A097083 Positive values of k such that there is exactly one permutation p of (1,2,3,...,k) such that i+p(i) is a Fibonacci number for 1<=i<=k.

1, 2, 3, 5, 9, 15, 24, 39, 64, 104, 168, 272, 441, 714, 1155, 1869, 3025, 4895, 7920, 12815, 20736, 33552, 54288, 87840, 142129, 229970, 372099, 602069, 974169, 1576239, 2550408, 4126647, 6677056, 10803704, 17480760, 28284464, 45765225

Offset: 1

Author

John W. Layman, Jul 23 2004

Keywords

• nonn

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

If the formula is correct, the bisections give A059840 and A064831. - 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).

Crossrefs

Cf. A073364, A000045.

Programs

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 *)

PARI

x='x+O('x^50); Vec(x/((x - 1)*(x^2 + 1)*(x^2 + x - 1))) \\ G. C. Greubel, Mar 05 2017

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) ).

a(n) = A000032(n+3)/5 -(-1)^n*A112030(n)/10 - 1/2. (End)

Conjecture: a(n) = floor(F(n+3)/sqrt(5)), where F(n) = A000045(n) are Fibonacci numbers. - Vladimir Reshetnikov, Nov 05 2015

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