A228091 Numbers n for which there exists such a natural number k < n that k + bitcount(k) = n + bitcount(n), where bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.
4, 12, 16, 17, 20, 28, 32, 34, 36, 44, 48, 49, 52, 60, 65, 68, 76, 80, 81, 84, 92, 96, 98, 100, 108, 112, 113, 116, 124, 128, 129, 130, 131, 132, 140, 144, 145, 148, 156, 160, 162, 164, 172, 176, 177, 180, 188, 193, 196, 204, 208, 209, 212, 220, 224, 226, 228
Offset: 1
Keywords
Examples
For cases 0 + A000120(0) = 0, 1 + A000120(1) = 2, 2 + A000120(2) = 3, 3 + A000120(3) = 5 there are no smaller solutions yielding the same result. However, for 4 + A000120(4) = 5, we already saw the case 3+A000120(3) giving the same result, thus 4 is the first term of this sequence. Next time this occurs for 12, as 12 + A000120(12) = 14 = 11 + A000120(11), and 11 < 12.
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