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

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

Offset: 1

Author

Amiram Eldar, Jul 25 2024

Keywords

• nonn
• base
• more

Comments

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

Examples

			  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
		

Links

• Lucas A. Brown, Python program.

Crossrefs

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

Programs

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] &]

PARI

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));