login
A339674
Irregular triangle T(n, k), n, k >= 0, read by rows; for any number m with runs in binary expansion (r_1, ..., r_j), let R(m) = {r_1 + ... + r_j, r_2 + ... + r_j, ..., r_j}; row n corresponds to the numbers k such that R(k) is included in R(n), in ascending order.
1
0, 0, 1, 0, 1, 2, 3, 0, 3, 0, 3, 4, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 6, 7, 0, 7, 0, 7, 8, 15, 0, 1, 6, 7, 8, 9, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 3, 4, 7, 8, 11, 12, 15, 0, 3, 12, 15, 0, 1, 2, 3, 12, 13, 14, 15, 0, 1, 14, 15, 0
OFFSET
0,6
COMMENTS
For any m > 0, R(m) contains the partial sums of the m-th row of A227736; by convention, R(0) = {}.
The underlying idea is to take some or all of the rightmost runs of a number, and possibly merge some of them.
For any n >= 0, the n-th row:
- has 2^A000120(A003188(n)) terms,
- has first term 0 and last term A003817(n),
- has n at position A090079(n),
- corresponds to the distinct terms in n-th row of table A341840.
FORMULA
T(n, 0) = 0.
T(n, A090079(n)) = n.
T(n, 2^A000120(A003188(n))-1) = A003817(n).
EXAMPLE
The triangle starts:
0;
0, 1;
0, 1, 2, 3;
0, 3;
0, 3, 4, 7;
0, 1, 2, 3, 4, 5, 6, 7;
0, 1, 6, 7;
0, 7;
0, 7, 8, 15;
0, 1, 6, 7, 8, 9, 14, 15;
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15;
0, 3, 4, 7, 8, 11, 12, 15;
0, 3, 12, 15;
0, 1, 2, 3, 12, 13, 14, 15;
0, 1, 14, 15;
0, 15;
...
PROG
(PARI) See Links section.
CROSSREFS
KEYWORD
nonn,base,tabf
AUTHOR
Rémy Sigrist, Feb 21 2021
STATUS
approved