A218254 Irregular table, where row n (n >= 0) starts with n, the next term is n-A000120(n), and the successive terms are obtained by repeatedly subtracting the number of 1's in the previous term's binary expansion, until zero is reached, after which the next row starts with one larger n.

0, 1, 0, 2, 1, 0, 3, 1, 0, 4, 3, 1, 0, 5, 3, 1, 0, 6, 4, 3, 1, 0, 7, 4, 3, 1, 0, 8, 7, 4, 3, 1, 0, 9, 7, 4, 3, 1, 0, 10, 8, 7, 4, 3, 1, 0, 11, 8, 7, 4, 3, 1, 0, 12, 10, 8, 7, 4, 3, 1, 0, 13, 10, 8, 7, 4, 3, 1, 0, 14, 11, 8, 7, 4, 3, 1, 0, 15, 11, 8, 7, 4, 3, 1, 0

Offset: 0

Author

Nico Brown, Oct 24 2012

Keywords

• nonn
• tabf
• easy

Examples

			The n-th row (starting indexing from zero) in this irregular table consists of block of length A071542(n)+1: 1,2,3,3,4,4,5,5,... which always ends with zero, as:
0
1,0
2,1,0
3,1,0
4,3,1,0
5,3,1,0
6,4,3,1,0
7,4,3,1,0
The 17th term is 6, which in binary is 110. The 18th term is then 6-2=4.
		

Links

• Antti Karttunen, Rows 0..255, flattened

Crossrefs

Cf. A218252, A218253. A213707 gives the positions of zeros (i.e. the ending index of each row). A071542, A000120.

The reversed tails of the rows converge towards A179016.

Programs

PARI

for(n=0,9,k=n;while(k, print1(k", "); k-=hammingweight(k)); print1("0, ")) \\ Charles R Greathouse IV, Oct 30 2012