OFFSET
5,2
COMMENTS
Let B(n) be the set of distinct numbers n!/(p_1!*...*p_r!), where (p_1, ..., p_r) ranges over the partitions of n. For an integer s, let r_n(s) = |{(x, y) : x, y in B(n), x <= y and x + y = s}|.
Thus T(n, k) = |{s : r_n(s) >= k}|.
The rows are weakly decreasing, since every sum having at least k + 1 representations also has at least k representations.
Appending a part 1 to every partition of n shows that (n + 1)*B(n) is a subset of B(n + 1). Hence T(n + 1, k) >= T(n, k) whenever 2 <= k <= A398263(n).
The first entry in row n is A398238(n), the row sum is A398233(n), and A398263(n) is the largest k for which T(n, k) > 0.
A sum having r representations contributes 1 to T(n, 2), ..., T(n, r), and therefore contributes r - 1 to the row sum.
The weighted row sum Sum_{k = 2 .. A398263(n)} (k - 1)*T(n, k) equals Sum_s binomial(r_n(s), 2), the number of pairs of distinct unordered representations of the same sum.
For q >= 3, let N(q) = Sum_{p prime, p <= q} p*floor(log(q)/log(p)). The set B(N(q)) contains an arithmetic progression of length q. Its pairwise sums give T(n, k) >= 2*q - 4*k + 3 whenever n >= N(q) and 2 <= k <= ceiling(q/2).
More strongly, for n >= N(q) + q, the set B(n) contains n - N(q) - q + 1 scaled copies of this progression whose internal pair-sum intervals are disjoint. Hence T(n, k) >= (n - N(q) - q + 1)*(2*q - 4*k + 3). See the Huber link for proofs.
A398230 and A398231 give the corresponding collision statistics for trinomial coefficients, A398232 and A398237 are the associated triangles by number of variables, A398233 and A398238 are respectively the row sums and first entries of this triangle, and A398263 gives the maximum representation multiplicity.
LINKS
Felix Huber, Large Sidon Subsets and Pair-Sum Multiplicities of Distinct Multinomial Coefficients, arXiv:2608.02667 [math.CO], 2026.
FORMULA
T(n, k) = |{s : r_n(s) >= k}|, where r_n(s) = |{(x, y) : x, y in B(n), x <= y and x + y = s}|.
T(n, 2) = A398238(n).
A398263(n) = max{k : T(n, k) > 0}.
T(n, k + 1) <= T(n, k) for 2 <= k < A398263(n).
T(n + 1, k) >= T(n, k) for n >= 5 and 2 <= k <= A398263(n).
Sum_{k = 2 .. A398263(n)} (k - 1)*T(n, k) = Sum_s binomial(r_n(s), 2).
T(n, k) >= 2*q - 4*k + 3 whenever n >= N(q), q >= 3, and 2 <= k <= ceiling(q/2), where N(q) = Sum_{p prime, p <= q} p*floor(log(q)/log(p)).
T(n, k) >= (n - N(q) - q + 1)*(2*q - 4*k + 3) whenever n >= N(q) + q, q >= 3, and 2 <= k <= ceiling(q/2).
EXAMPLE
The irregular triangle begins:
n\k | 2 3 4 5 6 7 8 9 10
----+----------------------------------------------------------------
5 | 1
6 | 6
7 | 8
8 | 24
9 | 42 1
10 | 68 6
11 | 128 19 1
12 | 198 40 7
13 | 336 89 14 2
14 | 557 156 32 7 1
15 | 781 282 88 23 4
16 | 1119 398 159 49 15 2
17 | 1623 571 255 109 35 10 2
18 | 2311 825 380 178 76 28 6 1
19 | 3321 1213 556 270 136 64 23 6
20 | 4809 1807 844 406 205 100 55 25 5
T(9, 3) = 1, since B(9) = {1, 9, 36, 72, 84, 126, 252, 504, 630, 756, 1260, 1512, 1680, 2520, 3024, 3780, 5040, 7560, 10080, 15120, 22680, 30240, 45360, 60480, 90720, 181440, 362880}, and 1764 is the only pairwise sum of elements of B(9) having at least 3 unordered representations: 1764 = 84 + 1680 = 252 + 1512 = 504 + 1260.
MAPLE
r := proc(n) option remember;
local a, b, c, i, j, k, l, m, q, s, t, u;
t := table();
b := proc(v, p, d)
local i;
if v = 0 then
t[n!/d] := 1;
return
end if;
for i from min(p, v) by -1 to 1 do
b(v - i, i, d*i!)
end do
end proc;
b(n, n, 1);
l := [indices(t, 'nolist')];
m := nops(l);
s := table();
for i to m do
for j from i to m do
u := l[i] + l[j];
if assigned(s[u]) then
s[u] := s[u] + 1
else
s[u] := 1
end if
end do
end do;
q := [indices(s, 'nolist')];
a := max(seq(s[q[i]], i = 1 .. nops(q)));
c := Array(1 .. a - 1, 'fill' = 0);
for i to nops(q) do
for k to s[q[i]] - 1 do
c[k] := c[k] + 1
end do
end do;
[seq(c[k], k = 1 .. a - 1)]
end proc:
T := proc(n, k) option remember;
r(n)[k - 1]
end proc:
seq(seq(T(n, k), k = 2 .. nops(r(n)) + 1), n = 5 .. 20);
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Felix Huber, Aug 11 2026
STATUS
approved