A279033 Irregular triangular array: T(n,i) = number of strict partitions of n having crossover index k; see Comments.

1, 1, 2, 2, 3, 4, 5, 6, 7, 1, 9, 1, 10, 2, 13, 2, 14, 4, 18, 4, 19, 8, 24, 8, 25, 13, 32, 14, 33, 21, 42, 22, 43, 33, 54, 35, 55, 49, 69, 53, 70, 72, 87, 78, 88, 103, 1, 109, 112, 1, 110, 145, 1, 136, 160, 137, 200, 3, 168, 220, 2, 169, 275, 4, 206, 303, 3

Offset: 1

Author

Clark Kimberling, Dec 04 2016

Keywords

• nonn
• easy
• tabf

Comments

Suppose that P = [p(1),p(2),...,p(k)] is a partition of n, where p(1) >= p(2) >= ... >= p(k). The crossover index of P is the least h such that p(1) + ... + p(h) > = n/2. Equivalently for k > 1, p(1) + ... + p(h) >= p(h+1) + ... + p(k). A strict partition is a partition into distinct parts. The n-th row sum is the number of strict partitions of n, A000009. Column 1 counts "non-squashing partitions", as in A088567.

First 32 rows (indexed by column 1):

1... 1

2... 1

3... 2

4... 2

5... 3

6... 4

7... 5

8... 6

9... 7 1

10... 9 1

11... 10 2

12... 13 2

13... 14 4

14... 18 4

15... 19 8

16... 24 8

17... 25 13

18... 32 14

19... 33 21

20... 42 22

21... 43 33

22... 54 35

23... 55 49

24... 69 53

25... 70 72

26... 87 78

27... 88 103 1

28... 109 112 1

29... 110 145 1

30... 136 160

31... 137 200 3

32... 168 220 3

Crossrefs

Cf. A000009, A088567, A276468.

Programs

Mathematica

p[n_] := p[n] = Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];
t[n_, k_] := t[n, k] = p[n][[k]];
q[n_, k_] := q[n, k] = Select[Range[50], Sum[t[n, k][[i]], {i, 1, #}] >= n/2 &, 1];
u[n_] := u[n] = Flatten[Table[q[n, k], {k, 1, Length[p[n]]}]];
c1[n_, k_] := c1[n, k] = Count[u[n], k];
m[n_] := -1 + Min[Flatten[Position[Table[c1[n, k], {k, 1, n + 1}], 0]]]
u = Table[c1[n, k], {n, 1, 50}, {k, 1, m[n]}]
TableForm[u] (* A279033 array *)
Flatten[u]   (* A279033 sequence *)