|
|
A342219
|
|
a(1) = 1, a(2) = 2; for n > 2, a(n) = the number of terms in the maximal length sum of previous consecutive terms that equals n.
|
|
1
|
|
|
1, 2, 2, 2, 3, 3, 4, 3, 4, 5, 3, 5, 6, 5, 5, 6, 7, 3, 7, 8, 7, 6, 8, 9, 7, 8, 8, 9, 10, 8, 10, 11, 8, 10, 8, 11, 12, 10, 8, 11, 10, 12, 13, 8, 12, 11, 13, 14, 11, 13, 6, 14, 15, 13, 10, 14, 9, 15, 16, 11, 14, 14, 15, 15, 16, 17, 13, 17, 18, 10, 16, 9, 17, 15, 18, 19, 16, 15, 17, 15, 18, 13
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
The equivalent sequence for a minimal length sum is given by A003059.
|
|
LINKS
|
|
|
EXAMPLE
|
a(3) = 2 as the only way to sum previous consecutive terms to make 3 is 1 + 2 = 3, which contains two terms.
a(7) = 4 as the previous consecutive terms 1 + 2 + 2 + 2 = 7, which contains four terms. Note that 7 can also be made by consecutive terms 2 + 2 + 3 = 7, but the sequence is the maximal sum length.
a(10) = 5 as the previous consecutive terms 1 + 2 + 2 + 2 + 3 = 10, which contains five terms. Three other consecutive term sums also exist that sum to 10 but they contain fewer terms.
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|