A200649 Number of 1's in the Stolarsky representation of n.

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

Offset: 1

Author

Casey Mongoven, Nov 19 2011

Keywords

• nonn
• base

Comments

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the length of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the length of each remaining row. For sum instead of length we appear to have A200648. For runs minus 1 instead of anti-runs see A200650. - Gus Wiseman, Jul 21 2025

Examples

			The Stolarsky representation of 19 is 11101. This has 4 1's. So a(19) = 4.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Casey Mongoven, Description of Stolarsky Representations.

Crossrefs

Cf. A072649, A130312, A135818, A200651.

For length instead of number of 1's we have A200648.

For 0's instead of 1's we have A200650.

Stolarsky representation is listed by A385888, ranks A200714.

A000120 counts 1's in binary expansion.

A384877 lists anti-run lengths of binary indices, duplicates removed A385886.

A384890 counts maximal anti-runs of binary indices, ranked by A385816.

Cf. A023758, A044813, A048793, A069010, A164707, A245562, A245563, A328592, A384893.

Programs

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_] := Count[stol[n], 1]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

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) = vecsum(stol(n)); \\ Amiram Eldar, Jul 07 2023

Formula

a(n) = a(n - A130312(n-1)) + (A072649(n-1) - A072649(n - A130312(n-1) - 1)) mod 2 for n > 2 with a(1) = 0, a(2) = 1. - Mikhail Kurkov, Oct 19 2021 [verification needed]

a(n) = A200648(n) - A200650(n). - Amiram Eldar, Jul 07 2023

Extensions

More terms from Amiram Eldar, Jul 07 2023