|
| |
|
|
A345347
|
|
Find the largest k with F(k) <= n, where F(k) is the k-th Fibonacci number. a(n) = F(k+2) + n.
|
|
1
|
|
|
|
1, 4, 7, 11, 12, 18, 19, 20, 29, 30, 31, 32, 33, 47, 48, 49, 50, 51, 52, 53, 54, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 199, 200, 201, 202, 203, 204
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
The terms consist of 1 together with numbers that appear in row m of the Wythoff array (A035513) if m is in the sequence.
a(0) = 1, otherwise a(n) is the number whose Zeckendorf representation is "10" followed by the Zeckendorf representation of n.
If we define an extended Zeckendorf representation to be the Zeckendorf representation with "01" appended, then the numbers in the sequence are exactly those whose extended representation starts 101... . This extended representation is a valid Fibonacci base representation if we specify the rightmost digit to have weight F(0) = 0.
Equivalently, for positive integer m, find the largest k with F(k) <= m, where F(k) is the k-th Fibonacci number. m is in the sequence if and only if m >= F(k) + F(k-2).
Numbers given to rabbits on Rabbit 1's branch of the generation tree described in the A035513 examples.
Equivalently, take the positive integers in turn, placing runs of them alternatively into 2 sets, with run lengths from A053602/A051792 (self-interleaved Fibonacci sequence) as follows:
set A: 1 0 1 1 2 3 5 ...
set B: 1 1 2 3 5 8 ...
The sequence lists the numbers in set A.
|
|
|
LINKS
|
|
|
|
FORMULA
|
Union_{k >= 2} {m : F(k)+F(k-2) <= m < F(k+1)}, where F(k) = A000045(k).
|
|
|
EXAMPLE
|
The initial Fibonacci numbers are F(0)..F(5) = 0, 1, 1, 2, 3, 5.
For n = 0, the largest k with F(k) <= 0 is k = 0, so F(k+2) = F(2) = 1, so a(0) = 1 + 0 = 1.
For n = 1, the largest k with F(k) <= 1 is k = 2, so F(k+2) = F(4) = 3, so a(1) = 3 + 1 = 4.
For n = 4, the largest k with F(k) <= 4 is k = 4, so F(k+2) = F(6) = 8, so a(4) = 8 + 4 = 12.
In the paragraph that follows we use the Wythoff array-based definition from the start of the comments.
Every positive integer appears once (only) in the Wythoff array. 0 is not positive, so does not appear in the array, so is not in the sequence. 1 is in the sequence by definition. 2 appears in Wythoff row 0, and 0 is not in the sequence, so 2 is not in the sequence. 4 appears in Wythoff row 1, and 1 is in the sequence, so 4 is in the sequence.
|
|
|
MATHEMATICA
|
kmax=12; Flatten[Table[Range[Fibonacci[k]+Fibonacci[k-2], Fibonacci[k+1]-1], {k, 2, kmax}]] (* Paolo Xausa, Jan 02 2022 *)
A108852[n_]:=1+Floor[Log[GoldenRatio, 1+n*Sqrt[5]]];
nterms=100; Table[n+Fibonacci[1+A108852[n]], {n, 0, nterms-1}](* Paolo Xausa, Jan 02 2022 *)
|
|
|
PROG
|
(PARI) a(n) = my(k=0); while(fibonacci(k)<=n, k=k+1); n+fibonacci(k+1)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,base,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
