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A173714 Floor(Lucas(n+1)/2), Lucas(n) = A000032(n). 1
0, 1, 2, 3, 5, 9, 14, 23, 38, 61, 99, 161, 260, 421, 682, 1103, 1785, 2889, 4674, 7563, 12238, 19801, 32039, 51841, 83880, 135721, 219602, 355323, 574925, 930249, 1505174 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
Sequences of the form a(0)=1, a(1)=b,
a(n) = a(n-1) + a(n-2) + 1 if n mod 3 =2, else
a(n) = a(n-1) + a(n-2) have a closed form of
a(n) = F(n-1)*a + F(n)*b + floor(F(n+1)/2),
where F(n)= Fibonacci(n) = A000045(n), floor(F(n+1)/2) = A004695(n+1).
We can generalize the definition of this sequence by changing the added 1 to any value of k and changing the last term of the formula to floor(F(n+1)/2)*k.
Two variants: if we add the constant at n mod 3 = 0, then a(n)=F(n-1)*a + F(n)*b + floor(F(n)/2), and if for n mod 3 =1, then a(n)=F(n-1)*a + F(n)*b + floor(F(n-1)/2).
LINKS
FORMULA
a(0)= 0, a(1)=1, a(n)=a(n-1)+a(n-2)+1 if n mod 3 =2, else a(n)=a(n-1)+a(n-2).
G.f.: x*(1+x-x^3)/[(1-x-x^2)*(1-x^3)].
a(n) = a(n-1) +a(n-2) +(1+(-1)^Fib(n+1))/2.
a(n) = A000204(n+1)/2 + A099837(n+1)/6 - 1/3. - R. J. Mathar, Nov 26 2010
a(n) = Fibonacci(n) + floor(Fibonacci(n+1)/2). - Gary Detlefs, Dec 10 2010
EXAMPLE
a(5) = a(4) + a(3) + 1 = 5 +3 +1 =9 because 5 mod 3 = 2.
a(6) = a(5) + a(4) = 9 +5 =14 because 6 mod 3 <>2.
MAPLE
with(combinat):
g:=(a, b, n)->fibonacci(n-1)*a+fibonacci(n)*b + floor(fibonacci(n+1)/2):
seq(g(0, 1, n), n=0..30)
MATHEMATICA
Table[Floor[LucasL[n + 1]/2], {n, 0, 50}] (* G. C. Greubel, Nov 24 2016 *)
PROG
(Magma) [Floor(Lucas(n+1)/2): n in [0..50]]; // Vincenzo Librandi, Apr 24 2011
CROSSREFS
Sequence in context: A293547 A124502 A251572 * A026746 A004699 A245800
KEYWORD
nonn,easy
AUTHOR
Gary Detlefs, Nov 25 2010
STATUS
approved

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Last modified April 24 12:40 EDT 2024. Contains 371937 sequences. (Running on oeis4.)