|
| |
|
|
A156044
|
|
A triangle sequence made symmetrical by reverse coefficients: t0(n,m)=2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m]; t(n,m)=(t0(n,m)+Reverse[t0(n,m)])/2
|
|
0
|
|
|
|
1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 3, 3, 3, 1, 1, 3, 4, 4, 3, 1, 1, 5, 6, 7, 6, 5, 1, 1, 5, 8, 9, 9, 8, 5, 1, 1, 8, 11, 14, 14, 14, 11, 8, 1, 1, 9, 15, 18, 20, 20, 18, 15, 9, 1, 1, 13, 20, 26, 28, 30, 28, 26, 20, 13, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,5
|
|
|
COMMENTS
|
Row sums are:
{1, 2, 4, 6, 11, 16, 31, 46, 82, 126, 206,...}.
The idea here was tom design a symmetrical analog of a Pascal's triangle using p
partitions and addition and subtraction instead of multiplication.
|
|
|
LINKS
|
Table of n, a(n) for n=0..65.
|
|
|
FORMULA
|
t0(n,m)=2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m];
t(n,m)=(t0(n,m)+Reverse[t0(n,m)])/2
|
|
|
EXAMPLE
|
{1},
{1, 1},
{1, 2, 1},
{1, 2, 2, 1},
{1, 3, 3, 3, 1},
{1, 3, 4, 4, 3, 1},
{1, 5, 6, 7, 6, 5, 1},
{1, 5, 8, 9, 9, 8, 5, 1},
{1, 8, 11, 14, 14, 14, 11, 8, 1},
{1, 9, 15, 18, 20, 20, 18, 15, 9, 1},
{1, 13, 20, 26, 28, 30, 28, 26, 20, 13, 1}
|
|
|
MATHEMATICA
|
Clear[t];
t[n_, m_] = 2 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m];
Table[(Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]])/2, {n, 0, 10}];
Flatten[%]
|
|
|
CROSSREFS
|
Sequence in context: A103691 A103441 A081206 * A180980 A048570 A090806
Adjacent sequences: A156041 A156042 A156043 * A156045 A156046 A156047
|
|
|
KEYWORD
|
nonn,tabl,uned
|
|
|
AUTHOR
|
Roger L. Bagula, Feb 02 2009
|
|
|
STATUS
|
approved
|
| |
|
|
