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

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

1,9

Comments

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

Links

Kenny Lau, Table of n, a(n) for n = 1..20000

Casey Mongoven, Description of Stolarsky Representations.

OEIS Wiki, Fibonacci rabbits per generation.

Formula

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

Example

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

Mathematica

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 *)

Prog

(PARI) 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

Crossrefs

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.

Sequence in context: A064272 A385814 A117479 * A281743 A118404 A089339

Adjacent sequences: A200647 A200648 A200649 * A200651 A200652 A200653

Keyword

nonn,base,easy

Author

Casey Mongoven, Nov 19 2011

Extensions

Corrected and extended by Kenny Lau, Jul 04 2016

Status

approved