A011373 -id:A011373 - cp's OEIS frontend

A350700 a(n) is the number of 1's minus the number of 0's in A004685(n).

-1, 1, 1, 0, 2, 1, -2, 2, 1, -2, 4, 1, -4, 2, 3, -2, 6, 3, -4, -3, 3, -2, 1, 7, -4, -5, 1, 4, 3, 5, -4, 1, -4, 4, 1, -2, 0, 3, -6, -2, 5, 6, 0, 3, 6, -1, 11, -6, -9, 3, 2, -1, -1, -2, -5, 6, 4, -7, 8, 0, -9, -4, 10, 3, -4, 6, -7, 6, -17, -1, -2, -5, 1, 4, -3

Offset: 0

Karl-Heinz Hofmann, Jan 18 2022

			A004685(0) = 0; this term has 0 ones and 1 zero. So a(0) = 0 - 1 = -1.
A004685(7) = 1101; this term has 3 ones and 1 zero. So a(7) = 3 - 1 = 2.

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000

Cf. A000045, A004685, A011373, A145037, A177718, A214852.

a[n_] := Subtract @@ DigitCount[Fibonacci[n], 2, {1, 0}]; Array[a, 75, 0] (* Amiram Eldar, Jan 22 2022 *)

from sympy import fibonacci
print([(bin(fibonacci(n))[2:].count("1") - bin(fibonacci(n))[2:].count("0")) for n in range (0,100)])

a(n) = A145037(A000045(n)) for n >= 1.

a(n) = 0 if and only if n is in A214852. - Amiram Eldar, Jan 22 2022

A353988 Numbers k such that Fibonacci(k) is a binary Niven number (A049445).

Original entry on oeis.org

1, 2, 3, 6, 8, 9, 10, 12, 18, 24, 30, 36, 48, 56, 60, 100, 120, 144, 150, 168, 240, 270, 288, 300, 324, 330, 336, 360, 444, 540, 594, 600, 624, 720, 750, 840, 864, 896, 900, 936, 1080, 1152, 1200, 1210, 1360, 1404, 1632, 1720, 1921, 2028, 2400, 2520, 2552, 2864

Offset: 1

Amiram Eldar, May 13 2022

Numbers k such that A011373(k) | A000045(k).

			1 is a term since A000045(1) = A011373(1) = 1 and 1 | 1.
10 is a term since A000045(10) = 55, A011373(1) = 5 and 5 | 55.

Amiram Eldar, Table of n, a(n) for n = 1..3000

Cf. A000045, A000120, A011373, A049445, A117774, A337448 (decimal analog).

Select[Range[3000], Divisible[(f = Fibonacci[#]), DigitCount[f, 2, 1]] &]

isok(k) = my(f=fibonacci(k)); ! (f % hammingweight(f)); \\ Michel Marcus, May 13 2022

A374962 Numbers k such that the number of terms in the Zeckendorf representation of 2^k equals the binary weight of Fibonacci(k).

Original entry on oeis.org

1, 3, 4, 7, 8, 13, 14, 20, 26, 50, 55, 58, 90, 140, 270, 314, 603

Offset: 1

Amiram Eldar, Jul 25 2024

Numbers k such that A007895(A000079(k)) = A000120(A000045(k)), or equivalently A020908(k) = A011373(k).
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.
a(18) > 333333, if it exists. - Lucas A. Brown, Aug 13 2024

			  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

Lucas A. Brown, Python program.

Cf. A000045, A000079, A000120, A007088, A007895, A011373, A014417, A020908.

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] &]

A007895(n)=if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
is(k) = A007895(2^k) == hammingweight(fibonacci(k));

A379835 Number of 1's in binary expansion of Lucas(n).

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 2, 4, 5, 3, 6, 5, 3, 3, 6, 5, 7, 9, 6, 5, 9, 10, 9, 11, 6, 9, 7, 10, 15, 11, 10, 15, 13, 11, 11, 15, 15, 12, 15, 17, 15, 11, 14, 15, 20, 15, 18, 17, 13, 11, 22, 20, 23, 23, 19, 22, 22, 22, 28, 25, 23, 19, 25, 27, 27, 24, 26, 25, 23, 27, 23, 27

Offset: 0

Vincenzo Librandi, Jan 05 2025

			a(10) = 6 because Lucas(10) = 123 is 1111011_2, which has 6 one bits.

Cf. A000032, A000120, A011373.

Magma
```
[&+Intseq(Lucas(n), 2): n in [0..100]];
```

Table[DigitCount[LucasL[n],2][[1]],{n,0,200}]

a(n) = A000120(A000032(n)).

cp's OEIS Frontend

A350700 a(n) is the number of 1's minus the number of 0's in A004685(n).

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A353988 Numbers k such that Fibonacci(k) is a binary Niven number (A049445).

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A374962 Numbers k such that the number of terms in the Zeckendorf representation of 2^k equals the binary weight of Fibonacci(k).

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A379835 Number of 1's in binary expansion of Lucas(n).

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Mathematica

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