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A008275
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Triangle read by rows of Stirling numbers of first kind, s(n,k), n >= 1, 1 <= k <= n.
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259
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1, -1, 1, 2, -3, 1, -6, 11, -6, 1, 24, -50, 35, -10, 1, -120, 274, -225, 85, -15, 1, 720, -1764, 1624, -735, 175, -21, 1, -5040, 13068, -13132, 6769, -1960, 322, -28, 1, 40320, -109584, 118124, -67284, 22449, -4536, 546, -36, 1, -362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45, 1
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OFFSET
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1,4
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COMMENTS
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The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.
The unsigned numbers (read from right to left) also give the number of permutations of 1..n with complexity k, where the complexity of a permutation is defined to be the sum of the lengths of the cycles minus the number of cycles. In other words, the complexity equals the sum of (length of cycle)-1 over all cycles. For n=5, the numbers of permutations with complexity 0,1,2,3,4 are 1, 10, 35, 50, 24. - N. J. A. Sloane, Feb 08 2019
The unsigned numbers are also the number of permutations of 1..n with k left to right maxima (see Khovanova and Lewis, Smith).
With P(n) = the number of integer partitions of n, T(i,n) = the number of parts of the i-th partition of n, D(i,n) = the number of different parts of the i-th partition of n, p(j,i,n) = the j-th part of the i-th partition of n, m(j,i,n) = multiplicity of the j-th part of the i-th partition of n, Sum_[T(i,n)=k]_{i=1}^{P(n)} = sum running from i=1 to i=p(n) but taking only partitions with T(i,n)=k parts into account, Product_{j=1..T(i,n)} = product running from j=1 to j=T(i,n), Product_{j=1..D(i,n)} = product running from j=1 to j=D(i,n) one has S1(n,k) = Sum_[T(i,n)=k]_{i=1}^{P(n)} (n!/Product_{j=1..T(i,n)} p(j,i,n))* (1/Product_{j=1..D(i,n)} m(j,i,n)!). For example, S1(6,3) = 225 because n=6 has the following partitions with k=3 parts: (114), (123), (222). Their complexions are: (114): (6!/1*1*4)*(1/2!*1!) = 90, (123): (6!/1*2*3)*(1/1!*1!*1!) = 120, (222): (6!/2*2*2)*(1/3!) = 15. The sum of the complexions is 90+120+15 = 225 = S1(6,3). - Thomas Wieder, Aug 04 2005
|s(n,k)| enumerates unordered n-vertex forests composed of k increasing non-plane (unordered) trees. Proof from the e.g.f. of the first column and the F. Bergeron et al. reference, especially Table 1, last row (non-plane "recursive"), given in A049029. - Wolfdieter Lang, Oct 12 2007
|s(n,k)| enumerates unordered increasing n-vertex k-forests composed of k unary trees (out-degree r from {0,1}) whose vertices of depth (distance from the root) j >= 0 come in j+1 colors (j=0 for the k roots). - Wolfdieter Lang, Oct 12 2007, Feb 22 2008
A refinement of the unsigned array is A036039. For an association to forests of "naturally grown" rooted non-planar trees, dispositions of flags on flagpoles, and colorings of the vertices of the complete graphs K_n, see A130534. - Tom Copeland, Mar 30 and Apr 05 2014
The Stirling numbers of the first kind were related to the falling factorial and the convolved, or generalized, Bernoulli numbers B_n by Norlund in 1924 through Sum_{k=1..n+1} T(n+1,k) * x^(k-1) = (x-1)!/(x-1-n)! = (x + B.(0))^n = B_n(x), umbrally evaluated with (B.(0))^k = B_k(0) and the associated Appell polynomial B_n(x) defined by the e.g.f. (t/(exp(t) - 1))^(n+1) * exp(x*t) = exp(B.(x)t). - Tom Copeland, Sep 29 2015
With x = e^z, D_x = d/dx, D_z = d/dz, and p_n(x) the row polynomials of this entry, x^n (D_x)^n = p_n(D_z) = (D_z)! / (D_z - n)! = (xD_x)! / (xD_x - n)!. - Tom Copeland, Nov 27 2015
