cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A112491 Sixth column of triangle A112486 used for e.g.f.s of |Stirling1|=|A008275| diagonals.

Original entry on oeis.org

945, 45045, 1406790, 37147110, 909508600, 21577556000, 508072685120, 12041790080320, 289769457978000, 7118294418969552, 179128660433168160, 4627741977903371040, 122904955276454869056, 3358132932795227294400
Offset: 5

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Examples

			1406790 = a(7) = 12*45045 + 11*78750.
		

Formula

a(n)= A112486(n, 5), n>=5. a(0), ..., a(4) = 0.
a(n)= (n+5)*a(n-1) + (n+4)* A112490(n-1), n>=5, a(4):=0.

A061115 Concatenation of numbers in n-th row of triangle of unsigned Stirling numbers of first kind (A008275).

Original entry on oeis.org

1, 11, 231, 61161, 245035101, 12027422585151, 72017641624735175211, 5040130681313267691960322281, 4032010958411812467284224494536546361, 36288010265761172700723680269325632739450870451
Offset: 1

Views

Author

Amarnath Murthy, Apr 21 2001

Keywords

References

  • Amarnath Murthy, Smarandache Star derived sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Programs

  • Maple
    with(combinat, stirling1): for n from 1 to 15 do for k from 1 to n do printf(`%d`, abs(stirling1(n,k))) od: printf(`,`): od:
  • Mathematica
    Table[FromDigits[Flatten[IntegerDigits/@Abs[Table[StirlingS1[n,m],{m,n}]]]],{n,10}] (* Harvey P. Dale, Feb 09 2016 *)

Extensions

More terms from James Sellers, Apr 23 2001

A322512 Triangle read by rows of the 2-adic valuation (A007814) of Stirling numbers of first kind (A008275).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 3, 1, 0, 1, 0, 3, 1, 0, 0, 0, 0, 4, 2, 3, 0, 0, 0, 0, 4, 2, 2, 0, 3, 1, 2, 0, 7, 4, 2, 2, 0, 3, 1, 2, 0, 7, 4, 2, 5, 0, 0, 1, 1, 0, 0, 8, 5, 3, 2, 1, 0, 0, 1, 3, 0, 0, 8, 5, 3, 2, 1, 0, 1, 0, 1, 0, 1, 0, 10, 7, 7, 3, 2, 1, 0, 1, 0, 1, 0, 1, 0
Offset: 1

Views

Author

Michel Marcus, Dec 13 2018

Keywords

Examples

			Triangle begins:
  0,
  0, 0,
  1, 0, 0,
  1, 0, 1, 0,
  3, 1, 0, 1, 0,
  3, 1, 0, 0, 0, 0,
  4, 2, 3, 0, 0, 0, 0,
  4, 2, 2, 0, 3, 1, 2, 0,
  ...
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := IntegerExponent[StirlingS1[n, k], 2]; Table[T[n, k], {n, 1, 20}, {k, 1, n}] // Flatten (* Amiram Eldar, Dec 13 2018 *)
  • PARI
    T(n,k) = valuation(stirling(n, k, 1), 2);
    row(n) = vector(n, k, T(n,k));
    tabl(nn) = vector(nn, k, row(k));(PARI) T(n,k) = valuation(stirling(n, k, 1), 2);

Formula

T(n,k) = A007814(A008275(n,k)).

A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0

Views

Author

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

Keywords

Comments

A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
+1
-1 +1
+1 -2 +1
-1 +3 -3 +1
+1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
From Milan Janjic, May 07 2008: (Start)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
See A193242. - Alexander R. Povolotsky, Feb 05 2013
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k, k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and = return to x axis} binomial(3,0)=1 (Uudududd); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
From Daniel Forgues, Mar 12 2015: (Start)
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
{0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111
(End)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the k-th row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024

Examples

			Triangle T(n,k) begins:
   n\k 0   1   2   3   4   5   6   7   8   9  10  11 ...
   0   1
   1   1   1
   2   1   2   1
   3   1   3   3   1
   4   1   4   6   4   1
   5   1   5  10  10   5   1
   6   1   6  15  20  15   6   1
   7   1   7  21  35  35  21   7   1
   8   1   8  28  56  70  56  28   8   1
   9   1   9  36  84 126 126  84  36   9   1
  10   1  10  45 120 210 252 210 120  45  10   1
  11   1  11  55 165 330 462 462 330 165  55  11   1
  ...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - _Dennis P. Walsh_, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - _Dennis P. Walsh_, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - _Gregory L. Simay_, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - _Wolfdieter Lang_, Nov 12 2018
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 68-74.
  • Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
  • A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
  • William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, pp. 140-141.
  • David Hök, Parvisa mönster i permutationer [Swedish], 2007.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
  • Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
  • Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 271-275.
  • A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
  • John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
  • Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, pages 43-52.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 13, 30-33.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
  • Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Axiom
    -- (start)
    )set expose add constructor OutputForm
    pascal(0,n) == 1
    pascal(n,n) == 1
    pascal(i,j | 0 < i and i < j) == pascal(i-1,j-1) + pascal(i,j-1)
    pascalRow(n) == [pascal(i,n) for i in 0..n]
    displayRow(n) == output center blankSeparate pascalRow(n)
    for i in 0..20 repeat displayRow i -- (end)
    
  • GAP
    Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018
  • Haskell
    a007318 n k = a007318_tabl !! n !! k
    a007318_row n = a007318_tabl !! n
    a007318_list = concat a007318_tabl
    a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
    -- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
    -- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010
    
  • Magma
    /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
    
  • Maple
    A007318 := (n,k)->binomial(n,k);
  • Mathematica
    Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
    Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)
  • Maxima
    create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011
    
  • Python
    # See Hobson link. Further programs:
    from math import prod,factorial
    def C(n,k): return prod(range(n,n-k,-1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
    
  • Python
    from math import comb, isqrt
    def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024
    
  • Sage
    def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014
    

Formula

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n
C(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011: (Start)
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012: (Start)
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.: E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0 P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Extensions

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1
Offset: 1

Comments

Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
From Manfred Boergens, Apr 07 2025: (Start)
With regard to the preceding comment:
For disjoint collections of subsets see A256894.
For arbitrary collections of subsets see A163353.
For arbitrary collections of nonempty subsets see A055154. (End)
Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005
Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005
Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008
From Peter Bala, Oct 03 2008: (Start)
For tables of restricted Stirling numbers of the second kind see A143494 - A143496.
S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009
S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010
S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011
If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by: E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014
S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015
The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017
With offsets 0 for n and k this is the Sheffer product matrix A007318*A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017
Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019
Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020
k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022

Examples

			The triangle S2(n, k) begins:
\ k    1       2       3        4         5         6         7         8        9
n \   10      11      12       13        14        15       ...
----------------------------------------------------------------------------------
1  |   1
2  |   1       1
3  |   1       3       1
4  |   1       7       6        1
5  |   1      15      25       10         1
6  |   1      31      90       65        15         1
7  |   1      63     301      350       140        21         1
8  |   1     127     966     1701      1050       266        28         1
9  |   1     255    3025     7770      6951      2646       462        36        1
10 |   1     511    9330    34105     42525     22827      5880       750       45
       1
11 |   1    1023   28501   145750    246730    179487     63987     11880     1155
      55       1
12 |   1    2047   86526   611501   1379400   1323652    627396    159027    22275
    1705      66       1
13 |   1    4095  261625  2532530   7508501   9321312   5715424   1899612   359502
   39325    2431      78        1
14 |   1    8191  788970 10391745  40075035  63436373  49329280  20912320  5135130
  752752   66066    3367       91         1
15 |   1   16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478  106470     4550       105         1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - _Daniel Forgues_, Jan 16 2016
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 103ff.
  • B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • G. Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
  • C. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, 2002, Theorem 8.11, pp. 298-299.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • S.N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
  • H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • Frank Avery Haight, Handbook of the Poisson distribution, John Wiley, 1967. See pages 6,7.
  • A. D. Korshunov, Asymptotic behavior of Stirling numbers of the second kind. (Russian) Metody Diskret. Analiz. No. 39 (1983), 24-41.
  • E. Kuz'min and A. I. Shirshov: On the number e, pp. 111-119, eq.(6), in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999, p. 116, eq. (11).
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.
  • J. Stirling, The Differential Method, London, 1749; see p. 7.

Crossrefs

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
See also A331155.
Cf. A000110 (row sums), A102661 (partial row sums).

