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A059023
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Triangle of Stirling numbers of order 4.
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6
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1, 1, 1, 1, 1, 35, 1, 126, 1, 336, 1, 792, 1, 1749, 5775, 1, 3718, 45045, 1, 7722, 231231, 1, 15808, 981981, 1, 32071, 3741738, 2627625, 1, 64702, 13307294, 35735700, 1, 130084, 45172842, 300179880, 1, 260984, 148417854, 2002016016, 1, 522937, 476330361
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OFFSET
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4,6
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COMMENTS
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The number of partitions of the set N, |N|=n, into k blocks, all of cardinality greater than or equal to 4. This is the 4-associated Stirling number of the second kind.
This is entered as a triangular array. The entries S_4(n,k) are zero for 4k>n, so these values are omitted. Initial entry in sequence is S_4(4,1).
Rows are of lengths 1,1,1,1,2,2,2,2,3,3,3,3,...
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REFERENCES
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L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.
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LINKS
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FORMULA
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S_r(n+1, k) = k*S_r(n, k) + binomial(n, r-1)*S_r(n-r+1, k-1); for this sequence, r=4.
G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t-sum(t^i/i!, i=0..r-1))).
T(n,k) = Sum_{j=0..min(n/3,k)} (-1)^j*n!/(6^j*j!*(n-3j)!)*S_3(n-3j,k-j), where S_3 are the 3-associated Stirling numbers of the second kind A059022. - Fabián Pereyra, Feb 21 2022
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EXAMPLE
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There are 35 ways of partitioning a set N of cardinality 8 into 2 blocks each of cardinality at least 4, so S_4(8,2) = 35.
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MAPLE
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b:= proc(n) option remember; `if`(n=0, 1, add(
expand(x*b(n-j))*binomial(n-1, j-1), j=4..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
# alternative
option remember;
if n<4 then
0;
elif n < 8 and k=1 then
1 ;
else
k*procname(n-1, k)+binomial(n-1, 3)*procname(n-4, k-1) ;
end if;
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MATHEMATICA
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s4[n_, k_] := k*s4[n-1, k] + Binomial[n-1, 3]*s4[n-4, k-1]; s4[n_, k_] /; 4 k > n = 0; s4[_, k_ /; k <= 0] = 0; s4[0, 0] = 1;
Flatten[Table[s4[n, k], {n, 4, 20}, {k, 1, Floor[n/4]}]][[1 ;; 42]] (* Jean-François Alcover, Jun 16 2011 *)
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CROSSREFS
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KEYWORD
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nonn,tabf,nice
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AUTHOR
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Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
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STATUS
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approved
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