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A210619
Triangle of numbers with n 1's and n 0's in their representation in base of Fibonacci numbers (A014417).
5
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
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
KEYWORD
nonn,tabl
AUTHOR
Alex Ratushnyak, May 07 2012
STATUS
approved