Programs

  • Haskell
    a008277 n k = a008277_tabl !! (n-1) !! (k-1)
    a008277_row n = a008277_tabl !! (n-1)
    a008277_tabl = map tail $ a048993_tabl  -- Reinhard Zumkeller, Mar 26 2012
    
  • J
    n ((] (1 % !)) * +/@((^~ * (] (1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi, Apr 06 2016
    
  • Magma
    [[StirlingSecond(n,k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019
  • Maple
    seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007
    stirling_2 := (n,k) -> (1/k!) * add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k);
  • Mathematica
    Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)
    BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    rows = 12;
    B = BellMatrix[1&, rows];
    Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
    a[n_, n_] := 1; a[n_, 1] := 1;
    a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
    Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *)
    With[{m = 11},
     Flatten@MapIndexed[Take[#, #2[[1]]] &,
       Transpose@
        Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
    Range[1, m]], {k, 1, m}]]] (* Oliver Seipel, Jun 12 2024 *)
  • Maxima
    create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
    
  • PARI
    for(n=1,22,for(k=1,n,print1(stirling(n,k,2),", "));print()); \\ Joerg Arndt, Apr 21 2013
    
  • PARI
    Stirling2(n,k)=sum(i=0,k,(-1)^i*binomial(k,i)*i^n)*(-1)^k/k!  \\ M. F. Hasler, Mar 06 2012
    
  • Sage
    stirling_number2 # Danny Rorabaugh, Oct 11 2015
    

Formula

S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.
S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.
A019538(n, k) = k! * S2(n, k).
A028248(n, k) = (k-1)! * S2(n, k).
For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005
Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007
Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011
Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
S2(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012
O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012
G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
From Tom Copeland, Apr 17 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).
With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and
C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes.
(End)
As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014
O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014
Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015
From Daniel Forgues, Jan 16 2016: (Start)
Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.
For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)
Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022
S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

A048993 Triangle of Stirling numbers of 2nd kind, S(n,k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 7, 6, 1, 0, 1, 15, 25, 10, 1, 0, 1, 31, 90, 65, 15, 1, 0, 1, 63, 301, 350, 140, 21, 1, 0, 1, 127, 966, 1701, 1050, 266, 28, 1, 0, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 0, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1
Offset: 0

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Also known as Stirling set numbers.
S(n,k) enumerates partitions of an n-set into k nonempty subsets.
The o.g.f. for the sequence of diagonal k (k=0 for the main diagonal) is G(k,x) = ((x^k)/(1-x)^(2*k+1))*Sum_{m=0..k-1} A008517(k,m+1)*x^m. A008517 is the second-order Eulerian triangle. - Wolfdieter Lang, Oct 14 2005
From Philippe Deléham, Nov 14 2007: (Start)
Sum_{k=0..n} S(n,k)*x^k = B_n(x), where B_n(x) = Bell polynomials.
The first few Bell polynomials are:
B_0(x) = 1;
B_1(x) = 0 + x;
B_2(x) = 0 + x + x^2;
B_3(x) = 0 + x + 3x^2 + x^3;
B_4(x) = 0 + x + 7x^2 + 6x^3 + x^4;
B_5(x) = 0 + x + 15x^2 + 25x^3 + 10x^4 + x^5;
B_6(x) = 0 + x + 31x^2 + 90x^3 + 65x^4 + 15x^5 + x^6;
(End)
This is the Sheffer triangle (1, exp(x) - 1), an exponential (binomial) convolution triangle. The a-sequence is given by A006232/A006233 (Cauchy sequence). The z-sequence is the zero sequence. See the link under A006232 for the definition and use of these sequences. The row sums give A000110 (Bell), and the alternating row sums give A000587 (see the Philippe Deléham formulas and crossreferences below). - Wolfdieter Lang, Oct 16 2014
Also the inverse Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265604. - Peter Luschny, Dec 31 2015
From Wolfdieter Lang, Feb 21 2017: (Start)
The transposed (trans) of this lower triagonal Sheffer matrix of the associated type S = (1, exp(x) - 1) (taken as N X N matrix for arbitrarily large N) provides the transition matrix from the basis {x^n/n!}, n >= 0, to the basis {y^n/n!}, n >= 0, with y^n/n! = Sum_{m>=n} S^{trans}(n, m) x^m/m! = Sum_{m>=0} x^m/m!*S(m, n).
The Sheffer transform with S = (g, f) of a sequence {a_n} to {b_n} for n >= 0, in matrix notation vec(b) = S vec(a), satisfies, with e.g.f.s A and B, B(x) = g(x)*A(f(x)) and B(x) = A(y(x)) identically, with vec(xhat) = S^{trans,-1} vec(yhat) in symbolic notation with vec(xhat)_n = x^n/n! (similarly for vec(yhat)).
(End)
Number of partitions of {1, 2, ..., n+1} into k+1 nonempty subsets such that no subset contains two adjacent numbers. - Thomas Anton, Sep 26 2022

Examples

			The triangle S(n,k) begins:
  n\k 0 1    2     3      4       5       6      7      8     9   10 11 12
  0:  1
  1:  0 1
  2:  0 1    1
  3:  0 1    3     1
  4:  0 1    7     6      1
  5:  0 1   15    25     10       1
  6:  0 1   31    90     65      15       1
  7:  0 1   63   301    350     140      21      1
  8:  0 1  127   966   1701    1050     266     28      1
  9:  0 1  255  3025   7770    6951    2646    462     36     1
 10:  0 1  511  9330  34105   42525   22827   5880    750    45    1
 11:  0 1 1023 28501 145750  246730  179487  63987  11880  1155   55  1
 12:  0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66  1
 ... reformatted and extended - _Wolfdieter Lang_, Oct 16 2014
Completely symmetric function S(4, 2) = h^{(2)}_2 = 1^2 + 2^2 + 1^1*2^1 = 7; S(5, 2) = h^{(2)}_3 = 1^3 + 2^3 + 1^2*2^1 + 1^1*2^2 = 15. - _Wolfdieter Lang_, May 26 2017
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: S(5, 3) = S(4, 2) + 2*S(4, 3) = 7 + 3*6 = 25.
Boas-Buck recurrence for column m = 3, and n = 5: S(5, 3) = (3/2)*((5/2)*S(4, 3) + 10*Bernoulli(2)*S(3, 3)) = (3/2)*(15 + 10*(1/6)*1) = 25. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

Crossrefs

See especially A008277 which is the main entry for this triangle.
A000110(n) = sum(S(n, k)) k=0..n, n >= 0. Cf. A085693.
Cf. A084938, A106800 (mirror image), A138378, A213061 (mod 2).

Programs

  • Haskell
    a048993 n k = a048993_tabl !! n !! k
    a048993_row n = a048993_tabl !! n
    a048993_tabl = iterate (\row ->
       [0] ++ (zipWith (+) row $ zipWith (*) [1..] $ tail row) ++ [1]) [1]
    -- Reinhard Zumkeller, Mar 26 2012
  • Maple
    for n from 0 to 10 do seq(Stirling2(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, Nov 01 2006
  • Mathematica
    t[n_, k_] := StirlingS2[n, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v *)
  • Maxima
    create_list(stirling2(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    for(n=0, 22, for(k=0, n, print1(stirling(n, k, 2), ", ")); print()); \\ Joerg Arndt, Apr 21 2013
    

Formula

S(n, k) = k*S(n-1, k) + S(n-1, k-1), n > 0; S(0, k) = 0, k > 0; S(0, 0) = 1.
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Sum_{k = 0..n} x^k*S(n, k) = A213170(n), A000587(n), A000007(n), A000110(n), A001861(n), A027710(n), A078944(n), A144180(n), A144223(n), A144263(n) respectively for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7. - Philippe Deléham, May 09 2004, Feb 16 2013
S(n, k) = Sum_{i=0..k} (-1)^(k+i)binomial(k, i)i^n/k!. - Paul Barry, Aug 05 2004
Sum_{k=0..n} k*S(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Equals the inverse binomial transform of A008277. - Gary W. Adamson, Jan 29 2008
G.f.: 1/(1-xy/(1-x/(1-xy/(1-2x/(1-xy/1-3x/(1-xy/(1-4x/(1-xy/(1-5x/(1-... (continued fraction equivalent to Deléham DELTA construction). - Paul Barry, Dec 06 2009
G.f.: 1/Q(0), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
Inverse of padded A008275 (padded just as A048993 = padded A008277). - Tom Copeland, Apr 25 2014
E.g.f. for the row polynomials s(n,x) = Sum_{k=0..n} S(n,k)*x^k is exp(x*(exp(z)-1)) (Sheffer property). E.g.f. for the k-th column sequence with k leading zeros is ((exp(x)-1)^k)/k! (Sheffer property). - Wolfdieter Lang, Oct 16 2014
G.f. for column k: x^k/Product_{j=1..k} (1-j*x), k >= 0 (with the empty product for k = 0 put to 1). See Abramowitz-Stegun, p. 824, 24.1.4 B. - Wolfdieter Lang, May 26 2017
Boas-Buck recurrence for column sequence m: S(n, k) = (k/(n - k))*(n*S(n-1, k)/2 + Sum_{p=k..n-2} (-1)^(n-p)*binomial(n,p)*Bernoulli(n-p)*S(p, k)), for n > k >= 0, with input T(k,k) = 1. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the white diamond multiplication operator defined in Bala - see Example E4. - Peter Bala, Jan 07 2018
Sum_{k=1..n} k*S(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
S(n,k) = Sum_{j=k..n} (-1)^(j-k)*A059297(n,j)*A354794(j,k). - Mélika Tebni, Jan 27 2023

A000262 Number of "sets of lists": number of partitions of {1,...,n} into any number of lists, where a list means an ordered subset.

