cp's OEIS Frontend

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

%I A200650 #75 Aug 10 2025 15:09:18
%S A200650 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,
%T A200650 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,
%U A200650 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
%N A200650 Number of 0's in Stolarsky representation of n.
%C A200650 For the Stolarsky representation of n, see the C. Mongoven link.
%C A200650 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
%C A200650 From _Daniel Forgues_, May 07 2015: (Start)
%C A200650 Concatenation of:
%C A200650   0:                      1,
%C A200650   1:                      0,
%C A200650   2:                      0,
%C A200650   3:                   1, 0,
%C A200650   4:                1, 1, 0,
%C A200650   5:          2, 1, 1, 1, 0,
%C A200650   6: 2, 2, 1, 2, 1, 1, 1, 0,
%C A200650 (...),
%C A200650 where row n, n >= 3, is row n-1 prepended by incremented row n-2. (End)
%C A200650 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
%C A200650 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
%C A200650 How is this related to A117479? - _R. J. Mathar_, Aug 10 2025
%H A200650 Kenny Lau, <a href="/A200650/b200650.txt">Table of n, a(n) for n = 1..20000</a>
%H A200650 Casey Mongoven, <a href="/A200648/a200648.txt">Description of Stolarsky Representations</a>.
%H A200650 OEIS Wiki, <a href="/wiki/Fibonacci_rabbits_per_generation">Fibonacci rabbits per generation</a>.
%F A200650 a(n) = A200648(n) - A200649(n). - _Amiram Eldar_, Jul 07 2023
%e A200650 The Stolarsky representation of 19 is 11101. This has one 0. So a(19) = 1.
%t A200650 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}]]];
%t A200650 a[n_] := If[n == 1, 1, Count[stol[n], 0]]; Array[a, 100] (* _Amiram Eldar_, Jul 07 2023 *)
%o A200650 (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])));}
%o A200650 a(n) = if(n == 1, 1,  my(s = stol(n)); #s - vecsum(s)); \\ _Amiram Eldar_, Jul 07 2023
%Y A200650 For length instead of number of 0's we have A200648.
%Y A200650 For sum (or number of 1's) instead of number of 0's we have A200649.
%Y A200650 Stolarsky representation is listed by A385888, ranks A200714.
%Y A200650 A000120 counts 1's in binary expansion, 0's A023416.
%Y A200650 A069010 counts maximal runs of binary indices, ranked by A385889.
%Y A200650 A245563 lists maximal run lengths of binary indices, duplicates removed A385817.
%Y A200650 Cf. A000045, A023758, A135818, A200651, A245562, A328592, A385886.
%K A200650 nonn,base,easy
%O A200650 1,9
%A A200650 _Casey Mongoven_, Nov 19 2011
%E A200650 Corrected and extended by _Kenny Lau_, Jul 04 2016