0,2

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

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

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.

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

Index entries for sequences related to linear recurrences with constant coefficients

Tanya Khovanova, Recursive Sequences

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

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

a(n)=sum{k=0..n, binomial(n+k, 2k)5^k} - Paul Barry (pbarry(AT)wit.ie), 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 (g_m(AT)mcraefamily.com), 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 (pbarry(AT)wit.ie), Mar 13 2005

[a(n), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 21 2008

(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]

Row 7 of array A094954.

Cf. A004187.

Sequence in context: A043069 A135232 A015551 this_sequence A122371 A083067 A000402

Adjacent sequences: A049682 A049683 A049684 this_sequence A049686 A049687 A049688

nonn,new

Clark Kimberling (ck6(AT)evansville.edu)