Search: id:A081016 Results 1-1 of 1 results found. %I A081016 %S A081016 1,6,40,273,1870,12816,87841,602070,4126648,28284465,193864606, %T A081016 1328767776,9107509825,62423800998,427859097160,2932589879121, %U A081016 20100270056686,137769300517680,944284833567073,6472224534451830 %N A081016 (Lucas(4n+3)+1)/5, or Fibonacci(2n+1)*Fibonacci(2n+2), or A081015/5. %C A081016 a(n-1) is, together with b(n) := A089508(n), n>=1, the solution to a binomial problem - see A089508. %C A081016 Numbers k such that 1 - 2 x + 5 x^2 is a square [From Artur Jasinski (grafix(AT)csl.pl), Oct 26 2008] %D A081016 Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75 %D A081016 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26. %F A081016 a(n) = 8a(n-1)-8a(n-2)+a(n-3) %F A081016 G.f.: (1-2*x)/((1-x)*(1-7*x+x^2)). %F A081016 F(1) + F(5) + F(9) +...+ F(4n+1) = F(2n)*F(2n+3) + 1. %F A081016 a(n)=(1/5)+(2/5)*{[(7/2)-(3/2)*sqrt(5)]^n+[(7/2)+(3/2)*sqrt(5)]^n+(1/ 5)*sqrt(5)*{[(7/2)+(3/2) *sqrt(5)]^n-[(7/2)-(3/2)*sqrt(5)]^n}, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Dec 01 2008] %p A081016 luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d, `,(luc(4*n+3)+1)/5) od: %Y A081016 Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015. %Y A081016 Partial sums of A033889. Bisection of A001654. Equals A003482 + 1. %Y A081016 A145995 [From Artur Jasinski (grafix(AT)csl.pl), Oct 26 2008] %Y A081016 Sequence in context: A135032 A122074 A123357 this_sequence A083426 A122471 A090041 %Y A081016 Adjacent sequences: A081013 A081014 A081015 this_sequence A081017 A081018 A081019 %K A081016 nonn,easy %O A081016 0,2 %A A081016 R. K. Guy (rkg(AT)cpsc.ucalgary.ca), Mar 01, 2003 %E A081016 More terms and Maple code from James A. Sellers (sellersj(AT)math.psu.edu), Mar 03, 2003 Search completed in 0.001 seconds