A117479 -id:A117479 - cp's OEIS frontend

A112310 Number of terms in lazy Fibonacci representation of n.

0, 1, 1, 2, 2, 2, 3, 2, 3, 3, 3, 4, 3, 3, 4, 3, 4, 4, 4, 5, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5, 5, 6, 4, 4, 5, 4, 5, 5, 5, 6, 4, 5, 5, 5, 6, 5, 5, 6, 5, 6, 6, 6, 7, 4, 5, 5, 5, 6, 5, 5, 6, 5, 6, 6, 6, 7, 5, 5, 6, 5, 6, 6, 6, 7, 5, 6, 6, 6, 7, 6, 6, 7, 6, 7, 7, 7, 8, 5, 5, 6, 5, 6, 6, 6, 7, 5, 6, 6, 6, 7, 6, 6, 7, 6

Offset: 0

N. J. A. Sloane, Dec 01 2005

Equivalently, the number of ones in the maximal Fibonacci bit-representation (A104326) of n.
Conjecture: if we split the sequence in groups that contain Fibonacci(k) terms like (0), (1), (1, 2), (2, 2, 3), (2, 3, 3, 3, 4), (3, 3, 4, 3, 4, 4, 4, 5) etc, the sums in the groups are the terms of A023610. - Gary W. Adamson, Nov 02 2010
Equivalently, the number of periods in the length-n prefix of the infinite Fibonacci word (A003849). An integer p, 1 <= p <= n, is a period of a length-n word x if x[i] = x[i+p] for 1 <= i <= n-p. - Jeffrey Shallit, May 23 2020

			a(10) = 3 because A104326(10) = 1110 contains three ones.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
J. L. Brown, Jr., A new characterization of the Fibonacci numbers, Fibonacci Quarterly 3, No. 1 (1965), 1-8.
Ron Knott, Using the Fibonacci numbers to represent whole numbers.
Wolfgang Steiner, The joint distribution of greedy and lazy Fibonacci expansions, Fib. Q., 43 (No. 1, 2005), 60-69.

Number of terms in row n of A112309.
Cf. A003849, A035517, A007895, A007953, A104326, A117479.
Record positions are in A001911. - Ray Chandler, Dec 01 2005

a112310 n = a112310_list !! n
a112310_list = concat fss where
   fss = [0] : [1] : (map (map (+ 1))) (zipWith (++) fss $ tail fss)
-- Reinhard Zumkeller, Oct 26 2013

A112310 := proc(n)
    convert(A104326(n),base,10) ;
    add(d,d=%) ;
end proc:
seq(A112310(n),n=0..120) ; # R. J. Mathar, Aug 28 2025

DeleteCases[IntegerDigits[Range[200], 2], {_, 0, 0, _}]
A112309 = Map[DeleteCases[Reverse[#] Fibonacci[Range[Length[#]] + 1], 0] &, DeleteCases[IntegerDigits[-1 + Range[200], 2], {_, 0, 0, _}]]
A112310 = Map[Length, A112309]
(* Peter J. C. Moses, Mar 03 2015 *)

a(n) = A007953(A104326(n)). - Amiram Eldar, Oct 10 2023

Extended by Ray Chandler, Dec 01 2005

Merged with a sequence from Casey Mongoven, Mar 20 2006, by Franklin T. Adams-Watters, Dec 19 2006

A200650 Number of 0's in Stolarsky representation of n.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 1, 0, 2, 1, 1, 1, 0, 2, 2, 1, 2, 1, 1, 1, 0, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 1, 0, 3, 3, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 1, 0, 4, 3, 3, 3, 2, 3, 3, 2, 3, 2, 2, 2, 1, 3, 3, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 1, 0, 4, 4, 3, 4, 3, 3, 3, 2, 4, 3, 3

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.
a(n+1), n >= 1, gives the size of the n-th generation of each of the "[male-female] pair of Fibonacci rabbits" in the Fibonacci rabbits tree read right-to-left by row, the first pair (the root) being the 0th generation. (Cf. OEIS Wiki link below.) - Daniel Forgues, May 07 2015
From Daniel Forgues, May 07 2015: (Start)
Concatenation of:
  0:                      1,
  1:                      0,
  2:                      0,
  3:                   1, 0,
  4:                1, 1, 0,
  5:          2, 1, 1, 1, 0,
  6: 2, 2, 1, 2, 1, 1, 1, 0,
(...),
where row n, n >= 3, is row n-1 prepended by incremented row n-2. (End)
For n >= 3, this algorithm yields the next F_n terms of the sequence, where F_n is the n-th Fibonacci number (A000045). Since it is asymptotic to (phi^n)/sqrt(5), the number of terms thus obtained grows exponentially at each step! - Daniel Forgues, May 22 2015
Conjecture: a(n) is one less than the length of row n-1 of A385817. To obtain it, first take maximal run lengths of binary indices of each nonnegative integer (giving A245563), then remove all duplicate rows (giving A385817), and finally take the length of each remaining row and subtract 1. For sum instead of length we appear to have A200648. For anti-runs instead of runs we appear to have A341259 = A200649-1. - Gus Wiseman, Jul 21 2025
How is this related to A117479? - R. J. Mathar, Aug 10 2025

			The Stolarsky representation of 19 is 11101. This has one 0. So a(19) = 1.

