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A011373
Number of 1's in binary expansion of Fibonacci(n).
10
0, 1, 1, 1, 2, 2, 1, 3, 3, 2, 5, 4, 2, 5, 6, 4, 8, 7, 4, 5, 8, 6, 8, 11, 6, 6, 9, 11, 11, 12, 8, 11, 9, 13, 12, 11, 12, 14, 10, 12, 16, 17, 14, 16, 18, 15, 21, 13, 12, 18, 18, 17, 17, 17, 16, 22, 21, 16, 24, 20, 16, 19, 26, 23, 20, 25, 19, 26, 15, 23, 23, 22, 25, 27, 24, 23, 23, 22
OFFSET
0,5
FORMULA
a(n) = A000120(A000045(n)). - Michel Marcus, Dec 27 2014
a(n) = [x^Fibonacci(n)] (1/(1 - x))*Sum_{k>=0} x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Mar 27 2018
Conjecture: Limit_{n->oo} a(n)/n = log_2(phi)/2 = A242208/2 = 0.3471209568... . - Amiram Eldar, May 13 2022
EXAMPLE
a(8) = 3 because Fibonacci(8) = 21, which in binary is 11001 and that has 3 on bits.
a(9) = 2 because Fibonacci(9) = 34, which in binary is 100010 and that only has 2 on bits.
MAPLE
A000120 := proc(n) add(d, d=convert(n, base, 2)) ; end proc:
A011373 := proc(n) A000120(combinat[fibonacci](n)) ; end proc:
seq(A011373(n), n=0..50) ; # R. J. Mathar, Mar 22 2011
MATHEMATICA
DigitCount[#, 2, 1]&/@Fibonacci[Range[0, 79]] (* Harvey P. Dale, Mar 14 2011 *)
Table[Plus@@IntegerDigits[Fibonacci[n], 2], {n, 0, 79}]
PROG
(PARI) a(n)=hammingweight(fibonacci(n)) \\ Charles R Greathouse IV, Mar 02 2014
(Scala) def fibonacci(n: BigInt): BigInt = {
val zero = BigInt(0)
def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
case `zero` => a
case _ => fibTail(n - 1, b, a + b)
}
fibTail(n, 0, 1)
} // Based on tail recursion by Dario Carrasquel
(0 to 79).map(fibonacci(_).bitCount) // Alonso del Arte, Apr 13 2019
CROSSREFS
KEYWORD
nonn,base
STATUS
approved