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
Examples
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).
Links
- Lars Blomberg, Table of n, a(n) for n = 0..10000
Comments