From the operator relation z + Psi(1) + sum_{n > 0} (-1)^n (-1/n) binomial(D,n) = z + Psi(1+D) with D = d/dz and Psi the digamma function, Zeta(n+1) = Sum_{k > n-1} (1/k) |S(k,n)| / k! for n > 0 and Zeta the Riemann zeta function. - Tom Copeland, Aug 12 2016
Let X_1,...,X_n be i.i.d. random variables with exponential distribution having mean = 1. Let Y = max{X_1,...,X_n}. Then (-1)^n*n!/( Sum_{k=1..n+1} a(n+1,k) t^(k-1) ) is the moment generating function of Y. The expectation of Y is the n-th harmonic number. - Geoffrey Critzer, Dec 25 2018
In the Ewens sampling theory describing the multivariate probability distribution of the sizes of the allelic classes in a sample of size n under the Infinite Alleles Model, |s(n,k)| gives the coefficient in the formula for the probability that a sample of n alleles has exactly k distinct types. - Noah A Rosenberg, Feb 10 2019
Named after the Scottish mathematician James Stirling (1692-1770). - Amiram Eldar, Jun 11 2021
The first few row polynomials along with a recursion formula are found in a manuscript by Newton written in 1664 or 1665 (p. 169 of Turnbull) giving a geometric presentation of the binomial theorem for rational powers. - Tom Copeland, Dec 10 2022
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 93ff.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8. English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 32.
George Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
Louis Comtet, Advanced Combinatorics, Reidel, 1974; Chapter V, also p. 310.
John H. Conway and Richard K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 93.
Florence Nightingale David, Maurice George Kendall and David Elliot Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
Saber N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
Herman H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 245. In the second edition, see Chapter 6, especially p. 259.
M. Miyata and J. W. Son, On the complexity of permutations and the metric space of bijections, Tensor, 60 (1998), No. 1, 109-116 (MR1768839).
Isaac Newton, A Method whereby to find ye areas of Those Lines wch can be squared, pp. 168-171 of Turnbull below.
John Riordan, An Introduction to Combinatorial Analysis, p. 48.
Robert Sedgewick and Phillipe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
H. Turnbull (editor), The Correspondence of Isaac Newton Vol. II 1676-1687, Cambridge Univ. Press, 1960.
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Raymond Scurr and Gloria Olive, Stirling numbers revisited, Discrete Math., Vol. 189, No. 1-3 (1998), pp. 209-219. MR1637761 (99d:11019).
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FORMULA
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s(n, k) = s(n-1, k-1) - (n-1)*s(n-1, k), n, k >= 1; s(n, 0) = s(0, k) = 0; s(0, 0) = 1.
The unsigned numbers a(n, k)=|s(n, k)| satisfy a(n, k) = a(n-1, k-1) + (n-1)*a(n-1, k), n, k >= 1; a(n, 0) = a(0, k) = 0; a(0, 0) = 1.
E.g.f.: for m-th column (unsigned): ((-log(1-x))^m)/m!.
s(n, k) = T(n-1, k-1), n>1 and k>1, where T(n, k) is the triangle [ -1, -1, -2, -2, -3, -3, -4, -4, -5, -5, -6, -6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, ...] and DELTA is Deléham's operator defined in A084938. The unsigned numbers are also |s(n, k)| = T(n-1, k-1), for n>0 and k>0, where T(n, k) = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...].
Sum_{i=0..n} (-1)^(n-i) * StirlingS1(n, i) * binomial(i, k) = (-1)^(n-k) * StirlingS1(n+1, k+1). - Carlo Wood (carlo(AT)alinoe.com), Feb 13 2007
G.f.: S(n) = Product_{j=1..n} (x-j) (i.e., (x-1)*(x-2)*(x-3) = x^3 - 6*x^2 + 11*x - 6). - Jon Perry, Nov 14 2005
a(n,k) = s(n,k) = lim_{y -> 0} Sum_{j=0..k} (-1)^j*binomial(k,j)*((-j*y)!/(-j*y-n)!)*y^(-k)/k! = Sum_{j=0..k} (-1)^(n-j)*binomial(k,j)*((j*y - 1 + n)!/(j*y-1)!)*y^(-k)/k!. - Tom Copeland, Aug 28 2015
Let x_(0) := 1 (empty product), and for n >= 1:
x_(n) := Product_{k=0..n-1} (x-k), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), and also x_(-n) := 1 / [Product_{k=0..n-1} (x+k)].
