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A254881
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Triangle read by rows, T(n,k) = sum(j=0..k-1, S(n+1,j+1)*S(n,k-j)) where S denotes the Stirling cycle numbers A132393, T(0,0)=1, n>=0, 0<=k<=2n.
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3
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1, 0, 1, 1, 0, 2, 5, 4, 1, 0, 12, 40, 51, 31, 9, 1, 0, 144, 564, 904, 769, 376, 106, 16, 1, 0, 2880, 12576, 23300, 24080, 15345, 6273, 1650, 270, 25, 1, 0, 86400, 408960, 840216, 991276, 748530, 381065, 133848, 32523, 5370, 575, 36, 1, 0, 3628800, 18299520
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OFFSET
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0,6
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COMMENTS
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These are also the coefficients of the polynomials interpolating the sequence k -> n!*((n+k)!/k!)*binomial(n+k-1,k-1) (for fixed n>=0). Divided by n! these polynomials generate the rows of Lah numbers L(n+k, k) = ((n+k)!/k!)* binomial(n+k-1,k-1).
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LINKS
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FORMULA
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EXAMPLE
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[1]
[0, 1, 1]
[0, 2, 5, 4, 1]
[0, 12, 40, 51, 31, 9, 1]
[0, 144, 564, 904, 769, 376, 106, 16, 1]
[0, 2880, 12576, 23300, 24080, 15345, 6273, 1650, 270, 25, 1]
For example in the case n=3 the polynomial (k^6+9*k^5+31*k^4+51*k^3+40*k^2+12*k)/3! generates the Lah numbers 0, 24, 240, 1200, 4200, 11760, 28224, ... (A253285).
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MAPLE
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# This is a special case of the recurrence given in A246117.
t := proc(n, k) option remember; if n=0 and k=0 then 1 elif
k <= 0 or k>n then 0 else iquo(n, 2)*t(n-1, k)+t(n-1, k-1) fi end:
seq(print(seq(A254881(n, k), k=0..2*n)), n=0..5);
# Illustrating the comment:
restart: with(PolynomialTools): with(CurveFitting): for N from 0 to 5 do
CoefficientList(PolynomialInterpolation([seq([k, N!*((N+k)!/k!)*binomial(N+k-1, k-1)], k=0..2*N)], n), n) od;
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MATHEMATICA
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Flatten[{1, Table[Table[Sum[Abs[StirlingS1[n+1, j+1]] * Abs[StirlingS1[n, k-j]], {j, 0, k-1}], {k, 0, 2*n}], {n, 1, 10}]}] (* Vaclav Kotesovec, Feb 10 2015 *)
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PROG
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(Sage)
def T(n, k):
if n == 0: return 1
return sum(stirling_number1(n+1, j+1)*stirling_number1(n, k-j) for j in range(k))
for n in range (6): [T(n, k) for k in (0..2*n)]
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CROSSREFS
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The sequences A000012, A002378, A083374, A253285 are the Lah number rows generated by the polynomials divided by n! for n=0, 1, 2, 3 respectivly.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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