
%F 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
%F 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
%F [a(n), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0]. - _Gary W. Adamson_, Mar 21 2008
%F 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
%e a(3) = L(4*3 + 2)/3 = 843/3 = 281. - _Indranil Ghosh_, Feb 06 2017
%t Table[LucasL[4*n+2]/3, {n,0,50}] (* or *) LinearRecurrence[{7,-1}, {1,6}, 50] (* _G. C. Greubel_, Dec 17 2017 *)
%o (Sage) [lucas_number1(n,7,1)-lucas_number1(n-1,7,1) for n in range(1, 20)] # _Zerinvary Lajos_, Nov 10 2009
%o (PARI) a(n)=(fibonacci(4*n+1)+fibonacci(4*n+3))/3 \\ _Charles R Greathouse IV_, Jun 16 2014
%o (Magma) [Lucas(4*n+2)/3: n in [0..30]]; // _G. C. Greubel_, Dec 17 2017
%Y Row 7 of array A094954. First differences of A004187.
%Y Cf. A002310, A002320, A049310, A049685, A056854.
%Y Cf. similar sequences listed in A238379.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_
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