OFFSET
2,2
COMMENTS
To clarify a slight ambiguity in the definition, the heaviest box in such an arrangement should contain exactly n-1 balls. - Gus Wiseman, Sep 22 2016
Conjecture: For n > 2, a(n) = 2*A000217(n+1). - Bruce J. Nicholson, Apr 05 2017
Clearly a(2)=1. Moreover, for n>2, a(n) = n*(n-1), since one can choose the box with n-1 balls in n ways, and the remaining ball can be put in one of the remaining n-1 boxes. So the conjecture above and the empirical formulas below are all correct. - Luca Ferrigno, Jul 13 2023
LINKS
R. H. Hardin, Table of n, a(n) for n = 2..59
FORMULA
Empirical: a(n) = n*binomial(n-1,n-2) for n > 2.
Empirical: a(n) = A002378(n-1) for n > 2. - R. J. Mathar, Sep 06 2010
Empirical: a(n) = n^2 - n for n > 2. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 5. G.f.: x^2*(1 + 3*x - 3*x^2 + x^3)/(1-x)^3. - Colin Barker, Mar 18 2012
a(n) = n*(n-1) for n > 2. - Luca Ferrigno, Jul 13 2023
MATHEMATICA
b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i == 0, 0, Sum[b[n - j, i - 1, k], {j, 0, Min[n, k]}]]];
a[n_] := b[n, n, n - 1] - b[n, n, n - 2];
a[n_] := If[n == 2, 1, n*(n - 1)] (* Luca Ferrigno, Jul 13 2023 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, formula from Robert Gerbicz in the Sequence Fans Mailing List, Aug 24 2010
STATUS
approved