A249144 a(0) = 0, after which a(n) gives the total number of runs of the same length as the rightmost run in the binary representation of a(n-1) [i.e., A136480(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.
0, 1, 2, 4, 1, 6, 7, 1, 8, 2, 11, 3, 4, 5, 17, 19, 7, 4, 8, 5, 25, 26, 29, 31, 1, 32, 2, 35, 12, 14, 37, 41, 45, 49, 50, 52, 22, 57, 58, 61, 63, 1, 64, 2, 67, 25, 69, 73, 76, 32, 3, 33, 80, 4, 34, 87, 14, 92, 35, 36, 38, 99, 42, 105, 108, 47, 5, 114, 116, 49, 119, 23, 24, 25, 123, 54, 126, 127, 1, 128, 2
Offset: 0
Examples
a(0) = 0 (by definition), and 0 is also '0' in binary.
For n = 1, we see that in a(0) there is one run of length 1, which is total number of runs of length 1 so far in terms a(0) .. a(n-1), thus a(1) = 1.
For n = 2, we see that the rightmost run of a(1) = 1 ('1' also in binary) has occurred two times in total (once in a(0) and a(1)), thus a(2) = 2.
For n = 3, we see that the rightmost run of a(2) = 2 ('10' in binary) is one bit long, and so far there has occurred four such runs in total (namely once in a(0) and a(1), twice in a(2)), thus a(3) = 4.
For n = 4, we see that the rightmost run of a(3) = 4 ('100' in binary) is two bits long, and it is so far the first and only two-bit run in the sequence, thus a(4) = 1.
For n = 5, we see that the rightmost run of a(4) = 1 ('1' in binary) is one bit long, and so far there has occurred 6 such one-bit runs in terms a(0) .. a(4), thus a(5) = 6.
For n = 6, we see that the rightmost run of a(5) = 6 ('110' in binary) is one bit long, and so far there has occurred 7 such one bit runs in terms a(0) .. a(5), thus a(6) = 7.
Links
- Antti Karttunen, Table of n, a(n) for n = 0..10000
Comments