Then, for n >= 1: x_(n) = Sum_{k=1..n} T(n,k) * x^k, 1 / [x_(-n)] = Sum_{k=1..n} |T(n,k)| * x^k, x^n = Sum_{k=1..n} A008277(n,k) * x_(k), where A008277(n,k) are Stirling numbers of the second kind.
The row sums (of either signed or absolute values) are Sum_{k=1..n} T(n,k) = 0^(n-1), Sum_{k=1..n} |T(n,k)| = T(n+1,1) = n!. (End)
Orthogonal relation: Sum_{i=0..n} i^p*Sum_{j=k..n} (-1)^(i+j) * binomial(j,i) * Stirling1(j,k)/j! = delta(p,k), i,k,p <= n, n >= 1. - Leonid Bedratyuk, Jul 27 2020
Sum_{k=1..n} (-1)^k * k * T(n, k) = -T(n+1, 2).
Sum_{k=1..n} k * T(n, k) = (-1)^n * (n-2)! = T(n-1, 1) for n>=2. (End)
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EXAMPLE
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|s(3,2)| = 3 for the three unordered 2-forest with 3 vertices and two increasing (nonplane) trees: ((1),(2,3)), ((2),(1,3)), ((3),(1,2)).
Triangle begins:
1
-1, 1
2, -3, 1
-6, 11, -6, 1
24, -50, 35, -10, 1
-120, 274, -225, 85, -15, 1
720, -1764, 1624, -735, 175, -21, 1
-5040, 13068, -13132, 6769, -1960, 322, -28, 1
40320, -109584, 118124, -67284, 22449, -4536, 546, -36, 1
Another version of the same triangle, from Joerg Arndt, Oct 05 2009: (Start)
s(n,k) := number of permutations of n elements with exactly k cycles ("Stirling cycle numbers")
n| total m=1 2 3 4 5 6 7 8 9
-+-----------------------------------------------------
1| 1 1
2| 2 1 1
3| 6 2 3 1
4| 24 6 11 6 1
5| 120 24 50 35 10 1
6| 720 120 274 225 85 15 1
7| 5040 720 1764 1624 735 175 21 1
8| 40320 5040 13068 13132 6769 1960 322 28 1
9| 362880 40320 109584 118124 67284 22449 4536 546 36 1
(End)
|s(4,2)| = 11 for the eleven unordered 2-forest with 4 vertices, composed of two increasing (nonplane) trees: ((1),((23)(24))), ((2),((13)(14)), ((3),((12)(14)), ((4),((12)(13)); ((1),(2,3,4)),((2),(1,2,3)), ((3), (1,2,4)), ((4),(1,2,3)); ((1,2),(3,4)), ((1,3),(2,4)), ((1,4),(2, 3)). - Wolfdieter Lang, Feb 22 2008
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MAPLE
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with (combinat):seq(seq(stirling1(n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 03 2007
for i from 0 to 9 do seq(stirling1(i, j), j = 1 .. i) od; # Zerinvary Lajos, Nov 29 2007
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MATHEMATICA
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PROG
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(PARI) T(n, k)=if(n<1, 0, n!*polcoeff(binomial(x, n), k))
(PARI) T(n, k)=if(n<1, 0, n!*polcoeff(polcoeff((1+x+x*O(x^n))^y, n), k))
(PARI) vecstirling(n)=Vec(factorback(vector(n-1, i, 1-i*'x))) /* (A function that returns all the s(n, k) as a vector) */ \\ Bill Allombert (Bill.Allombert(AT)math.u-bordeaux1.fr), Mar 16 2009
(Maxima) create_list(stirling1(n+1, k+1), n, 0, 30, k, 0, n); /* Emanuele Munarini, Jun 01 2012 */
(Haskell)
a008275 n k = a008275_tabl !! (n-1) !! (k-1)
a008275_row n = a008275_tabl !! (n-1)
a008275_tabl = map tail $ tail a048994_tabl
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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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