|
|
A272982
|
|
a(n) is the number of ways of putting n labeled balls into 3 indistinguishable boxes such that each box contains at least 3 balls.
|
|
5
|
|
|
280, 2100, 10395, 42735, 158301, 549549, 1827826, 5903898, 18682014, 58257810, 179765973, 550478241, 1676305723, 5083927299, 15372843684, 46383762084, 139730030100, 420448298400, 1264071094975, 3798101973315, 11406989362185, 34248214131465, 102803026929030, 308533903071390
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
9,1
|
|
LINKS
|
|
|
FORMULA
|
G.f.: x^9*(280 - 1820*x + 4795*x^2 - 6615*x^3 + 5106*x^4 - 2100*x^5 + 360*x^6)/((1 - 3*x)*(1 - 2*x)^3*(1 - x)^5).
a(n) = (1/3)*(1/16)*(6*n^4 - 12*n^3 - 3*2^n*n^2 + 42*n^2 - 9*2^n*n + 12*n + 8*3^n - 3*2^(n+3) + 24).
a(n) = 3*a(n-1) + C(n-1,2)*(2^(n-4) + 2 - n - C(n-3, 2)), a(n)=0, n < 9. - Vladimir Kruchinin, Oct 04 2018
|
|
EXAMPLE
|
For n=9, label the balls A through I. The box containing ball A can contain 8*7/2 = 28 combinations of other balls. There are 6 balls for the other two boxes, so there are A272352(6) = 10 combinations for those two boxes. Thus, a(9) = 28*10 = 280. - Michael B. Porter, Jul 01 2016
|
|
MATHEMATICA
|
Table[(1/3) (1/16) (6 n^4 - 12 n^3 - 3 2^n n^2 + 42 n^2 - 9 2^n n + 12 n + 8 3^n - 3 2^(n + 3) + 24), {n, 9, 40}]
CoefficientList[Series[(280 - 1820*x + 4795*x^2 - 6615*x^3 + 5106*x^4 - 2100*x^5 + 360*x^6)/((1 - 3*x)*(1 - 2*x)^3*(1 - x)^5), {x, 0, 40}], x] (* Stefano Spezia, Oct 04 2018 *)
|
|
PROG
|
(Magma) [(1/3)*(1/16)*(6*n^4-12*n^3-3*2^n*n^2+42*n^2-9*2^n*n+12*n+8*3^n-3*2^(n+3)+24): n in [9..40]];
(PARI) Vec(x^9*(280 - 1820*x + 4795*x^2 - 6615*x^3 + 5106*x^4 - 2100*x^5 + 360*x^6)/((1 - 3*x)*(1 - 2*x)^3*(1 - x)^5) + O(x^40)) \\ Stefano Spezia, Oct 04 2018
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
Data, formulas and programs corrected for erroneous formula in Mezo's paper by Bruno Berselli, May 21 2016
|
|
STATUS
|
approved
|
|
|
|