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A368565
a(n) = number of pairs (p,q) of partitions of n such that d(p,q) < o(p,q), where d and o are distance functions; see Comments.
3
0, 0, 0, 0, 0, 30, 106, 316, 652, 1388, 2618, 5170, 9164, 16790, 29046, 50714, 84732, 143588, 234048, 385210, 617050, 990868, 1558310, 2459300, 3806838, 5900184
OFFSET
1,6
COMMENTS
The definition of d depends on the greedy ordering of the partitions p(i) of n; that is, p(1) >= p(2) >= ... >= p(k), where k = A000041(n); see A366156. The ordinal distance o is defined by o(p(i),p(j)) = |i-j|.
FORMULA
A368564(n) + a(n) + A368566(n) = A001255(n) for n >= 1.
EXAMPLE
The 5 partitions of 4 are (p(1),p(2),p(3),p(4),p(5)) = (4,21,22,211,1111). The following table shows the 25 pairs d(p(i),q(j)) and o(p(i),q(j)):
| 4 31 22 211 1111
------------------------------------------------
4 d | 0 2 4 4 6
o | 0 1 2 3 4
31 d | 2 0 2 2 4
o | 1 0 1 2 3
22 d | 4 2 0 2 4
o | 2 1 0 1 2
211 d | 4 2 2 0 2
o | 3 2 1 0 1
1111 d | 6 4 4 2 0
o | 4 3 2 1 0
The table shows 0 pairs (p,q) for which d(p,q) < o(p,q), so a(4) = 0.
MATHEMATICA
c[n_] := PartitionsP[n];
q[n_, k_] := q[n, k] = IntegerPartitions[n][[k]];
r[n_, k_] := r[n, k] = Join[q[n, k], ConstantArray[0, n - Length[q[n, k]]]];
d[u_, v_] := Total[Abs[u - v]];
p[n_] := Flatten[Table[d[r[n, j], r[n, k]] - Abs[j - k], {j, 1, c[n]}, {k, 1, c[n]}]];
Table[Count[p[n], 0], {n, 1, 16}] (* A368565 *)
Table[Length[Select[p[n], Sign[#] == -1 &]], {n, 1, 16}] (* A368566 *)
Table[Length[Select[p[n], Sign[#] == 1 &]], {n, 1, 16}] (* A368567 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Clark Kimberling, Dec 31 2023
STATUS
approved