A361644 Irregular triangle T(n, k), n >= 0, k = 1..max(1, 2^(A005811(n)-1)), read by rows; the n-th row lists the integers with the same binary length as n and whose partial sums of run lengths are included in those of n.
0, 1, 2, 3, 3, 4, 7, 4, 5, 6, 7, 6, 7, 7, 8, 15, 8, 9, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 8, 11, 12, 15, 12, 15, 12, 13, 14, 15, 14, 15, 15, 16, 31, 16, 17, 30, 31, 16, 17, 18, 19, 28, 29, 30, 31, 16, 19, 28, 31, 16, 19, 20, 23, 24, 27, 28, 31
Offset: 0
Examples
Triangle begins (in decimal and in binary):
n n-th row bin(n) n-th row in binary
-- ------------ ------ ------------------
0 0 0 0
1 1 1 1
2 2, 3 10 10, 11
3 3 11 11
4 4, 7 100 100, 111
5 4, 5, 6, 7 101 100, 101, 110, 111
6 6, 7 110 110, 111
7 7 111 111
8 8, 15 1000 1000, 1111
9 8, 9, 14, 15 1001 1000, 1001, 1110, 1111
.
For n = 9:
- the binary expansion of 9 is "1001",
- the corresponding run lengths are 1, 2, 1,
- so the 9th row contains the values with the following run lengths:
1, 2, 1 -> 9 ("1001" in binary)
1, 2+1 -> 8 ("1000" in binary)
1+2, 1 -> 14 ("1110" in binary)
1+2+1 -> 15 ("1111" in binary)
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..9841 (rows for n = 0..511 flattened)
- Index entries for sequences related to binary expansion of n
Crossrefs
Programs
-
PARI
row(n) = { my (r = []); while (n, my (v = valuation(n+n%2, 2)); n \= 2^v; r = concat(v, r)); my (s = [if (#r, 2^r[1]-1, 0)]); for (k = 2, #r, s = concat(s * 2^r[k], [(h+1)*2^r[k]-1|h<-s]);); vecsort(s); }
Comments