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

Author

Alex Ratushnyak, May 07 2012

Keywords

• nonn
• tabl

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

Examples

			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.
		

Links

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

Crossrefs

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

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

Programs

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 *)

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)