Original entry on oeis.org

1, 1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, 58941091, 824073141, 12470162233, 202976401213, 3535017524403, 65573803186921, 1290434218669921, 26846616451246353, 588633468315403843, 13564373693588558173, 327697927886085654441, 8281153039765859726341
Offset: 0

Keywords

Comments

Determinant of n X n matrix M=[m(i,j)] where m(i,i)=i, m(i,j)=1 if i > j, m(i,j)=i-j if j > i. - Vladeta Jovovic, Jan 19 2003
With p(n) = the number of integer partitions of n, d(i) = the number of different parts of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, Sum_{i=1..p(n)} = sum over i and Product_{j=1..d(i)} = product over j, one has: a(n) = Sum_{i=1..p(n)} n!/(Product_{j=1..d(i)} m(i,j)!). - Thomas Wieder, May 18 2005
Consider all n! permutations of the integer sequence [n] = 1,2,3,...,n. The i-th permutation, i=1,2,...,n!, consists of Z(i) permutation cycles. Such a cycle has the length lc(i,j), j=1,...,Z(i). For a given permutation we form the product of all its cycle lengths Product_{j=1..Z(i)} lc(i,j). Furthermore, we sum up all such products for all permutations of [n] which gives Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = A000262(n). For n=4 we have Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = 1*1*1*1 + 2*1*1 + 3*1 + 2*1*1 + 3*1 + 2*1 + 3*1 + 4 + 3*1 + 4 + 2*2 + 2*1*1 + 3*1 + 4 + 3*1 + 2*1*1 + 2*2 + 4 + 2*2 + 4 + 3*1 + 2*1*1 + 3*1 + 4 = 73 = A000262(4). - Thomas Wieder, Oct 06 2006
For a finite set S of size n, a chain gang G of S is a partially ordered set (S,<=) consisting only of chains. The number of chain gangs of S is a(n). For example, with S={a, b}and n=2, there are a(2)=3 chain gangs of S, namely, {(a,a),(b,b)}, {(a,a),(a,b),(b,b)} and {(a,a),(b,a),(b,b)}. - Dennis P. Walsh, Feb 22 2007
(-1)*A000262 with the first term set to 1 and A084358 form a reciprocal pair under the list partition transform and associated operations described in A133314. Cf. A133289. - Tom Copeland, Oct 21 2007
Consider the distribution of n unlabeled elements "1" onto n levels where empty levels are allowed. In addition, the empty levels are labeled. Their names are 0_1, 0_2, 0_3, etc. This sequence gives the total number of such distributions. If the empty levels are unlabeled ("0"), then the answer is A001700. Let the colon ":" separate two levels. Then, for example, for n=3 we have a(3)=13 arrangements: 111:0_1:0_2, 0_1:111:0_2, 0_1:0_2:111, 111:0_2:0_1, 0_2:111:0_1, 0_2:0_1:111, 11:1:0, 11:0:1, 0:11:1, 1:11:0, 1:0:11, 0:1:11, 1:1:1. - Thomas Wieder, May 25 2008
Row sums of exponential Riordan array [1,x/(1-x)]. - Paul Barry, Jul 24 2008
a(n) is the number of partitions of [n] (A000110) into lists of noncrossing sets. For example, a(3)=3 counts 12, 1-2, 2-1 and a(4) = 73 counts the 75 partitions of [n] into lists of sets (A000670) except for 13-24, 24-13 which fail to be noncrossing. - David Callan, Jul 25 2008
a(i-j)/(i-j)! gives the value of the non-null element (i, j) of the lower triangular matrix exp(S)/exp(1), where S is the lower triangular matrix - of whatever dimension - having all its (non-null) elements equal to one. - Giuliano Cabrele, Aug 11 2008, Sep 07 2008
a(n) is also the number of nilpotent partial one-one bijections (of an n-element set). Equivalently, it is the number of nilpotents in the symmetric inverse semigroup (monoid). - Abdullahi Umar, Sep 14 2008
A000262 is the exp transform of the factorial numbers A000142. - Thomas Wieder, Sep 10 2008
If n is a positive integer then the infinite continued fraction (1+n)/(1+(2+n)/(2+(3+n)/(3+...))) converges to the rational number A052852(n)/A000262(n). - David Angell (angell(AT)maths.unsw.edu.au), Dec 18 2008
Vladeta Jovovic's formula dated Sep 20 2006 can be restated as follows: a(n) is the n-th ascending factorial moment of the Poisson distribution with parameter (mean) 1. - Shai Covo (green355(AT)netvision.net.il), Jan 25 2010
The umbral exponential generating function for a(n) is (1-x)^{-B}. In other words, write (1-x)^{-B} as a power series in x whose coefficients are polynomials in B, and then replace B^k with the Bell number B_k. We obtain a(0) + a(1)x + a(2)x^2/2! + ... . - Richard Stanley, Jun 07 2010
a(n) is the number of Dyck n-paths (A000108) with its peaks labeled 1,2,...,k in some order, where k is the number of peaks. For example a(2)=3 counts U(1)DU(2)D, U(2)DU(1)D, UU(1)DD where the label at each peak is in parentheses. This is easy to prove using generating functions. - David Callan, Aug 23 2011
a(n) = number of permutations of the multiset {1,1,2,2,...,n,n} such that for 1 <= i <= n, all entries between the two i's exceed i and if any such entries are present, they include n. There are (2n-1)!! permutations satisfying the first condition, and for example: a(3)=13 counts all 5!!=15 of them except 331221 and 122133 which fail the second condition. - David Callan, Aug 27 2014
a(n) is the number of acyclic, injective functions from subsets of [n] to [n]. Let subset D of [n] have size k. The number of acyclic, injective functions from D to [n] is (n-1)!/(n-k-1)! and hence a(n) = Sum_{k=0..n-1} binomial(n,k)*(n-1)!/(n-k-1)!. - Dennis P. Walsh, Nov 05 2015
a(n) is the number of acyclic, injective, labeled directed graphs on n vertices with each vertex having outdegree at most one. - Dennis P. Walsh, Nov 05 2015
For n > 0, a(n) is the number of labeled-rooted skinny-tree forests on n nodes. A skinny tree is a tree in which each vertex has at most one child. Let k denote the number of trees. There are binomial(n,k) ways to choose the roots, binomial(n-1,k-1) ways to choose the number of descendants for each root, and (n-k)! ways to permute those descendants. Summing over k, we obtain a(n) = Sum_{k=1..n} C(n,k)*C(n-1,k-1)*(n-k)!. - Dennis P. Walsh, Nov 10 2015
This is the Sheffer transform of the Bell numbers A000110 with the Sheffer matrix S = |Stirling1| = (1, -log(1-x)) = A132393. See the e.g.f. formula, a Feb 21 2017 comment on A048993, and R. Stanley's Jun 07 2010 comment above. - Wolfdieter Lang, Feb 21 2017
All terms = {1, 3} mod 10. - Muniru A Asiru, Oct 01 2017
We conjecture that for k = 2,3,4,..., the difference a(n+k) - a(n) is divisible by k: if true, then for each k the sequence a(n) taken modulo k is periodic with period dividing k. - Peter Bala, Nov 14 2017
The above conjecture is true - see the Bala link. - Peter Bala, Jan 20 2018
The terms of this sequence can be derived from the numerators of the fractions, produced by the recursion: b(0) = 1, b(n) = 1 + ((n-1) * b(n-1)) / (n-1 + b(n-1)) for n > 0. The denominators give A002720. - Dimitris Valianatos, Aug 01 2018
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 123. It is also the number of rooted labeled forests that avoid 312, 213, and 132, as well as the number of rooted labeled forests that avoid 132, 213, and 321. - Kassie Archer, Aug 30 2018
For n > 0, a(n) is the number of partitions of [2n-1] whose nontrivial blocks are of type {a,b}, with a <= n and b > n. In fact, for n > 0, a(n) = A056953(2n-1). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) is the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select one person from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). - Enrique Navarrete, Oct 01 2023

Examples

			Illustration of first terms as sets of ordered lists of the first n integers:
  a(1) = 1  : (1)
  a(2) = 3  : (12), (21), (1)(2).
  a(3) = 13 : (123) (6 ways), (12)(3) (2*3 ways) (1)(2)(3) (1 way)
  a(4) = 73 : (1234) (24 ways), (123)(4) (6*4 ways), (12)(34) (2*2*3 ways), (12)(3)(4) (2*6 ways), (1)(2)(3)(4) (1 way).
G.f. = 1 + x + 3*x^2 + 13*x^3 + 73*x^4 + 501*x^4 + 4051*x^5 + 37633*x^6 + 394353*x^7 + ...
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 194.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of A271703 and for n >= 1 of A008297. Unsigned row sums of A111596.
A002868 is maximal element of the n-th row of A271703 and for n >= 1 of A008297.
Main diagonal of A257740 and of A319501.

