A350091 a(n) = a(floor(n/4)) for n == 2 (mod 4), otherwise n.

0, 1, 0, 3, 4, 5, 1, 7, 8, 9, 0, 11, 12, 13, 3, 15, 16, 17, 4, 19, 20, 21, 5, 23, 24, 25, 1, 27, 28, 29, 7, 31, 32, 33, 8, 35, 36, 37, 9, 39, 40, 41, 0, 43, 44, 45, 11, 47, 48, 49, 12, 51, 52, 53, 13, 55, 56, 57, 3, 59, 60, 61, 15, 63, 64, 65, 16, 67, 68, 69, 17, 71, 72, 73

Offset: 0

Author

Ruud H.G. van Tol, Dec 14 2021

Keywords

• nonn
• easy

Comments

a(n) deletes any trailing '10' bit pairs from n. So in base 4, it removes all trailing '2' digits.

Examples

			Numbers between '' are in base 2: '0'->'0', so a(0)=0. '110'->'1', so a(6)=1. '1010'->'10' -> '0', so a(10)=0. a(floor((2^1000001)/3))=0.
		

Links

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

Crossrefs

Cf. A006519, A139391, A001511, A065883 (chop trailing 00 bit pairs), A347840.

Programs

Mathematica

a[n_] := a[n] = If[Mod[n, 4] == 2, a[(n - 2)/4], n]; Array[a, 100, 0] (* Amiram Eldar, Dec 14 2021 *)

PARI

a(n) = if(2!=(n%4), n, my(m=3*n+2); m=m/4^valuation(m,4);(m+1)/3-1)

Formula

A139391(2*a(n)+1) = A139391(2*n+1).

Sum_{k=1..n} a(k) ~ 2 * n^2 / 5. - Amiram Eldar, Aug 30 2024

a(n) = (A347840(n+1) - 1)/2. - Alan Michael Gómez Calderón, Dec 08 2024