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A188378
Partial sums of A005248.
4
2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240, 54018522, 141422325, 370248452, 969323030, 2537720637, 6643838880, 17393796002, 45537549125, 119218851372, 312119004990, 817138163597, 2139295485800
OFFSET
0,1
COMMENTS
Different from A024851.
Luo proves that these integers cannot be uniquely decomposed as the sum of distinct and nonconsecutive terms of the Lucas number sequence. - Michel Marcus, Apr 20 2020
LINKS
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
David C. Luo, Nonuniqueness Properties of Zeckendorf Related Decompositions, arXiv:2004.08316 [math.NT], 2020.
FORMULA
a(n) = A000032(2n+1)+1 = A002878(n)+1 = 2*A027941(n+1)-3*A027941(n).
G.f.: ( -2+3*x ) / ( (x-1)*(x^2-3*x+1) ). - R. J. Mathar, Mar 30 2011
a(n) = 5*A001654(n) + 1 + (-1)^n, n>=0. [Wolfdieter Lang, Jul 23 2012]
(a(n)^3 + (a(n)-2)^3) / 2 = A000032(A016945(n)) = Lucas(6*n+3) = A267797(n), for n>0. - Altug Alkan, Jan 31 2016
a(n) = 2^(-1-n)*(2^(1+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n). - Colin Barker, Nov 02 2016
MAPLE
f:= gfun:-rectoproc({a(n+3)-4*a(n+2)+4*a(n+1)-a(n), a(0) = 2, a(1) = 5, a(2) = 12}, a(n), remember):
map(f, [$0..60]); # Robert Israel, Feb 02 2016
MATHEMATICA
LinearRecurrence[{4, -4, 1}, {2, 5, 12}, 30] (* Harvey P. Dale, Oct 05 2015 *)
Accumulate@ LucasL@ Range[0, 58, 2] (* Michael De Vlieger, Jan 24 2016 *)
PROG
(PARI) a(n) = 5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n; \\ Michel Marcus, Aug 26 2013
(PARI) Vec((-2+3*x)/((x-1)*(x^2-3*x+1)) + O(x^100)) \\ Altug Alkan, Jan 24 2016
(Magma) [5*Fibonacci(n)*Fibonacci(n+1)+1+(-1)^n: n in [0..40]]; // Vincenzo Librandi, Jan 24 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Gabriele Fici, Mar 29 2011
STATUS
approved