Search: id:A049685
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%I A049685
%S A049685 1,6,41,281,1926,13201,90481,620166,4250681,29134601,199691526,
%T A049685 1368706081,9381251041,64300051206,440719107401,3020733700601,
%U A049685 20704416796806,141910183877041,972666870342481
%N A049685 a(n)=L(4n+2)/3, where L=A000032 (the Lucas sequence).
%C A049685 In general, sum{k=0..n, binomial(2n-k,k)j^(n-k)}=(-1)^n*U(2n,I*sqrt(j)/
               2), I=sqrt(-1); - Paul Barry (pbarry(AT)wit.ie), Mar 13 2005
%C A049685 a(n) = L(n,7), where L is defined as in A108299; see also A033890 for 
               L(n,-7). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Jun 01 2005
%C A049685 Take 7 numbers consisting of 5 ones together with any two successive 
               terms from this ssequence. 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 A049685 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 (tanyakh(AT)yahoo.com), Jan 10 
               2007
%H A049685 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%H A049685 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%F A049685 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 (benoit7848c(AT)orange.fr), 
               Nov 10 2002
%F A049685 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). 
               - R. Stephan, May 29 2004
%F A049685 a(n)=sum{k=0..n, binomial(n+k, 2k)5^k} - Paul Barry (pbarry(AT)wit.ie), 
               Aug 30 2004
%F A049685 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 (g_m(AT)mcraefamily.com), 
               Jan 30 2005
%F A049685 a(n)=(-1)^n*U(2n, I*sqrt(5)/2), U(n, x) Chebyshev polynomial of second 
               kind, I=sqrt(-1); - Paul Barry (pbarry(AT)wit.ie), Mar 13 2005
%F A049685 [a(n), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), 
               Mar 21 2008
%o A049685 (Other) sage: [lucas_number1(n,7,1)-lucas_number1(n-1,7,1) for n in xrange(1, 
               20)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 10 
               2009]
%Y A049685 Row 7 of array A094954.
%Y A049685 Cf. A004187.
%Y A049685 Sequence in context: A043069 A135232 A015551 this_sequence A122371 A083067 
               A000402
%Y A049685 Adjacent sequences: A049682 A049683 A049684 this_sequence A049686 A049687 
               A049688
%K A049685 nonn,new
%O A049685 0,2
%A A049685 Clark Kimberling (ck6(AT)evansville.edu)

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