Programs

  • GAP
    a:=[1,1];; for n in [3..10^2] do a[n]:=(2*n-3)*a[n-1]-(n-2)*(n-3)*a[n-2]; od; A000262:=a;  # Muniru A Asiru, Oct 01 2017
    
  • Haskell
    a000262 n = a000262_list !! n
    a000262_list = 1 : 1 : zipWith (-)
                   (tail $ zipWith (*) a005408_list a000262_list)
                          (zipWith (*) a002378_list a000262_list)
    -- Reinhard Zumkeller, Mar 06 2014
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (2*n-1)*Self(n-1) - (n-1)*(n-2)*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 14 2019
    
  • Magma
    [Factorial(n)*Evaluate(LaguerrePolynomial(n, -1), -1): n in [0..30]]; // G. C. Greubel, Feb 23 2021
    
  • Maple
    A000262 := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(1) fi: (2*n-1)*procname(n-1) - (n-1)*(n-2)*procname(n-2) end proc:
    for n from 0 to 20 do printf(`%d,`,a(n)) od:
    spec := [S, {S = Set(Prod(Z,Sequence(Z)))}, labeled]; [seq(combstruct[count](spec, size=n), n=0..40)];
    with(combinat):seq(sum(abs(stirling1(n, k))*bell(k), k=0..n), n=0..18); # Zerinvary Lajos, Nov 26 2006
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=13)},labelled]: seq(combstruct[count](B, size=n), n=0..19);# Zerinvary Lajos, Mar 21 2009
    a := n -> `if`(n=0, 1, n!*hypergeom([1 - n], [2], -1)): seq(simplify(a(n)), n=0..19); # Peter Luschny, Jun 05 2014
  • Mathematica
    Range[0, 19]! CoefficientList[ Series[E^(x/(1-x)), {x, 0, 19}], x] (* Robert G. Wilson v, Apr 04 2005 *)
    a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Exp[x/(1-x)], {x, 0, n}]]; (* Michael Somos, Jul 19 2005 *)
    a[n_]:=If[n==0, 1, n! Sum[Binomial[n-1, j]/(j+1)!, {j, 0, n-1}]]; Table[a[n], {n, 0, 30}] (* Wilf *)
    a[0] = 1; a[n_]:= n!*Hypergeometric1F1[n+1, 2, 1]/E; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jun 18 2012, after Shai Covo *)
    Table[Sum[BellY[n, k, Range[n]!], {k, 0, n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
    a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Product[ QPochhammer[x^k]^(-MoebiusMu[k]/k), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 02 2019 *)
    Table[n!*LaguerreL[n, -1, -1], {n, 0, 30}] (* G. C. Greubel, Feb 23 2021 *)
    RecurrenceTable[{a[n] == (2*n - 1)*a[n-1] - (n-1)*(n-2)*a[n-2], a[0] == 1, a[1] == 1}, a, {n, 0, 30}] (* Vaclav Kotesovec, Jul 21 2022 *)
  • Maxima
    makelist(sum(abs(stirling1(n,k))*belln(k),k,0,n),n,0,24); /* Emanuele Munarini, Jul 04 2011 */
    
  • Maxima
    makelist(hypergeometric([-n+1,-n],[],1),n,0,12); /* Emanuele Munarini, Sep 27 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( x / (1 - x) + x * O(x^n)), n))}; /* Michael Somos, Feb 10 2005 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, eta(x^k + x * O(x^n))^( -moebius(k) / k)), n))}; /* Michael Somos, Feb 10 2005 */
    
  • PARI
    {a(n) = s = 1; for(k = 1, n-1, s = 1 + k * s / (k + s)); return( numerator(s))}; /* Dimitris Valianatos, Aug 03 2018 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, (1 - x^k + x * O(x^n))^(-eulerphi(k) / k)), n))}; /* Michael Somos, Jun 02 2019 */
    
  • PARI
    a(n) = if (n==0, 1, (n-1)!*pollaguerre(n-1,1,-1)); \\ Michel Marcus, Feb 23 2021; Jul 13 2024
    
  • Python
    from sympy import hyper, hyperexpand
    def A000262(n): return hyperexpand(hyper((-n+1, -n), [], 1)) # Chai Wah Wu, Jan 14 2022
  • Sage
    A000262 = lambda n: hypergeometric([-n+1, -n], [], 1)
    [simplify(A000262(n)) for n in (0..19)] # Peter Luschny, Apr 08 2015
    

Formula

D-finite with recurrence: a(n) = (2*n-1)*a(n-1) - (n-1)*(n-2)*a(n-2).
E.g.f.: exp( x/(1-x) ).
a(n) = Sum_{k=0..n} |A008275(n,k)| * A000110(k). - Vladeta Jovovic, Feb 01 2003
a(n) = (n-1)!*LaguerreL(n-1,1,-1) for n >= 1. - Vladeta Jovovic, May 10 2003
Representation as n-th moment of a positive function on positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x-1)*BesselI(1, 2*x^(1/2))/x^(1/2) dx, n >= 1. - Karol A. Penson, Dec 04 2003
a(n) = Sum_{k=0..n} A001263(n, k)*k!. - Philippe Deléham, Dec 10 2003
a(n) = n! Sum_{j=0..n-1} binomial(n-1, j)/(j+1)!, for n > 0. - Herbert S. Wilf, Jun 14 2005
Asymptotic expansion for large n: a(n) -> (0.4289*n^(-1/4) + 0.3574*n^(-3/4) - 0.2531*n^(-5/4) + O(n^(-7/4)))*(n^n)*exp(-n + 2*sqrt(n)). - Karol A. Penson, Aug 28 2002
Minor part of this asymptotic expansion is wrong! Right is (in closed form): a(n) ~ n^(n-1/4)*exp(-1/2+2*sqrt(n)-n)/sqrt(2) * (1 - 5/(48*sqrt(n)) - 95/(4608*n)), numerically a(n) ~ (0.42888194248*n^(-1/4) - 0.0446752023417*n^(-3/4) - 0.00884196713*n^(-5/4) + O(n^(-7/4))) *(n^n)*exp(-n+2*sqrt(n)). - Vaclav Kotesovec, Jun 02 2013
a(n) = exp(-1)*Sum_{m>=0} [m]^n/m!, where [m]^n = m*(m+1)*...*(m+n-1) is the rising factorial. - Vladeta Jovovic, Sep 20 2006
Recurrence: D(n,k) = D(n-1,k-1) + (n-1+k) * D(n-1,k) n >= k >= 0; D(n,0)=0. From this, D(n,1) = n! and D(n,n)=1; a(n) = Sum_{i=0..n} D(n,i). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Proof: Notice two distinct subsets of the lists for [n]: 1) n is in its own list, then there are D(n-1,k-1); 2) n is in a list with other numbers. Denoting the separation of lists by |, it is not hard to see n has (n-1+k) possible positions, so (n-1+k) * D(n-1,k). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Define f_1(x), f_2(x), ... such that f_1(x) = exp(x), f_{n+1}(x) = (d/dx)(x^2*f_n(x)), for n >= 2. Then a(n-1) = exp(-1)*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! * Sum_{k=1..n} (a(n-k)*k!)/((n-k)!*(k-1)!), where a(0)=1. - Thomas Wieder, Sep 10 2008
a(n) = exp(-1)*n!*M(n+1,2,1), n >= 1, where M (=1F1) is the confluent hypergeometric function of the first kind. - Shai Covo (green355(AT)netvision.net.il), Jan 20 2010
a(n) = n!* A067764(n) / A067653(n). - Gary Detlefs, May 15 2010
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A000110, A049118, A049119 and A049120. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, Nov 17 2011, Aug 02 2012, Dec 11 2012, Jan 27 2013, Jul 31 2013, Dec 25 2013: (Start)
Continued fractions:
E.g.f.: Q(0) where Q(k) = 1+x/((1-x)*(2k+1)-x*(1-x)*(2k+1)/(x+(1-x)*(2k+2)/Q(k+1))).
E.g.f.: 1 + x/(G(0)-x) where G(k) = (1-x)*k + 1 - x*(1-x)*(k+1)/G(k+1).
E.g.f.: exp(x/(1-x)) = 4/(2-(x/(1-x))*G(0))-1 where G(k) = 1 - x^2/(x^2 + 4*(1-x)^2*(2*k+1)*(2*k+3)/G(k+1) ).
E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) = 1 + 1/(k+1)/(1-x)/(1-x/(x+1/E(k+1) )).
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - x/(x + (k+1)*(1-x)/E(k+1) )).
E.g.f.: E(0)-1, where E(k) = 2 + x/( (2*k+1)*(1-x) - x/E(k+1) ).
(End)
E.g.f.: Product {n >= 1} ( (1 + x^n)/(1 - x^n) )^( phi(2*n)/(2*n) ), where phi(n) = A000010(n) is the Euler totient function. Cf. A088009. - Peter Bala, Jan 01 2014
a(n) = n!*hypergeom([1-n],[2],-1) for n >= 1. - Peter Luschny, Jun 05 2014
a(n) = (-1)^(n-1)*KummerU(1-n,2,-1). - Peter Luschny, Sep 17 2014
a(n) = hypergeom([-n+1, -n], [], 1) for n >= 0. - Peter Luschny, Apr 08 2015
E.g.f.: Product_{k>0} exp(x^k). - Franklin T. Adams-Watters, May 11 2016
0 = a(n)*(18*a(n+2) - 93*a(n+3) + 77*a(n+4) - 17*a(n+5) + a(n+6)) + a(n+1)*(9*a(n+2) - 80*a(n+3) + 51*a(n+4) - 6*a(n+5)) + a(n+2)*(3*a(n+2) - 34*a(n+3) + 15*a(n+4)) + a(n+3)*(-10*a(n+3)) if n >= 0. - Michael Somos, Feb 27 2017
G.f. G(x)=y satisfies a differential equation: (1-x)*y-2*(1-x)*x^2*y'+x^4*y''=1. - Bradley Klee, Aug 13 2018
a(n) = n! * LaguerreL(n, -1, -1) = c_{n}(n-1; -1) where c_{n}(x; a) are the Poisson - Charlier polynomials. - G. C. Greubel, Feb 23 2021
3 divides a(3*n-1); 9 divides a(9*n-1); 11 divides a(11*n-1). - Peter Bala, Mar 26 2022
For n > 0, a(n) = Sum_{k=0..n-1}*k!*C(n-1,k)*C(n,k). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) = (n-1)! * (Sum_{i=0..n-1} A002720(i) / i!). - Werner Schulte, Mar 29 2024
a(n+1) = numerator of (1 + n/(1 + n/(1 + (n-1)/(1 + (n-1)/(1 + ... + 1/(1 + 1/(1))))))). See A002720 for the denominators. - Peter Bala, Feb 11 2025

