

A194703


Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (3 + m).


2



3, 2, 1, 0, 1, 2, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 3, 2, 1, 0, 1, 2, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0
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OFFSET

1,1


COMMENTS

Subtriangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 3. For further information see A182703 and A135010.


LINKS

Table of n, a(n) for n=1..105.


FORMULA

T(k,m) = A182703(3+m,k), with T(k,m) = 0 if k > 3+m.
T(k,m) = A194812(3+m,k).


EXAMPLE

Triangle begins:
3,
2, 1,
0, 1, 2,
1, 0, 1, 1,
0, 1, 0, 1, 1,
0, 0, 1, 0, 1, 1,
0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1, 1,
...
For k = 1 and m = 1; T(1,1) = 3 because there are three parts of size 1 in the last section of the set of partitions of 4, since 3 + m = 4, so a(1) = 3. For k = 2 and m = 1; T(2,1) = 2 because there two parts of size 2 in the last section of the set of partitions of 4, since 3 + m = 4, so a(2) = 2.


CROSSREFS

Always the sum of row k = p(3) = A000041(3) = 3.
The first (010) members of this family of triangles are A023531, A129186, A194702, this sequence, A194704A194710.
Cf. A135010, A138121, A182712A182714, A194812.
Sequence in context: A204176 A062160 A301296 * A219175 A022959 A023445
Adjacent sequences: A194700 A194701 A194702 * A194704 A194705 A194706


KEYWORD

nonn,tabl


AUTHOR

Omar E. Pol, Feb 05 2012


STATUS

approved



