A210619 - OEIS

%I #29 Jul 25 2021 11:18:23

%S 2,6,7,17,19,20,46,51,53,54,122,135,140,142,143,321,355,368,373,375,

%T 376,842,931,965,978,983,985,986,2206,2439,2528,2562,2575,2580,2582,

%U 2583,5777,6387,6620,6709,6743,6756,6761,6763,6764,15126,16723,17333,17566,17655,17689,17702,17707,17709,17710

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

%C 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

%H T. D. Noe, <a href="/A210619/b210619.txt">Table of rows 1 to 49, flattened</a>

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

%F From _Kevin Ryde_, Jul 24 2021: (Start)

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

%F 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) ).

%F (End)

%e Representation of 20 is 101010, three 1's and three 0's, so 20 is in the sequence.

%e Representation of 22 is 1000001, two 1's and five 0's, so 22 is not in the sequence.

%t 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 *)

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

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

%K nonn,tabl

%O 1,1

%A _Alex Ratushnyak_, May 07 2012