A059022
Triangle of Stirling numbers of order 3.
Original entry on oeis.org
1, 1, 1, 1, 10, 1, 35, 1, 91, 1, 210, 280, 1, 456, 2100, 1, 957, 10395, 1, 1969, 42735, 15400, 1, 4004, 158301, 200200, 1, 8086, 549549, 1611610, 1, 16263, 1827826, 10335325, 1401400, 1, 32631, 5903898, 57962905, 28028000, 1, 65382, 18682014, 297797500
Offset: 3
Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
There are 10 ways of partitioning a set N of cardinality 6 into 2 blocks each of cardinality at least 3, so S_3(6,2) = 10.
From _Wesley Ivan Hurt_, Feb 24 2022: (Start)
Triangle starts:
1;
1;
1;
1, 10;
1, 35;
1, 91;
1, 210, 280;
1, 456, 2100;
1, 957, 10395;
1, 1969, 42735, 15400;
...
(End)
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
- J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.
- Gilles Bonnet and Anna Gusakova, Concentration inequalities for Poisson U-statistics, arXiv:2404.16756 [math.PR], 2024. See p. 17.
- Antal E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.
- Gergő Nemes, On the Coefficients of the Asymptotic Expansion of n!, J. Int. Seq. 13 (2010), 10.6.6.
-
b:= proc(n) option remember; `if`(n=0, 1, add(
expand(x*b(n-j))*binomial(n-1, j-1), j=3..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
seq(T(n), n=3..20); # Alois P. Heinz, Feb 21 2022
# alternative
A059022 := proc(n, k)
option remember;
if n<3 then
0;
elif n < 6 and k=1 then
1 ;
else
k*procname(n-1, k)+binomial(n-1, 2)*procname(n-3, k-1) ;
end if;
end proc: # R. J. Mathar, Apr 15 2022
-
S3[3, 1] = S3[4, 1] = S3[5, 1] = 1; S3[n_, k_] /; 1 <= k <= Floor[n/3] := S3[n, k] = k*S3[n-1, k] + Binomial[n-1, 2]*S3[n-3, k-1]; S3[, ] = 0; Flatten[ Table[ S3[n, k], {n, 3, 20}, {k, 1, Floor[n/3]}]] (* Jean-François Alcover, Feb 21 2012 *)
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.
Original entry on oeis.org
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
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
-
[(2^n-2-2*n-2*Binomial(n,2))/2: n in [6..50]];
-
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 *)
A261724
a(n) is the number of ways of putting n labeled balls into 4 indistinguishable boxes such that each box contains at least 3 balls.
Original entry on oeis.org
15400, 200200, 1611610, 10335325, 57962905, 297797500, 1439774336, 6662393738, 29844199346, 130445781284, 559533979466, 2365296391535, 9885290914059, 40944327590760, 168389163468240, 688631376550260, 2803570746766140, 11373212443859760, 46006062639998890
Offset: 12
- Colin Barker, Table of n, a(n) for n = 12..1000
- I. Mezo, Periodicity of the last digits of some combinatorial sequences, arXiv preprint arXiv:1308.1637 [math.CO], 2013 (third formula on page 16 is incorrect).
- Index entries for linear recurrences with constant coefficients, signature (30, -415, 3514, -20386, 85924, -272198, 661180, -1244717, 1822478, -2068955, 1802474, -1181760, 563888, -184752, 37152, -3456).
-
[(1/12)*(-3^(n-2)*(n^2+5*n+18)+(1/64)*(2^(2*n+5)+3*2^n*(n^4+2*n^3+19*n^2+42*n+64)-16*(n^6-9*n^5+43*n^4-91*n^3+112*n^2-32*n+8))): n in [12..40]];
-
Table[(1/12) (-(3^(n - 2) (n^2 + 5 n + 18)) + (1/64) (2^(2 n + 5) + 3 2^n (n^4 + 2 n^3 + 19 n^2 + 42 n + 64) - 16 (n^6 - 9 n^5 + 43 n^4 - 91 n^3 + 112 n^2 - 32 n + 8))), {n, 12, 40}]
-
Vec(x^12*(15400 -261800*x +1996610*x^2 -9045575*x^3 +27162905*x^4 -57079715*x^5 +86268721*x^6 -94696602*x^7 +75062256*x^8 -41952000*x^9 +15705360*x^10 -3538080*x^11 +362880*x^12) / ((1 -x)^7*(1 -2*x)^5*(1 -3*x)^3*(1 -4*x)) + O(x^30)) \\ Colin Barker, May 24 2016
A352474
a(n) is the number of different ways to partition the set of vertices of a convex n-gon into 3 intersecting polygons.
Original entry on oeis.org
268, 2055, 10285, 42515, 157911, 548912, 1826846, 5902458, 18679974, 58255005, 179762211, 550473301, 1676299353, 5083919214, 15372833564, 46383749572, 139730014800, 420448279875, 1264071072745, 3798101946855, 11406989330923, 34248214094780
Offset: 9
The set of vertices of a convex 11-gon can be partitioned into 3 polygons in 10395 different ways:
- as 2 triangles and 1 pentagon ((1/2!)*C(11,3)*C(8,3)*C(5,5) = 4620 different ways) or
- as 1 triangle and 2 quadrilaterals ((1/2!)*C(11,3)*C(8,4)*C(4,4) = 5775 different ways).
Subtracting the A350116(11-8) = 110 nonintersecting partitions leaves a(11)=10285.
-
b(n) = if (n==8, 0, 3*b(n-1)+binomial(n-1,2)*(2^(n-4)+2-n-binomial(n-3,2)));
a(n) = b(n) - n*(n-1)*(n-7)*(n-8)/12; \\ Michel Marcus, Mar 19 2022
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