Kenny Lau, Table of n, a(n) for n = 1..20000
Casey Mongoven, Description of Stolarsky Representations.
OEIS Wiki, Fibonacci rabbits per generation.

For length instead of number of 0's we have A200648.
For sum (or number of 1's) instead of number of 0's we have A200649.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion, 0's A023416.
A069010 counts maximal runs of binary indices, ranked by A385889.
A245563 lists maximal run lengths of binary indices, duplicates removed A385817.
Cf. A000045, A023758, A135818, A200651, A245562, A328592, A385886.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := If[n == 1, 1, Count[stol[n], 0]]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = if(n == 1, 1,  my(s = stol(n)); #s - vecsum(s)); \\ Amiram Eldar, Jul 07 2023

a(n) = A200648(n) - A200649(n). - Amiram Eldar, Jul 07 2023

Corrected and extended by Kenny Lau, Jul 04 2016

A341259 Number of 0's in n-th word defined at A341258.

Original entry on oeis.org

1, 0, 2, 1, 1, 3, 0, 2, 2, 2, 4, 1, 1, 3, 1, 3, 3, 3, 5, 0, 2, 2, 2, 4, 2, 2, 4, 2, 4, 4, 4, 6, 1, 1, 3, 1, 3, 3, 3, 5, 1, 3, 3, 3, 5, 3, 3, 5, 3, 5, 5, 5, 7, 0, 2, 2, 2, 4, 2, 2, 4, 2, 4, 4, 4, 6, 2, 2, 4, 2, 4, 4, 4, 6, 2, 4, 4, 4, 6, 4, 4, 6, 4, 6, 6, 6

Offset: 1

Clark Kimberling, Mar 16 2021

nonn

The number of 1's in the n-th word is given by A117479.

Cf. A117479, A341258.

Appears to be A200649 - 1.

z = 200; r = (1 + Sqrt[5])/2;
s = Table[Floor[r*n], {n, 1, z}]; t = Complement[Range[Max[s]], s];
s1[n_] := Length[Intersection[Range[n - 1], s]];
t1[n_] := n - 1 - s1[n];
w[1] = {0}; w[t[[1]]] = {1};
w[n_] := If[MemberQ[s, n], Join[{0}, w[s1[n]]], Join[{1}, w[t1[n]]]]
tt = Table[w[n], {n, 1, z}] (* A341258 *)
c[n_] := Count[tt[[n]], 0]; Table[c[n], {n, 1, z}]  (* A341259 *)
d[n_] := Count[tt[[n]], 1]; Table[d[n], {n, 1, z}]  (* A117479 *)

A356894 a(n) is the number of 0's in the maximal tribonacci representation of n (A352103).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 1, 0, 2, 2, 1, 2, 1, 1, 0, 3, 2, 3, 2, 2, 1, 2, 2, 1, 2, 1, 1, 0, 4, 3, 3, 2, 3, 3, 2, 3, 2, 2, 1, 3, 2, 3, 2, 2, 1, 2, 2, 1, 2, 1, 1, 0, 4, 4, 3, 4, 3, 3, 2, 4, 3, 4, 3, 3, 2, 3, 3, 2, 3, 2, 2, 1, 4, 3, 3, 2, 3, 3, 2, 3, 2, 2, 1, 3, 2, 3, 2

Offset: 0

Amiram Eldar, Sep 03 2022

			  n  a(n)  A352103(n)
  -  ----  ----------
  0     1           0
  1     0           1
  2     1          10
  3     0          11
  4     2         100
  5     1         101
  6     1         110
  7     0         111
  8     2        1001
  9     2        1010

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000073, A352103, A352104, A356895.

Similar sequences: A023416, A102364, A117479, A278042.

t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; trib[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; IntegerDigits[Total[2^(s - 1)], 2]]; a[n_] := Module[{v = trib[n]}, nv = Length[v]; i = 1; While[i <= nv - 3, If[v[[i ;; i + 3]] == {1, 0, 0, 0}, v[[i ;; i + 3]] = {0, 1, 1, 1}; If[i > 3, i -= 4]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, 1, Count[v[[i[[1, 1]] ;; -1]], 0]]]; Array[a, 100, 0]

a(n) = A356895(n) - A352104(n).

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A112310 Number of terms in lazy Fibonacci representation of n.

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A200650 Number of 0's in Stolarsky representation of n.

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A341259 Number of 0's in n-th word defined at A341258.

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A356894 a(n) is the number of 0's in the maximal tribonacci representation of n (A352103).

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