OFFSET
0,7
COMMENTS
Let B(n) be the set of distinct coefficients in the expansion of (x + y + z)^n, that is, B(n) = {multinomial(n; i, j, k) : i, j, k >= 0 and i + j + k = n}. Thus |B(n)| = A086753(n).
For an integer t, let r_n(t) = |{(x, y) : x, y in B(n), x <= y and x + y = t}|.
Thus a(n) is the number of distinct pairwise sums t having at least two unordered representations with repetition as x + y with x, y in B(n).
A398230(n) counts the total collision excess among these pairwise sums. Hence a(n) <= A398230(n), with equality if and only if no pairwise sum has more than two representations.
For n >= 6, a(n) > 0 since multinomial(n; n - 6, 6, 0) + multinomial(n; n - 6, 3, 3) = multinomial(n; n - 6, 5, 1) + multinomial(n; n - 6, 4, 2), which follows from 1 + 20 = 6 + 15 in row 6 of Pascal's triangle. Together with the collision 10 + 30 = 20 + 20 for n = 5 and direct inspection for 0 <= n <= 4, this shows that a(n) = 0 if and only if 0 <= n <= 4.
Every additive relation among binomial coefficients in row r of Pascal's triangle induces an additive relation among trinomial coefficients of order n for every n >= r, since multinomial(n; n - r, r - k, k) = binomial(n, r)*binomial(r, k).
For odd n = 2*s + 1, a four-term identity gives floor((s - 1)/2) distinct genuine collision sums, and hence a(n) >= floor((n - 3)/4) for odd n >= 7. See the Huber link for a proof.
Using embedded Pascal-row collisions with r = d^2 - 4 gives a(n) >= ceiling((floor(sqrt(n + 4)) - 3)/2) for n >= 12. See the Huber link for a proof.
Direct computation shows that the first pairwise sum having three representations occurs at n = 11: 2310 + 11550 = 4620 + 9240 = 6930 + 6930.
LINKS
Felix Huber, Table of n, a(n) for n = 0..242
Felix Huber, Proofs and lower bounds for A398230 and A398231
FORMULA
EXAMPLE
For n = 5, B(5) = {1, 5, 10, 20, 30}. The sum 40 has the two representations 10 + 30 and 20 + 20, while every other unordered pairwise sum has only one representation. Hence a(5) = 1.
MAPLE
A398231 := proc(n) option remember;
local a, i, j, l, m, r, s, t;
t := table();
for i from iquo(n + 2, 3) to n do
for j from iquo(n - i + 1, 2) to min(i, n - i) do
t[binomial(n, i)*binomial(n - i, j)] := 1
end do
end do;
l := [indices(t, 'nolist')];
m := nops(l);
s := table();
a := 0;
for i to m do
for j from i to m do
r := l[i] + l[j];
if assigned(s[r]) then
if s[r] = 1 then
a := a + 1;
s[r] := 2
end if
else
s[r] := 1
end if
end do
end do;
a
end proc:
seq(A398231(n), n = 0 .. 65);
CROSSREFS
KEYWORD
nonn
AUTHOR
Felix Huber, Jul 29 2026
STATUS
approved