A048994 Triangle of Stirling numbers of first kind, s(n,k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 2, -3, 1, 0, -6, 11, -6, 1, 0, 24, -50, 35, -10, 1, 0, -120, 274, -225, 85, -15, 1, 0, 720, -1764, 1624, -735, 175, -21, 1, 0, -5040, 13068, -13132, 6769, -1960, 322, -28, 1, 0, 40320, -109584, 118124, -67284, 22449, -4536, 546, -36, 1, 0, -362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45, 1
Offset: 0

Keywords

Comments

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.
Mirror image of the triangle A054654. - Philippe Deléham, Dec 30 2006
Also the triangle gives coefficients T(n,k) of x^k in the expansion of C(x,n) = (a(k)*x^k)/n!. - Mokhtar Mohamed, Dec 04 2012
From Wolfdieter Lang, Nov 14 2018: (Start)
This is the Sheffer triangle of Jabotinsky type (1, log(1 + x)). See the e.g.f. of the triangle below.
This is the inverse Sheffer triangle of the Stirling2 Sheffer triangle A008275.
The a-sequence of this Sheffer triangle (see a W. Lang link in A006232)
is from the e.g.f. A(x) = x/(exp(x) -1) a(n) = Bernoulli(n) = A027641(n)/A027642(n), for n >= 0. The z-sequence vanishes.
The Boas-Buck sequence for the recurrences of columns has o.g.f. B(x) = Sum_{n>=0} b(n)*x^n = 1/((1 + x)*log(1 + x)) - 1/x. b(n) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), b = {-1/2, 5/12, -3/8, 251/720, -95/288, 19087/60480,...}. For the Boas-Buck recurrence of Riordan and Sheffer triangles see the Aug 10 2017 remark in A046521, adapted to the Sheffer case, also for two references. See the recurrence and example below. (End)
Let G(n,m,k) be the number of simple labeled graphs on [n] with m edges and k components. Then T(n,k) = Sum (-1)^m*G(n,m,k). See the Read link below. Equivalently, T(n,k) = Sum mu(0,p) where the sum is over all set partitions p of [n] containing k blocks and mu is the Moebius function in the incidence algebra associated to the set partition lattice on [n]. - Geoffrey Critzer, May 11 2024

Examples

			Triangle begins:
  n\k 0     1       2       3      4      5      6    7    8   9 ...
  0   1
  1   0     1
  2   0    -1       1
  3   0     2      -3       1
  4   0    -6      11      -6      1
  5   0    24     -50      35    -10      1
  6   0  -120     274    -225     85    -15      1
  7   0   720   -1764    1624   -735    175    -21    1
  8   0 -5040   13068  -13132   6769  -1960    322  -28    1
  9   0 40320 -109584  118124 -67284  22449  -4536  546  -36   1
  ... - _Wolfdieter Lang_, Aug 22 2012
------------------------------------------------------------------
From _Wolfdieter Lang_, Nov 14 2018: (Start)
Recurrence: s(5,2)= s(4, 1) - 4*s(4, 2) = -6 - 4*11 = -50.
Recurrence from the a- and z-sequences: s(6, 3) = 2*(1*1*(-50) + 3*(-1/2)*35 + 6*(1/6)*(-10) + 10*0*1) = -225.
Boas-Buck recurrence for column k = 3, with b = {-1/2, 5/12, -3/8, ...}:
s(6, 3) = 6!*((-3/8)*1/3! + (5/12)*(-6)/4! + (-1/2)*35/5!) = -225. (End)
		

References

  • 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.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974; Chapter V, also p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 93.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 245.
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

Crossrefs

See especially A008275 which is the main entry for this triangle. A132393 is an unsigned version, and A008276 is another version.
A000142(n) = Sum_{k=0..n} |s(n, k)| for n >= 0.
Row sums give A019590(n+1).

Programs

  • Haskell
    a048994 n k = a048994_tabl !! n !! k
    a048994_row n = a048994_tabl !! n
    a048994_tabl = map fst $ iterate (\(row, i) ->
    (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 0)
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    A048994:= proc(n,k) combinat[stirling1](n,k) end: # R. J. Mathar, Feb 23 2009
    seq(print(seq(coeff(expand(k!*binomial(x,k)),x,i),i=0..k)),k=0..9); # Peter Luschny, Jul 13 2009
    A048994_row := proc(n) local k; seq(coeff(expand(pochhammer(x-n+1,n)), x,k), k=0..n) end: # Peter Luschny, Dec 30 2010
  • Mathematica
    Table[StirlingS1[n, m], {n, 0, 9}, {m, 0, n}] (* Peter Luschny, Dec 30 2010 *)
  • Maxima
    create_list(stirling1(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    a(n,k) = if(k<0 || k>n,0, if(n==0,1,(n-1)*a(n-1,k)+a(n-1,k-1)))
    
  • PARI
    trg(nn)=for (n=0, nn-1, for (k=0, n, print1(stirling(n,k,1), ", ");); print();); \\ Michel Marcus, Jan 19 2015
    

Formula

s(n, k) = A008275(n,k) for n >= 1, k = 1..n; column k = 0 is {1, repeat(0)}.
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.
Triangle (signed) = [0, -1, -1, -2, -2, -3, -3, -4, -4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; Triangle(unsigned) = [0, 1, 1, 2, 2, 3, 3, 4, 4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; where DELTA is Deléham's operator defined in A084938.
Sum_{k=0..n} (-m)^(n-k)*s(n, k) = A000142(n), A001147(n), A007559(n), A007696(n), ... for m = 1, 2, 3, 4, ... .- Philippe Deléham, Oct 29 2005
A008275*A007318 as infinite lower triangular matrices. - Gerald McGarvey, Aug 20 2009
T(n,k) = n!*[x^k]([t^n]exp(x*log(1+t))). - Peter Luschny, Dec 30 2010, updated Jun 07 2020
From Wolfdieter Lang, Nov 14 2018: (Start)
Recurrence from the Sheffer a-sequence (see a comment above): s(n, k) = (n/k)*Sum_{j=0..n-k} binomial(k-1+j, j)*Bernoulli(j)*s(n-1, k-1+j), for n >= 1 and k >= 1, with s(n, 0) = 0 if n >= 1, and s(0,0) = 1.
Boas-Buck type recurrence for column k: s(n, k) = (n!*k/(n - k))*Sum_{j=k..n-1} b(n-1-j)*s(j, k)/j!, for n >= 1 and k = 0..n-1, with input s(n, n) = 1. For sequence b see the Boas-Buck comment above. (End)
T(n,k) = Sum_{j=k..n} (-1)^(n-j)*A271705(n,j)*A216294(j,k). - Mélika Tebni, Feb 23 2023

Extensions

Offset corrected by R. J. Mathar, Feb 23 2009
Formula corrected by Philippe Deléham, Sep 10 2009

A000254 Unsigned Stirling numbers of first kind, s(n+1,2): a(n+1) = (n+1)*a(n) + n!.

