

A049685


a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).


28



1, 6, 41, 281, 1926, 13201, 90481, 620166, 4250681, 29134601, 199691526, 1368706081, 9381251041, 64300051206, 440719107401, 3020733700601, 20704416796806, 141910183877041, 972666870342481
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OFFSET

0,2


COMMENTS

In general, sum{k=0..n, binomial(2*nk,k)j^(nk)} = (1)^n*U(2n, I*sqrt(j)/2), I=sqrt(1).  Paul Barry, Mar 13 2005
a(n) = L(n,7), where L is defined as in A108299; see also A033890 for L(n,7).  Reinhard Zumkeller, Jun 01 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 01avoiding 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


LINKS

Indranil Ghosh, Table of n, a(n) for n = 0..1193
Tanya Khovanova, Recursive Sequences
J.C. Novelli, J.Y. Thibon, Hopf Algebras of mpermutations,(m+1)ary trees, and mparking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their Nnumbers, not their Anumbers.
Index entries for linear recurrences with constant coefficients, signature (7,1).


FORMULA

Let q(n, x)=sum(i=0, n, x^(ni)*binomial(2*ni, i)); then q(n, 5)=a(n); a(n) = 7a(n1)  a(n2).  Benoit Cloitre, Nov 10 2002
From Ralf Stephan, May 29 2004: (Start)
a(n+2) = 7a(n+1)  a(n).
G.f.: (1x)/(17x+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(n1).  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), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0].  Gary W. Adamson, Mar 21 2008


EXAMPLE

a(3) = L(4 * 3 + 2) / 3 = 843 / 3 = 281.  Indranil Ghosh, Feb 06 2017


MATHEMATICA

Table[LucasL[4*n+2]/3, {n, 0, 50}] (* or *) LinearRecurrence[{7, 1}, {1, 6}, 50] (* G. C. Greubel, Dec 17 2017 *)


PROG

(Sage) [lucas_number1(n, 7, 1)lucas_number1(n1, 7, 1) for n in xrange(1, 20)] # Zerinvary Lajos, Nov 10 2009
(PARI) a(n)=(fibonacci(4*n+1)+fibonacci(4*n+3))/3 \\ Charles R Greathouse IV, Jun 16 2014
(MAGMA) [Lucas(4*n+2)/3: n in [0..30]]; // G. C. Greubel, Dec 17 2017


CROSSREFS

Row 7 of array A094954.
Cf. A004187.
Cf. similar sequences listed in A238379.
Sequence in context: A015551 A291018 A227214 * A196954 A122371 A083067
Adjacent sequences: A049682 A049683 A049684 * A049686 A049687 A049688


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling


STATUS

approved



