A264800 Nearly-Fibonacci sequence.

1, 1, 2, 4, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155

Offset

1,3

Comments

Generate a tree T by these rules: 0 is in T, and if x is in T, then x+1 and -2x are in T, with duplicates deleted as they occur; see A264799. Let g(0) = {0}, g(1) = {1}, g(2) = {-2,2}, g(3) = {-4,-1,3,4}, etc. The number |g(n)| of numbers in the n-th generation of T is a Fibonacci number except for g(3).

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,1).

Formula

a(n) = F(n) for 1 <= n <= 3 and n >= 5, and a(4) = 4; where F = A000045, the Fibonacci numbers.

From Colin Barker, Nov 25 2015: (Start)

a(n) = a(n-1) - a(n-2) for n>6.

G.f.: x*(x-1)*(x+1)*(x^3+x^2+1) / (x^2+x-1).

(End)

Mathematica

z = 10; t = Expand[NestList[DeleteDuplicates[Flatten[Map[{# + 1, -2*#} &, #], 1]] &, {0}, z]]; s[0] = t[[1]]; s[n_] := s[n] = Union[t[[n]], s[n - 1]];
g[n_] := Complement[s[n], s[n - 1]]; g[1] = {0};
Table[Length[g[k]], {k, 1, z}]  (* A264800 *)
u = Table[g[k], {k, 1, z}]
Flatten[u] (* A264799 *)
LinearRecurrence[{1, 1}, {1, 1, 2, 4, 5, 8}, 40] (* Harvey P. Dale, Dec 09 2021 *)

Prog

(PARI) Vec(x*(x-1)*(x+1)*(x^3+x^2+1)/(x^2+x-1) + O(x^100)) \\ Colin Barker, Nov 25 2015

Crossrefs

Cf. A264799, A000045.

Sequence in context: A238589 A327045 A288668 * A293189 A278695 A105134

Adjacent sequences: A264797 A264798 A264799 * A264801 A264802 A264803

Keyword

nonn,easy

Author

Clark Kimberling, Nov 25 2015

Status

approved