A272352 a(n) is the number of ways of putting n labeled balls into 2 indistinguishable boxes such that each box contains at least 3 balls.
10, 35, 91, 210, 456, 957, 1969, 4004, 8086, 16263, 32631, 65382, 130900, 261953, 524077, 1048344, 2096898, 4194027, 8388307, 16776890, 33554080, 67108485, 134217321, 268435020, 536870446, 1073741327, 2147483119, 4294966734, 8589933996, 17179868553
Offset: 6
Examples
For n=6, label the balls A, B, C, D, E, and F. Then each box must contain exactly 3 balls, and the 10 ways are ABC/DEF, ABD/CEF, ABE/CDF, ABF/CDE, ACD/BEF, ACE/BDF, ACF/BDE, ADE/BCF, ADF/BCE, AEF/BCD. - _Michael B. Porter_, Jul 01 2016
Links
- Vincenzo Librandi, Table of n, a(n) for n = 6..1000
- I. Mezo, Periodicity of the last digits of some combinatorial sequences, arXiv preprint arXiv:1308.1637 [math.CO], 2013 (page 16).
- Index entries for linear recurrences with constant coefficients, signature (5,-9,7,-2).
Crossrefs
Programs
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Magma
[(2^n-2-2*n-2*Binomial(n,2))/2: n in [6..50]];
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Mathematica
Table[1/2 (2^n - 2 - 2 n - 2 Binomial[n, 2]), {n, 6, 40}] LinearRecurrence[{5,-9,7,-2},{10,35,91,210},30] (* Harvey P. Dale, Mar 29 2018 *)
Formula
G.f.: x^6*(10 - 15*x + 6*x^2)/((1 - x)^3*(1 - 2*x)).
a(n) = (2^n - 2 - 2*n - 2*binomial(n, 2))/2.
a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4), for n > 3.
E.g.f.: (2 - 2*exp(x) + 2*x + x^2)^2/8. - Stefano Spezia, Jul 25 2021