A210619 Triangle of numbers with n 1's and n 0's in their representation in base of Fibonacci numbers (A014417).

2, 6, 7, 17, 19, 20, 46, 51, 53, 54, 122, 135, 140, 142, 143, 321, 355, 368, 373, 375, 376, 842, 931, 965, 978, 983, 985, 986, 2206, 2439, 2528, 2562, 2575, 2580, 2582, 2583, 5777, 6387, 6620, 6709, 6743, 6756, 6761, 6763, 6764, 15126, 16723, 17333, 17566, 17655, 17689, 17702, 17707, 17709, 17710

Offset

1,1

Comments

There are n such 2n-bit numbers. For example, 17, 19, and 20 all require six bits: 100101, 101001, 101010. The least number in each group is Fib(2n+1) + Fib(2n-1) - 1, which is A005592(n). The greatest number in each group is Fib(2n+2) - 1, which is A035508(n). - T. D. Noe, May 08 2012

Links

T. D. Noe, Table of rows 1 to 49, flattened

Formula

Numbers with equal counts of 1's and 0's in their Zeckendorf representation.

From Kevin Ryde, Jul 24 2021: (Start)

T(n,k) = Fibonacci(2*n+2) - Fibonacci(2*(n-k)) - 1.

G.f.: x*y*(2 - 2*x + x^2 - (1 + x + x^2)*x*y + x^3*y^2) / ( (1-x) * (1 - 3*x + x^2) * (1 - x*y) * (1 - 3*x*y + (x*y)^2) ).

(End)

Example

Representation of 20 is 101010, three 1's and three 0's, so 20 is in the sequence.
Representation of 22 is 1000001, two 1's and five 0's, so 22 is not in the sequence.

Mathematica

nn = 10; f = Join[{0}, Accumulate[Fibonacci[Range[2, 2*nn, 2] - 1]]]; t = Table[hi = f[[n+1]] - 1; Reverse[Table[hi - f[[i]], {i, n - 1}]], {n, 2, nn}]; t = Flatten[t] (* T. D. Noe, May 08 2012 *)

Crossrefs

Cf. A014417, A003714, A000045 (Fibonacci numbers).

Cf. A005592 (column k=1), A035508 (main diagonal), A249450 (second diagonal), A346434 (in Fibonacci base).

Sequence in context: A049399 A060133 A107784 * A358578 A095036 A290379

Adjacent sequences: A210616 A210617 A210618 * A210620 A210621 A210622

Keyword

nonn,tabl

Author

Alex Ratushnyak, May 07 2012

Status

approved