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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.
5
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
0,3
COMMENTS
In other words, the n-th row contains the numbers k with the same binary length as n and for any i >= 0, if the i-th bit and the (i+1)-th bit in k are different then they are also different in n (i = 0 corresponding to the least significant bit).
The value m appears 2^A092339(m) times in the triangle (see A361674).
FORMULA
T(n, 1) = A342126(n).
T(n, max(1, 2^(A005811(n)-1))) = A003817(n).
EXAMPLE
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)
PROG
(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); }
KEYWORD
nonn,base,tabf
AUTHOR
Rémy Sigrist, Mar 19 2023
STATUS
approved