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A060475
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Triangular array formed from successive differences of factorial numbers, then with factorials removed.
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5
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1, 1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 3, 7, 11, 9, 1, 4, 13, 32, 53, 44, 1, 5, 21, 71, 181, 309, 265, 1, 6, 31, 134, 465, 1214, 2119, 1854, 1, 7, 43, 227, 1001, 3539, 9403, 16687, 14833, 1, 8, 57, 356, 1909, 8544, 30637, 82508, 148329, 133496, 1, 9, 73, 527, 3333, 18089, 81901, 296967, 808393, 1468457, 1334961
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OFFSET
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0,8
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COMMENTS
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T(n,k) is also the number of partial bijections (of an n-element set) with a fixed domain of size k and without fixed points. Equivalently, T(n,k) is the number of partial derangements with a fixed domain of size k in the symmetric inverse semigroup (monoid), I sub n. - Abdullahi Umar, Sep 14 2008
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LINKS
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L. Takacs, The Problem of Coincidences, Archive for History of Exact Sciences, Volume 21, No. 3, Sept. 1980. pp 229-244, paragraph 10 (Catalan).
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FORMULA
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T(n,k) = A047920(n,k)/(n-k)! = (n-1)*T(n-1,k-1) + (k-1)*T(n-2,k-2) = (n-k+1)*T(n, k-1) - T(n-1,k-1).
T(n,k) = k! * Sum_{j=0..k} C(n-j,k-j)*(-1)^j/j!.
C(n,k)*T(n,k) = A144089(n, k). (End)
E.g.f. as a square array: A(x,y) = exp(-x)/(1 - x - y) = (1 + y + y^2 + y^3 + ...) + (y + 2*y^2 + 3*y^3 + 4*y^4 + ...)*x + (1 + 3*y + 7*y^2 + 13*y^3 + ...)*x^2/2! + (2 + 11*y + 32*y^2 + 71*y^3 + ...)*x^3/3! + .... Observe that (1 - y)*A(x*(1 - y),y) = exp(x*(y - 1))/(1 - x) is the e.g.f. for A008290. - Peter Bala, Sep 25 2013
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EXAMPLE
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Triangle begins
1,
1, 0,
1, 1, 1,
1, 2, 3, 2,
1, 3, 7, 11, 9,
1, 4, 13, 32, 53, 44,
...
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MAPLE
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A060475 := proc(n, k): k! * add(binomial(n-j, k-j)*(-1)^j/j!, j=0..k) end:
T := (n, k) -> KummerU(-k, -n, -1):
seq(seq(simplify(T(n, k)), k = 0..n), n = 0..10); # Peter Luschny, Jul 07 2022
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MATHEMATICA
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t[n_, k_] := k!*Sum[Binomial[n - j, k - j]*(-1)^j/j!, {j, 0, k}]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v, Aug 08 2011 *)
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PROG
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(PARI) {T(n, k) = k!*sum(j=0, k, (-1)^j*binomial(n-j, k-j)/j!)};
for(n=0, 10, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Mar 04 2019
(Magma) [[Factorial(k)*(&+[(-1)^j*Binomial(n-j, k-j)/Factorial(j): j in [0..k]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 04 2019
(Sage) [[factorial(k)*sum((-1)^j*binomial(n-j, k-j)/factorial(j) for j in (0..k)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 04 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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