Search: id:A001254 Results 1-1 of 1 results found. %I A001254 %S A001254 4,1,9,16,49,121,324,841,2209,5776,15129,39601,103684,271441,710649, %T A001254 1860496,4870849,12752041,33385284,87403801,228826129,599074576, %U A001254 1568397609,4106118241,10749957124,28143753121,73681302249 %N A001254 Squares of Lucas numbers. %D A001254 Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001. %D A001254 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 36,60. %H A001254 T. D. Noe, Table of n, a(n) for n=0..200 %H A001254 Index entries for sequences related to linear recurrences with constant coefficients %H A001254 Tanya Khovanova, Recursive Sequences %H A001254 T. Mansour, A note on sum of k-th power of Horadam's sequence %H A001254 P. Stanica, Generating functions, weighted and non-weighted sums of powers... %F A001254 G.f.: g(x)=(4-7x-x^2)/(1-2x-2x^2+x^3). - Len Smiley (smiley(AT)math.uaa.alaska.edu), Nov 30 2001 %F A001254 a(n)=r^n+(1/r)^n+2*(-1)^n, with r=(3+sqrt(5))/2. a(n+3)=2*a(n+2)+2*a(n+1)-a(n). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Feb 08 2003 %F A001254 a(n) = L(2n) + 2(-1)^n = L(n-1)*L(n+1) + 5(-1)^n. %Y A001254 Cf. A000032, A000204. %Y A001254 Cf. A007598, A079291. %Y A001254 With alternating signs, cf. A075150. %Y A001254 Bisection of A001638 and A006499. First differences of A005970. %Y A001254 Second row of array A103324. %Y A001254 Sequence in context: A158199 A091885 A069606 this_sequence A075150 A143763 A128626 %Y A001254 Adjacent sequences: A001251 A001252 A001253 this_sequence A001255 A001256 A001257 %K A001254 nonn,easy %O A001254 0,1 %A A001254 N. J. A. Sloane (njas(AT)research.att.com). Search completed in 0.001 seconds