OFFSET
1,2
COMMENTS
Numbers k such that A097082(k) = 1. If f is a Fibonacci number and k < f <= 2k, then a permutation for f-k-1 may be extended to a permutation for k, with p(i) = f-i for f-k < i <= k. This explains the sparseness of this sequence. - David Wasserman, Dec 19 2007
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,0,-1).
FORMULA
It appears that {a(n)} satisfies a(1)=1, a(2)=2 and, for n>2, a(n) = F(n+2) - a(n-2) - 1, where {F(k)} is the sequence of Fibonacci numbers, i.e, that the sequence is the partial sums of A006498.
If the partial sum assumption is correct: a(n) = floor(phi^(n+3)/5), where phi=(1+sqrt(5))/2 = A001622, and a(n) = a(n-1) + a(n-2) + ( (n*(n+1)/2) mod 2). - Gary Detlefs, Mar 12 2011
From R. J. Mathar, Mar 13 2011: (Start)
If the partial sum assumption is correct: a(n)= +2*a(n-1) -a(n-2) +a(n-3) -a(n-5).
G.f.: x/( (x-1)*(x^2+1)*(x^2+x-1) ).
Conjecture: a(n) = floor(F(n+3)/sqrt(5)), where F(n) = A000045(n) are Fibonacci numbers. - Vladimir Reshetnikov, Nov 05 2015
MATHEMATICA
a=b=c=d=0; Table[e=a+b+d+1; a=b; b=c; c=d; d=e, {n, 100}] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2011 *)
CoefficientList[Series[x/((x - 1)*(x^2 + 1)*(x^2 + x - 1)), {x, 0, 50}], x] (* G. C. Greubel, Mar 05 2017 *)
LinearRecurrence[{2, -1, 1, 0, -1}, {1, 2, 3, 5, 9}, 50] (* Harvey P. Dale, Nov 09 2024 *)
PROG
(PARI) x='x+O('x^50); Vec(x/((x - 1)*(x^2 + 1)*(x^2 + x - 1))) \\ G. C. Greubel, Mar 05 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
John W. Layman, Jul 23 2004
EXTENSIONS
a(9) from Ray Chandler, Jul 29 2004
More terms from David Wasserman, Dec 19 2007
Terms > 90000 assuming the partial sums formula by Vladimir Joseph Stephan Orlovsky, Feb 26 2011
STATUS
approved