OFFSET
0,3
COMMENTS
Equals row sums of triangle A173284. - Gary W. Adamson, Feb 14 2010
The Kn21 sums (see A180662 for definition) of the 'Races with Ties' triangle A035317 produce this sequence. - Johannes W. Meijer, Jul 20 2011
a(n-1), for n >= 1, gives the number of compositions of n with relative prime parts, and parts not exceeding 2. See the row sums of triangle A030528 where for even n the leading 1 is missing. - Wolfdieter Lang, Jul 27 2023
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1023
K. Kuhapatanakul, On the Sums of Reciprocal Generalized Fibonacci Numbers, J. Int. Seq. 16 (2013) #13.7.1, eq (1).
Steven Linton, James Propp, Tom Roby, and Julian West, Equivalence Classes of Permutations under Various Relations Generated by Constrained Transpositions, Journal of Integer Sequences, Vol. 15 (2012), #12.9.1.
H. Ohtsuka and S. Nakamura, On the sum of reciprocal sums of Fibonacci numbers, Fibonacci Quart. 46/47 (2008/2009), 153-159.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).
FORMULA
G.f.: 1/((1-x-x^2)*(1-x^2)).
a(n) = A074331(n+1).
a(n) = A054450(n+1, 1) (second column of triangle).
a(n) = 2*a(n-2) + a(n-3) + 1, with a(0)=1, a(1)=1, a(2)=3.
a(n) = Sum_{alpha=RootOf(-1+z+z^2)} (3+alpha)*alpha^(-1-n)/3 - Sum_{beta=RootOf(-1+z^2)} beta^(-1-n)/2.
a(2*k) = Sum_{j=0..k} F(2*j+1) = F(2*(k+1)) for k >= 0; a(2*k-1) = Sum_{j=0..k} F(2*j) = F(2*k+1)-1 for k >= 1 (F = A000045, Fibonacci numbers).
a(n) = a(n-1) + a(n-2) + (1+(-1)^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+1, k). - Paul Barry, Oct 23 2004
a(n) = floor(phi^(n+2) / sqrt(5)), where phi is the golden ratio: phi = (1+sqrt(5))/2. - Reinhard Zumkeller, Apr 19 2005
a(n) = Fibonacci(n+1) + a(n-2) with n>1, a(0)=a(1)=1. - Zerinvary Lajos, Mar 17 2008
a(n) = floor(Fibonacci(n+3)^2/Fibonacci(n+4)). - Gary Detlefs, Nov 29 2010
a(4*n) = F(4*n+2); a(4*n+1) = F(4*n+3) - 1; a(4*n+2) = F(4*n+4); a(4*n+3) = F(4*n+5) - 1. - Johannes W. Meijer, Jul 20 2011
a(n+1) = a(n) + a(n-1) + A059841(n+1). - Reinhard Zumkeller, Jan 06 2012
a(n) = floor(|F((1+i)*(n+2))|), n >= 0, with the complex Fibonacci function F: C -> C, z -> F(z) with F(z) := (exp(log(phi)*z) - exp(i*Pi*z)*exp(-log(phi)*z))/(2*phi-1) with the modulus |z|, the imaginary unit i and the golden section phi:=(1+sqrt(5))/2. A Conjecture: For F(z) see, e.g., the T. Koshy reference. ch. 45, p. 523, where F is called f, given in A000045. - Wolfdieter Lang, Jul 24 2012
5*a(n) = (L(n+3)-1)*(L(n+4)+3) -14 -Sum_{k=0..n} L(k+1)*L(k+5) = (L(n+3)-1)*(L(n+4)+3) -L(2*n+7) +A168309(n), where L=A000032. - J. M. Bergot, Jun 13 2014
a(n) = floor(phi*Fibonacci(n+1)), where phi is the golden section. - Michel Dekking, Dec 02 2016
a(n) = -(-1)^n * a(-4-n) for all n in Z. - Michael Somos, Dec 03 2016
a(n) = Sum_{k=0..n} Sum_{i=0..n} C(n-k-1,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = floor(1/(Sum_{k>=n+4} 1/Fibonacci(k))) [Ohtsuka and Nakamura]. - Michel Marcus, Aug 09 2018
a(n) = floor(abs(chebyshevU(n/2, 3/2))). - Federico Provvedi, Feb 23 2022
E.g.f.: exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - sinh(x). - Stefano Spezia, Mar 09 2024
EXAMPLE
G.f. = 1 + x + 3*x^2 + 4*x^3 + 8*x^4 + 12*x^5 + 21*x^6 + 33*x^7 + ...
MAPLE
A052952 :=proc(n)
option remember;
local t1;
if n <= 1 then
return 1 ;
fi:
if n mod 2 = 1 then
t1:=0
else
t1:=1;
fi:
procname(n-1)+procname(n-2)+t1;
end proc;
seq(A052952(n), n=0..40) ; # N. J. A. Sloane, May 25 2008
MATHEMATICA
Table[Fibonacci[n+2] -(1-(-1)^n)/2, {n, 0, 40}] (* Vincenzo Librandi, Dec 02 2016 *)
Sum[(-1)^k*Fibonacci[Range[2, 41], 1-k], {k, 0, 1}] (* G. C. Greubel, Oct 21 2019 *)
CoefficientList[Series[1/((1-x-x^2)*(1-x^2)), {x, 0, 40}], x] (* Harvey P. Dale, Sep 12 2020 *)
PROG
(PARI) {a(n) = fibonacci(n+2) - n%2};
(Haskell)
a052952 n = a052952_list !! n
a052952_list = 1 : 1 : zipWith (+)
a059841_list (zipWith (+) a052952_list $ tail a052952_list)
-- Reinhard Zumkeller, Jan 06 2012
(Magma) [Fibonacci(n+2)-(1-(-1)^n)/2: n in [0..40]]; // Vincenzo Librandi, Dec 02 2016
(Sage) [fibonacci(n+2) -(1-(-1)^n)/2 for n in (0..40)] # G. C. Greubel, Jul 10 2019
(GAP) List([0..40], n-> Fibonacci(n+2) -(1-(-1)^n)/2); # G. C. Greubel, Jul 10 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
encyclopedia(AT)pommard.inria.fr, Jan 25 2000
EXTENSIONS
Additional formulas and more terms from Wolfdieter Lang, May 02 2000
Better description from Olivier Gérard, Jun 05 2001
STATUS
approved