Original entry on oeis.org

0, 1, 3, 11, 50, 274, 1764, 13068, 109584, 1026576, 10628640, 120543840, 1486442880, 19802759040, 283465647360, 4339163001600, 70734282393600, 1223405590579200, 22376988058521600, 431565146817638400, 8752948036761600000, 186244810780170240000
Offset: 0

Keywords

Comments

Number of permutations of n+1 elements with exactly two cycles.
Number of cycles in all permutations of [n]. Example: a(3) = 11 because the permutations (1)(2)(3), (1)(23), (12)(3), (13)(2), (132), (123) have 11 cycles altogether. - Emeric Deutsch, Aug 12 2004
Row sums of A094310: In the symmetric group S_n, each permutation factors into k independent cycles; a(n) = sum k over S_n. - Harley Flanders (harley(AT)umich.edu), Jun 28 2004
The sum of the top levels of the last column over all deco polyominoes of height n. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: a(2)=3 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, the levels of their last columns being 2 and 1, respectively. - Emeric Deutsch, Aug 12 2006
a(n) is divisible by n for all composite n >= 6. a(2*n) is divisible by 2*n + 1. - Leroy Quet, May 20 2007
For n >= 2 the determinant of the n-1 X n-1 matrix M(i,j) = i + 2 for i = j and 1 otherwise (i,j = 1..n-1). E.g., for n = 3 the determinant of [(3, 1), (1, 4)]. See 53rd Putnam Examination, 1992, Problem B5. - Franz Vrabec, Jan 13 2008, Mar 26 2008
The numerator of the fraction when we sum (without simplification) the terms in the harmonic sequence. (1 + 1/2 = 2/2 + 1/2 = 3/2; 3/2 + 1/3 = 9/6 + 2/6 = 11/6; 11/6 + 1/4 = 44/24 + 6/24 = 50/24;...). The denominator of this fraction is n!*A000142. - Eric Desbiaux, Jan 07 2009
The asymptotic expansion of the higher order exponential integral E(x,m=2,n=1) ~ exp(-x)/x^2*(1 - 3/x + 11/x^2 - 50/x^3 + 274/x^4 - 1764/x^5 + 13068/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
a(n) is the number of permutations of [n+1] containing exactly 2 cycles. Example: a(2) = 3 because the permutations (1)(23), (12)(3), (13)(2) are the only permutations of [3] with exactly 2 cycles. - Tom Woodward (twoodward(AT)macalester.edu), Nov 12 2009
It appears that, with the exception of n= 4, a(n) mod n = 0 if n is composite and = n-1 if n is prime. - Gary Detlefs, Sep 11 2010
a(n) is a multiple of A025527(n). - Charles R Greathouse IV, Oct 16 2012
Numerator of harmonic number H(n) = Sum_{i=1..n} 1/i when not reduced. See A001008 (Wolstenholme numbers) for the reduced numerators. - Rahul Jha, Feb 18 2015
The Stirling transform of this sequence is A222058(n) (Harmonic-geometric numbers). - Anton Zakharov, Aug 07 2016
a(n) is the (n-1)-st elementary symmetric function of the first n numbers. - Anton Zakharov, Nov 02 2016
The n-th iterated integral of log(x) is x^n * (n! * log(x) - a(n))/(n!)^2 + a polynomial of degree n-1 with arbitrary coefficients. This can be proven using the recurrence relation a(n) = (n-1)! + n*a(n-1). - Mohsen Maesumi, Oct 31 2018
Primes p such that p^3 | a(p-1) are the Wolstenholme primes A088164. - Amiram Eldar and Thomas Ordowski, Aug 08 2019
Total number of left-to-right maxima (or minima) in all permutations of [n]. a(3) = 11 = 3+2+2+2+1+1: (1)(2)(3), (1)(3)2, (2)1(3), (2)(3)1, (3)12, (3)21. - Alois P. Heinz, Aug 01 2020

Examples

			(1-x)^-1 * (-log(1-x)) = x + 3/2*x^2 + 11/6*x^3 + 25/12*x^4 + ...
G.f. = x + x^2 + 5*x^3 + 14*x^4 + 94*x^5 + 444*x^6 + 3828*x^7 + 25584*x^8 + ...
		

References

  • 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.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identities 186-190.
  • N. Bleistein and R. A. Handelsman, Asymptotic Expansions of Integrals, Dover Publications, 1986, see page 2. MR0863284 (89d:41049)
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 217.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
  • Shanzhen Gao, Permutations with Restricted Structure (in preparation).
  • K. Javorszky, Natural Orders: De Ordinibus Naturalibus, 2016, ISBN 978-3-99057-139-2.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    a:=[]; for n in [1..22] do a:=a cat [Abs(StirlingFirst(n,2))]; end for; a; // Marius A. Burtea, Jan 01 2020
  • Maple
    A000254 := proc(n) option remember; if n<=1 then n else n*A000254(n-1)+(n-1)!; fi; end: seq(A000254(n),n=0..21);
    a := n -> add(n!/k, k=1..n): seq(a(n), n=0..21); # Zerinvary Lajos, Jan 22 2008
  • Mathematica
    Table[ (PolyGamma[ m ]+EulerGamma) (m-1)!, {m, 1, 24} ] (* Wouter Meeussen *)
    Table[ n!*HarmonicNumber[n], {n, 0, 19}] (* Robert G. Wilson v, May 21 2005 *)
    Table[Sum[1/i,{i,1,n}]/Product[1/i,{i,1,n}],{n,1,30}] (* Alexander Adamchuk, Jul 11 2006 *)
    Abs[StirlingS1[Range[20],2]] (* Harvey P. Dale, Aug 16 2011 *)
    Table[Gamma'[n + 1] /. EulerGamma -> 0, {n, 0, 30}] (* Li Han, Feb 14 2024*)
  • Maxima
    a(n):=(-1)^(n+1)/2*(n+1)*sum(k*bern(k-1)*stirling1(n,k),k,1,n); /* Vladimir Kruchinin, Nov 20 2016 */
    
  • MuPAD
    A000254 := proc(n) begin n*A000254(n-1)+fact(n-1) end_proc: A000254(1) := 1:
    
  • PARI
    {a(n) = if( n<0, 0, (n+1)! / 2 * sum( k=1, n, 1 / k / (n+1-k)))} /* Michael Somos, Feb 05 2004 */
    
  • Sage
    [stirling_number1(i, 2) for i in range(1, 22)]  # Zerinvary Lajos, Jun 27 2008
    

Formula

Let P(n,X) = (X+1)*(X+2)*(X+3)*...*(X+n); then a(n) is the coefficient of X; or a(n) = P'(n,0). - Benoit Cloitre, May 09 2002
Sum_{k > 0} a(k) * x^k/ k!^2 = exp(x) *(Sum_{k>0} (-1)^(k+1) * x^k / (k * k!)). - Michael Somos, Mar 24 2004; corrected by Warren D. Smith, Feb 12 2006
a(n) is the coefficient of x^(n+2) in (-log(1-x))^2, multiplied by (n+2)!/2.
a(n) = n! * Sum_{i=1..n} 1/i = n! * H(n), where H(n) = A001008(n)/A002805(n) is the n-th harmonic number.
a(n) ~ 2^(1/2)*Pi^(1/2)*log(n)*n^(1/2)*e^-n*n^n. - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
E.g.f.: log(1 - x) / (x-1). (= (log(1 - x))^2 / 2 if offset 1). - Michael Somos, Feb 05 2004
D-finite with recurrence: a(n) = a(n-1) * (2*n - 1) - a(n-2) * (n - 1)^2, if n > 1. - Michael Somos, Mar 24 2004
a(n) = A081358(n)+A092691(n). - Emeric Deutsch, Aug 12 2004
a(n) = n!*Sum_{k=1..n} (-1)^(k+1)*binomial(n, k)/k. - Vladeta Jovovic, Jan 29 2005
p^2 divides a(p-1) for prime p > 3. a(n) = (Sum_{i=1..n} 1/i) / Product_{i=1..n} 1/i. - Alexander Adamchuk, Jul 11 2006
a(n) = 3* A001710(n) + 2* A001711(n-3) for n > 2; e.g., 11 = 3*3 + 2*1, 50 = 3*12 + 2*7, 274 = 3*60 + 2*47, ... - Gary Detlefs, May 24 2010
a(n) = A138772(n+1) - A159324(n). - Gary Detlefs, Jul 05 2010
a(n) = A121633(n) + A002672(n). - Gary Detlefs, Jul 18 2010
a(n+1) = Sum_{i=1..floor((n-1)/2)} n!/((n-i)*i) + Sum_{i=ceiling(n/2)..floor(n/2)} n!/(2*(n-i)*i). - Shanzhen Gao, Sep 14 2010
From Gary Detlefs, Sep 11 2010: (Start)
a(n) = (a(n-1)*(n^2 - 2*n + 1) + (n + 1)!)/(n - 1) for n > 2.
It appears that, with the exception of n = 2, (a(n+1)^2 - a(n)^2) mod n^2 = 0 if n is composite and 4*n if n is prime.
It appears that, with the exception of n = 2, (a(n+1)^3 - a(n)^2) mod n = 0 if n is composite and n - 2 if n is prime.
It appears that, with the exception of n = 2, (a(n)^2 + a(n+1)^2) mod n = 0 if n is composite and = 2 if n is prime. (End)
a(n) = Integral_{x=0..oo} (x^n - n!)*log(x)*exp(-x) dx. - Groux Roland, Mar 28 2011
a(n) = 3*n!/2 + 2*(n-2)!*Sum_{k=0..n-3} binomial(k+2,2)/(n-2-k) for n >= 2. - Gary Detlefs, Sep 02 2011
a(n)/(n-1)! = ml(n) = n*ml(n-1)/(n-1) + 1 for n > 1, where ml(n) is the average number of random draws from an n-set with replacement until the total set has been observed. G.f. of ml: x*(1 - log(1 - x))/(1 - x)^2. - Paul Weisenhorn, Nov 18 2011
a(n) = det(|S(i+2, j+1)|, 1 <= i,j <= n-2), where S(n,k) are Stirling numbers of the second kind. - Mircea Merca, Apr 06 2013
E.g.f.: x/(1 - x)*E(0)/2, where E(k) = 2 + E(k+1)*x*(k + 1)/(k + 2). - Sergei N. Gladkovskii, Jun 01 2013 [Edited by Michael Somos, Nov 28 2013]
0 = a(n) * (a(n+4) - 6*a(n+3) + 7*a(n+2) - a(n+1)) - a(n+1) * (4*a(n+3) - 6*a(n+2) + a(n+1)) + 3*a(n+2)^2 unless n=0. - Michael Somos, Nov 28 2013
For a simple way to calculate the sequence, multiply n! by the integral from 0 to 1 of (1 - x^n)/(1 - x) dx. - Rahul Jha, Feb 18 2015
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Inverse binomial transform of A073596.
a(n) ~ sqrt(2*Pi*n) * n^n * (log(n) + gamma)/exp(n), where gamma is the Euler-Mascheroni constant A001620. (End)
a(n) = ((-1)^(n+1)/2*(n+1))*Sum_{k=1..n} k*Bernoulli(k-1)*Stirling1(n,k). - Vladimir Kruchinin, Nov 20 2016
a(n) = (n)! * (digamma(n+1) + gamma), where gamma is the Euler-Mascheroni constant A001620. - Pedro Caceres, Mar 10 2018
From Andy Nicol, Oct 21 2021: (Start)
Gamma'(x) = a(x-1) - (x-1)!*gamma, where Gamma'(x) is the derivative of the gamma function at positive integers and gamma is the Euler-Mascheroni constant. E.g.:
Gamma'(1) = -gamma, Gamma'(2) = 1-gamma, Gamma'(3) = 3-2*gamma,
Gamma'(22) = 186244810780170240000 - 51090942171709440000*gamma. (End)
From Peter Bala, Feb 03 2022: (Start)
The following are all conjectural:
E.g.f.: for nonzero m, (1/m)*Sum_{n >= 1} (-1)^(n+1)*(1/n)*binomial(m*n,n)* x^n/(1 - x)^(m*n+1) = x + 3*x^2/2! + 11*x^3/3! + 50*x^4/4! + ....
For nonzero m, a(n) = (1/m)*n!*Sum_{k = 1..n} (-1)^(k+1)*(1/k)*binomial(m*k,k)* binomial(n+(m-1)*k,n-k).
a(n)^2 = (1/2)*n!^2*Sum_{k = 1..n} (-1)^(k+1)*(1/k^2)*binomial(n,k)* binomial(n+k,k). (End)
From Mélika Tebni, Jun 20 2022: (Start)
a(n) = -Sum_{k=0..n} k!*A021009(n, k+1).
a(n) = Sum_{k=0..n} k!*A094587(n, k+1). (End)
a(n) = n! * 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n - 1)^2/((2*n - 1)))))). - Peter Bala, Mar 16 2024

