cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A180291 Number of arrangements of n indistinguishable balls in n boxes with the maximum number of balls in any box equal to n-1.

Original entry on oeis.org

1, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450, 2550
Offset: 2

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Author

R. H. Hardin, formula from Robert Gerbicz in the Sequence Fans Mailing List, Aug 24 2010

Keywords

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
a(n) is the maximum number of squares that can be marked on an n X n bingo board without getting bingo (i.e., without completing any row, column, or diagonal). - Rasmus Lindahl, Feb 23 2025

Crossrefs

(n-1)-th entry in rows of A180281.
Essentially the same as A060798 and A002378.

Programs

  • 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];
    Table[a[n], {n, 2, 51}] (* Jean-François Alcover, Aug 28 2022, after Alois P. Heinz in A180281 *)
    a[n_] := If[n == 2, 1, n*(n - 1)] (* Luca Ferrigno, Jul 13 2023 *)

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