1, 2, 6, 14, 33, 70, 149, 298, 591, 1132, 2139, 3948, 7199, 12894, 22836, 39894, 68982, 117948, 199852, 335426, 558429, 922112, 1511610, 2460208, 3977963, 6390942, 10206862, 16207444, 25596941, 40214896, 62868772, 97814358
Offset: 0
We represent each summand, k, in a partition of n as k identical objects. Then we color each object. We have no regard for the order of the colored objects.
a(3) = 14 because we have: www; wwb; wbb; bbb; ww + w; ww + b; wb + w; wb + b; bb + w; bb + b; w + w + w; w + w + b; w + b + b; b + b + b, where the 2 colors are black b and white w. - _Geoffrey Critzer_, Sep 27 2012
a(3) = 14 because we have: 3; 3'; 3''; 3'''; 2 + 1; 2 + 1'; 2' + 1; 2' + 1'; 2'' + 1; 2'' + 1'; 1 + 1 + 1; 1 + 1 + 1'; 1 + 1' + 1'; 1' + 1' + 1', where a part k of different sorts is given as k, k', k'', etc. - _Joerg Arndt_, Mar 09 2015
From _Alois P. Heinz_, Mar 09 2015: (Start)
The a(4) = 33 = 5 + 9 + 6 + 8 + 5 partitions of 4 objects of 2 colors are:
5 partitions for the integer partition of 4 = 1 + 1 + 1 + 1:
01: {{b}, {b}, {b}, {b}}
02: {{b}, {b}, {b}, {w}}
03: {{b}, {b}, {w}, {w}}
04: {{b}, {w}, {w}, {w}}
05: {{w}, {w}, {w}, {w}}
9 partitions for the integer partition of 4 = 1 + 1 + 2:
06: {{b}, {b}, {b,b}}
07: {{b}, {w}, {b,b}}
08: {{w}, {w}, {b,b}}
09: {{b}, {b}, {w,b}}
10: {{b}, {w}, {w,b}}
11: {{w}, {w}, {w,b}}
12: {{b}, {b}, {w,w}}
13: {{b}, {w}, {w,w}}
14: {{w}, {w}, {w,w}}
6 partitions for the integer partition of 4 = 2 + 2:
15: {{b,b}, {b,b}}
16: {{b,b}, {w,b}}
17: {{b,b}, {w,w}}
18: {{w,b}, {w,b}}
19: {{w,b}, {w,w}}
20: {{w,w}, {w,w}}
8 partitions for the integer partition of 4 = 1 + 3:
21: {{b}, {b,b,b}}
22: {{w}, {b,b,b}}
23: {{b}, {w,b,b}}
24: {{w}, {w,b,b}}
25: {{b}, {w,w,b}}
26: {{w}, {w,w,b}}
27: {{b}, {w,w,w}}
28: {{w}, {w,w,w}}
5 partitions for the integer partition of 4 = 4:
29: {{b,b,b,b}}
30: {{w,b,b,b}}
31: {{w,w,b,b}}
32: {{w,w,w,b}}
33: {{w,w,w,w}}
Some see number partitions, others see set partitions, ...
(End)
It is obvious from the example of _Alois P. Heinz_ that a(n) enumerates multi-set partitions of a multi-set of n elements of two kinds. In the case that there is only one kind, this reduces to the usual case of numerical partitions. In the case that all the n elements are distinct, then this reduces to the case of set partitions. - _Michael Somos_, Mar 09 2015
There are a(3) = 14 plane partitions of 6 with trace 3; of 7 with trace 4; of 8 with trace 5; etc. See a formula above with the Stanley Exercise 7.99. - _Wolfdieter Lang_, Mar 09 2015
From _Daniel Forgues_, Mar 09 2015: (Start)
The a(3) = 14 = 4 + 6 + 4 partitions of 3 objects of 2 colors are:
4 partitions for the integer partition of 3 = 1 + 1 + 1:
01: {{b}, {b}, {b}}
02: {{b}, {b}, {w}}
03: {{b}, {w}, {w}}
04: {{w}, {w}, {w}}
6 partitions for the integer partition of 3 = 1 + 2:
05: {{b}, {b,b}}
06: {{w}, {b,b}}
07: {{b}, {w,b}}
08: {{w}, {w,b}}
09: {{b}, {w,w}}
10: {{w}, {w,w}}
4 partitions for the integer partition of 3 = 3:
11: {{b,b,b}}
12: {{w,b,b}}
13: {{w,w,b}}
14: {{w,w,w}}
The a(2) = 6 = 3 + 3 partitions of 2 objects of 2 colors are:
3 partitions for the integer partition of 2 = 1 + 1:
01: {{b}, {b}}
02: {{b}, {w}}
03: {{w}, {w}}
3 partitions for the integer partition of 2 = 2:
04: {{b,b}}
05: {{w,b}}
06: {{w,w}}
The a(1) = 2 partitions of 1 object of 2 colors are:
2 partitions for the integer partition of 1 = 1:
01: {{b}}
02: {{w}}
a(0) = 1: the empty partition, since empty sum is 0.
Triangle(sort of, since n_th row has p(n) = A000041 terms):
1: 2
2: 3, 3
3: 4, 6, 4
4: 5, 9, 6, 8, 5
5: 6, ?, ?, ?, ?, ?, 6
6: 7, ?, ?, ?, ?, ?, ?, ?, ?, ?, 7
Can we find a recurrence relation? (End)
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