A019538 Triangle of numbers T(n,k) = k!*Stirling2(n,k) read by rows (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 14, 36, 24, 1, 30, 150, 240, 120, 1, 62, 540, 1560, 1800, 720, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880, 1, 1022, 55980, 818520, 5103000, 16435440, 29635200, 30240000, 16329600, 3628800
Offset: 1

Author

N. J. A. Sloane and Manfred Goebel (goebel(AT)informatik.uni-tuebingen.de), Dec 11 1996

Comments

Number of ways n labeled objects can be distributed into k nonempty parcels. Also number of special terms in n variables with maximal degree k.
In older terminology these are called differences of 0. - Michael Somos, Oct 08 2003
Number of surjections (onto functions) from an n-element set to a k-element set.
Also coefficients (in ascending order) of so-called ordered Bell polynomials.
(k-1)!*Stirling2(n,k-1) is the number of chain topologies on an n-set having k open sets [Stephen].
Number of set compositions (ordered set partitions) of n items into k parts. Number of k dimensional 'faces' of the n dimensional permutohedron (see Simion, p. 162). - Mitch Harris, Jan 16 2007
Correction of comment before: Number of (n-k)-dimensional 'faces' of the permutohedron of order n (an (n-1)-dimensional polytope). - Tilman Piesk, Oct 29 2014
This array is related to the reciprocal of an e.g.f. as sketched in A133314. For example, the coefficient of the fourth-order term in the Taylor series expansion of 1/(a(0) + a(1) x + a(2) x^2/2! + a(3) x^3/3! + ...) is a(0)^(-5) * {24 a(1)^4 - 36 a(1)^2 a(2) a(0) + [8 a(1) a(3) + 6 a(2)^2] a(0)^2 - a(4) a(0)^3}. The unsigned coefficients characterize the P3 permutohedron depicted on page 10 in the Loday link with 24 vertices (0-D faces), 36 edges (1-D faces), 6 squares (2-D faces), 8 hexagons (2-D faces) and 1 3-D permutohedron. Summing coefficients over like dimensions gives A019538 and A090582. Compare to A133437 for the associahedron. - Tom Copeland, Sep 29 2008, Oct 07 2008
Further to the comments of Tom Copeland above, the permutohedron of type A_3 can be taken as the truncated octahedron. Its dual is the tetrakis hexahedron, a simplicial polyhedron, with f-vector (1,14,36,24) giving the fourth row of this triangle. See the Wikipedia entry and [Fomin and Reading p. 21]. The corresponding h-vectors of permutohedra of type A give the rows of the triangle of Eulerian numbers A008292. See A145901 and A145902 for the array of f-vectors for type B and type D permutohedra respectively. - Peter Bala, Oct 26 2008
Subtriangle of triangle in A131689. - Philippe Deléham, Nov 03 2008
Since T(n,k) counts surjective functions and surjective functions are "consistent", T(n,k) satisfies a binomial identity, namely, T(n,x+y) = Sum_{j=0..n} C(n,j)*T(j,x)*T(n-j,y). For definition of consistent functions and a generalized binomial identity, see "Toy stories and combinatorial identities" in the link section below. - Dennis P. Walsh, Feb 24 2012
T(n,k) is the number of labeled forests on n+k vertices satisfying the following two conditions: (i) each forest consists of exactly k rooted trees with roots labeled 1, 2, ..., k; (ii) every root has at least one child vertex. - Dennis P. Walsh, Feb 24 2012
The triangle is the inverse binomial transform of triangle A028246, deleting the left column and shifting up one row. - Gary W. Adamson, Mar 05 2012
See A074909 for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
E.g.f. for the shifted signed polynomials is G(x,t) = (e^t-1)/[1+(1+x)(e^t-1)] = 1-(1+x)(e^t-1) + (1+x)^2(e^t-1)^2 - ... (see also A008292 and A074909), which has the infinitesimal generator g(x,u)d/du = [(1-x*u)(1-(1+x)u)]d/du, i.e., exp[t*g(x,u)d/du]u eval. at u=0 gives G(x,t), and dG(x,t)/dt = g(x,G(x,t)). The compositional inverse is log((1-xt)/(1-(1+x)t)). G(x,t) is a generating series associated to the generalized Hirzebruch genera. See the G. Rzadowski link for the relation of the derivatives of g(x,u) to solutions of the Riccatt differential equation, soliton solns. to the KdV equation, and the Eulerian and Bernoulli numbers. In addition A145271 connects products of derivatives of g(x,u) and the refined Eulerian numbers to the inverse of G(x,t), which gives the normalized, reverse face polynomials of the simplices (A135278, divided by n+1). See A028246 for the generator g(x,u)d/dx. - Tom Copeland, Nov 21 2014
For connections to toric varieties and Eulerian polynomials, see the Dolgachev and Lunts and the Stembridge links. - Tom Copeland, Dec 31 2015
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra (this entry) and stellahedra. - Tom Copeland, Nov 14 2016
T(n, k) appears in a Worpitzky identity relating monomials to binomials: x^n = Sum_{k=1..n} T(n, k)*binomial(x,k), n >= 1. See eq. (11.) of the Worpitzky link on p. 209. The relation to the Eulerian numbers is given there in eqs. (14.) and (15.). See the formula below relating to A008292. See also Graham et al. eq. (6.10) (relating monomials to falling factorials) on p. 248 (2nd ed. p. 262). The Worpitzky identity given in the Graham et al. reference as eq. (6.37) (2nd ed. p. 269) is eq. (5.), p. 207, of Worpitzky. - Wolfdieter Lang, Mar 10 2017
T(n, m) is also the number of minimum clique coverings and minimum matchings in the complete bipartite graph K_{m,n}. - Eric W. Weisstein, Apr 26 2017
From the Hasan and Franco and Hasan papers: The m-permutohedra for m=1,2,3,4 are the line segment, hexagon, truncated octahedron and omnitruncated 5-cell. The first three are well-known from the study of elliptic models, brane tilings and brane brick models. The m+1 torus can be tiled by a single (m+2)-permutohedron. Relations to toric Calabi-Yau Kahler manifolds are also discussed. - Tom Copeland, May 14 2020
From Manfred Boergens, Jul 25 2021: (Start)
Number of n X k binary matrices with row sums = 1 and no zero columns. These matrices are a subset of the matrices defining A183109.
The distribution into parcels in the leading comment can be regarded as a covering of [n] by tuples (A_1,...,A_k) in P([n])^k with nonempty and disjoint A_j, with P(.) denoting the power set (corrected for clarity by Manfred Boergens, May 26 2024). For the non-disjoint case see A183109 and A218695.
For tuples with "nonempty" dropped see A089072. For tuples with "nonempty and disjoint" dropped see A092477 and A329943 (amendment by Manfred Boergens, Jun 24 2024). (End)

