A248893 a(0)=0; for n>0, choose a(n) to be the smallest number > a(n-1) such that the condition a(n) > Sum_{k=0..n} (number of 1's in binary expansion of n a(k)) holds

0, 2, 4, 5, 8, 9, 10, 12, 16, 17, 18, 20, 22, 24, 28, 32, 33, 34, 35, 38, 40, 44, 48, 49, 52, 56, 58, 64, 65, 66, 67, 68, 72, 73, 76, 80, 81, 84, 88, 92, 96, 97, 100, 104, 106, 112, 113, 118, 124, 128, 129, 130, 131, 134, 136, 140, 144, 145, 148, 152, 154, 160

Offset

0,2

Comments

See A244510 for a decimal version.

Links

Lars Blomberg, Table of n, a(n) for n = 0..10000

Example

a(7)=12 because the sum of 1's in the binary expansions of a(0)..a(6) is 9, and 12 is the smallest number greater than a(6)=10 and also greater than 9 + sum of 1's in the binary expansion of 12 (which is 2).

Crossrefs

Cf. A000120, A244510

Sequence in context: A344845 A166110 A087815 * A022549 A045704 A368796

Adjacent sequences: A248890 A248891 A248892 * A248894 A248895 A248896

Keyword

nonn,base

Author

Lars Blomberg, Mar 06 2015

Status

approved