OFFSET
1,8
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). The crossover part of P is p(h). The n-th row sum is the number of partitions of n, A000041. The bisections of column 1 are also given by A000041. The limit of the reversal of row n is given by A000041.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
EXAMPLE
First 14 rows (indexed by column 1):
1... 1
2... 1 1
3... 1 1 1
4... 1 2 1 1
5... 2 1 2 1 1
6... 2 2 3 2 1 1
7... 3 4 1 3 2 1 1
8... 3 5 2 5 3 2 1 1
9... 5 7 5 1 5 3 2 1 1
10... 5 9 7 2 7 5 3 2 1 1
11... 7 12 12 5 1 7 5 4 2 1 1
12... 7 14 15 8 2 11 7 5 3 2 1 1
13... 11 20 20 14 5 1 11 7 5 3 2 1 1
14... 11 24 25 20 8 2 15 11 7 5 3 2 1 1
MATHEMATICA
p[n_] := p[n] = IntegerPartitions[n]; 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[p[n][[k]][[q[n, k]]], {k, 1, Length[p[n]]}]];
c[n_, k_] := c[n, k] = Count[u[n], k];
v = Table[c[n, k], {n, 1, 25}, {k, 1, n}];
TableForm[v] (*A279044 array*)
Flatten[v] (*A279044 sequence*)
CROSSREFS
KEYWORD
nonn,easy,tabf
AUTHOR
Clark Kimberling, Dec 04 2016
STATUS
approved