|
|
A374962
|
|
Numbers k such that the number of terms in the Zeckendorf representation of 2^k equals the binary weight of Fibonacci(k).
|
|
1
|
|
|
1, 3, 4, 7, 8, 13, 14, 20, 26, 50, 55, 58, 90, 140, 270, 314, 603
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
The corresponding values of A020908(k) = A011373(k) are 1, 1, 2, 3, 3, 5, 6, 8, 9, 18, 22, 24, 33, 53, 106, 122, 232, ... .
a(18) > 63000, if it exists.
|
|
LINKS
|
|
|
EXAMPLE
|
n | k = a(n) | 2^k | A014417(2^k) | F(k) | A007088(F(k)) | Number of 1's
--+----------+-----+--------------+------+---------------+--------------
1 | 1 | 2 | 10 | 1 | 1 | 1
2 | 3 | 8 | 10000 | 2 | 10 | 1
3 | 4 | 16 | 100100 | 3 | 11 | 2
4 | 7 | 128 | 1010001000 | 13 | 1101 | 3
5 | 8 | 256 | 100001000010 | 21 | 10101 | 3
|
|
MATHEMATICA
|
z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
Select[Range[700], z[2^#] == DigitCount[Fibonacci[#], 2, 1] &]
|
|
PROG
|
is(k) = A007895(2^k) == hammingweight(fibonacci(k));
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,more
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|