Examples

			The triangle T(n, k) begins:
  n\k 1    2     3      4       5        6        7        8        9      10
  1:  1
  2:  1    2
  3:  1    6     6
  4:  1   14    36     24
  5:  1   30   150    240     120
  6:  1   62   540   1560    1800      720
  7:  1  126  1806   8400   16800    15120     5040
  8:  1  254  5796  40824  126000   191520   141120    40320
  9:  1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
---------------------------------------------------------------------------
T(4,1) = 1: {1234}. T(4,2) = 14: {1}{234} (4 ways), {12}{34} (6 ways), {123}{4} (4 ways). T(4,3) = 36: {12}{3}{4} (12 ways), {1}{23}{4} (12 ways), {1}{2}{34} (12 ways). T(4,4) = 1: {1}{2}{3}{4} (1 way).
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 89, ex. 1; also p. 210.
  • Miklos Bona, Combinatorics of Permutations, Chapman and Hall,2004, p.12.
  • G. Boole, A Treatise On The Calculus of Finite Differences, Dover Publications, 1960, p. 20.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, 1989, p. 155. Also eqs.(6.10) and (6.37).
  • Kiran S. Kedlaya and Andrew V. Sutherland, Computing L -Series of Hyperelliptic Curves in Algorithmic Number Theory Lecture Notes in Computer Science Volume 5011/2008.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.6.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • J. F. Steffensen, Interpolation, 2nd ed., Chelsea, NY, 1950, see p. 54.
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.
  • E. Whittaker and G. Robinson, The Calculus of Observations, Blackie, London, 4th ed., 1949; p. 7.

Crossrefs

Row sums give A000670. Maximal terms in rows give A002869. Central terms T(2k-1,k) give A233734.
Diagonal is n! (A000142). 2nd diagonal is A001286. 3rd diagonal is A037960.
Reflected version of A090582. A371568 is another version.
See also the two closely related triangles: A008277(n, k) = T(n, k)/k! (Stirling numbers of second kind) and A028246(n, k) = T(n, k)/k.
Cf. A033282 'faces' of the associahedron.
Cf. A008292, A047969, A145901, A145902. - Peter Bala, Oct 26 2008
Visible in the 3-D array in A249042.
See also A000182.

Programs

  • Haskell
    a019538 n k = a019538_tabl !! (n-1) !! (k-1)
    a019538_row n = a019538_tabl !! (n-1)
    a019538_tabl = iterate f [1] where
       f xs = zipWith (*) [1..] $ zipWith (+) ([0] ++ xs) (xs ++ [0])
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    with(combinat): A019538 := (n,k)->k!*stirling2(n,k);
  • Mathematica
    Table[k! StirlingS2[n, k], {n, 9}, {k, n}] // Flatten
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, sum(i=0, k, (-1)^i * binomial(k, i) * (k-i)^n))}; /* Michael Somos, Oct 08 2003 */
    
  • Sage
    def T(n, k): return factorial(k)*stirling_number2(n,k) # Danny Rorabaugh, Oct 10 2015

Formula

T(n, k) = k*(T(n-1, k-1)+T(n-1, k)) with T(0, 0) = 1 [or T(1, 1) = 1]. - Henry Bottomley, Mar 02 2001
E.g.f.: (y*(exp(x)-1) - exp(x))/(y*(exp(x)-1) - 1). - Vladeta Jovovic, Jan 30 2003
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*j^n*binomial(k, j). - Mario Catalani (mario.catalani(AT)unito.it), Nov 28 2003. See Graham et al., eq. (6.19), p. 251. For a proof see Bert Seghers, Jun 29 2013.
Sum_{k=0..n} T(n, k)(-1)^(n-k) = 1, Sum_{k=0..n} T(n, k)(-1)^k = (-1)^n. - Mario Catalani (mario.catalani(AT)unito.it), Dec 11 2003
O.g.f. for n-th row: polylog(-n, x/(1+x))/(x+x^2). - Vladeta Jovovic, Jan 30 2005
E.g.f.: 1 / (1 + t*(1-exp(x))). - Tom Copeland, Oct 13 2008
From Peter Bala, Oct 26 2008: (Start)
O.g.f. as a continued fraction: 1/(1 - x*t/(1 - (x + 1)*t/(1 - 2*x*t/(1 - 2*(x + 1)*t/(1 - ...))))) = 1 + x*t + (x + 2*x^2)*t^2 + (x + 6*x^2 + 6*x^3)*t^3 + ... .
The row polynomials R(n,x), which begin R(1,x) = x, R(2,x) = x + 2*x^2, R(3,x) = x + 6*x^2 + 6*x^3, satisfy the recurrence x*d/dx ((x + 1)*R(n,x)) = R(n+1,x). It follows that the zeros of R(n,x) are real and negative (apply Corollary 1.2 of [Liu and Wang]).
Since this is the triangle of f-vectors of the (simplicial complexes dual to the) type A permutohedra, whose h-vectors form the Eulerian number triangle A008292, the coefficients of the polynomial (x-1)^n*R(n,1/(x-1)) give the n-th row of A008292. For example, from row 3 we have x^2 + 6*x + 6 = 1 + 4*y + y^2, where y = x + 1, producing [1,4,1] as the third row of A008292. The matrix product A008292 * A007318 gives the mirror image of this triangle (see A090582).
For n,k >= 0, T(n+1,k+1) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*[(j+1)^(n+1) - j^(n+1)]. The matrix product of Pascal's triangle A007318 with the current array gives (essentially) A047969. This triangle is also related to triangle A047969 by means of the S-transform of [Hetyei], a linear transformation of polynomials whose value on the basis monomials x^k is given by S(x^k) = binomial(x,k). The S-transform of the shifted n-th row polynomial Q(n,x) := R(n,x)/x is S(Q(n,x)) = (x+1)^n - x^n. For example, from row 3 we obtain S(1 + 6*x + 6*x^2) = 1 + 6*x + 6*x*(x-1)/2 = 1 + 3*x + 3*x^2 = (x+1)^3 - x^3. For fixed k, the values S(Q(n,k)) give the nonzero entries in column (k-1) of the triangle A047969 (the Hilbert transform of the Eulerian numbers). (End)
E.g.f.: (exp(x)-1)^k = sum T(n,k)x^n/n!. - Vladimir Kruchinin, Aug 10 2010
T(n,k) = Sum_{i=1..k} A(n,i)*Binomial(n-i,k-i) where A(n,i) is the number of n-permutations that have i ascending runs, A008292.
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = -1 + 1/(1+t*(1-exp(x))), the comp. inverse in x is B(x,t) = log(((1+t)/t) - 1/(t(1+x))).
With h(x,t) = 1/(dB/dx)= (1+x)((1+t)(1+x)-1), the row polynomial P(n,t) is given by (h(x,t)*d/dx)^n x, eval. at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(0,t)=0.
(A factor of -1/n! was removed by Copeland on Aug 25 2016.) (End)
The term linear in x of [x*h(d/dx,t)]^n 1 gives the n-th row polynomial. (See A134685.) - Tom Copeland, Nov 07 2011
Row polynomials are given by D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. - Peter Bala, Nov 25 2011
T(n,x+y) = Sum_{j=0..n} binomial(n,j)*T(j,x)*T(n-j,y). - Dennis P. Walsh, Feb 24 2012
Let P be a Rota-Baxter operator of weight 1 satisfying the identity P(x)*P(y) = P(P(x)*y) + P(x*P(y)) + P(x*y). Then P(1)^2 = P(1) + 2*P^2(1). More generally, Guo shows that P(1)^n = Sum_{k=1..n} T(n,k)*P^k(1). - Peter Bala, Jun 08 2012
Sum_{i=1..n} (-1)^i*T(n,i)/i = 0, for n > 1. - Leonid Bedratyuk, Aug 09 2012
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(k-j)^n. [M. Catalani's re-indexed formula from Nov 28 2003] Proof: count the surjections of [n] onto [k] with the inclusion-exclusion principle, as an alternating sum of the number of functions from [n] to [k-j]. - Bert Seghers, Jun 29 2013
n-th row polynomial = 1/(1 + x)*( Sum_{k>=0} k^n*(x/(1 + x))^k ), valid for x in the open interval (-1/2, inf). See Tanny link. Cf. A145901. - Peter Bala, Jul 22 2014
T(n,k) = k * A141618(n,k-1) / binomial(n,k-1). - Tom Copeland, Oct 25 2014
Sum_{n>=0} n^k*a^n = Sum_{i=1..k} (a / (1 - a))^i * T(k, i)/(1-a) for |a| < 1. - David A. Corneth, Mar 09 2015
From Peter Bala, May 26 2015: (Start)
The row polynomials R(n,x) satisfy (1 + x)*R(n,x) = (-1)^n*x*R(n,-(1 + x)).
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = BINOMIAL(A(k,z))^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). Cf. A145901. For cases see A084784 (k = 1), A090352 (k = 2), A090355 (k = 3), A090357 (k = 4), A090362 (k = 5) and A084785 (k = -2 with z -> -z).
A(k,z)^(k + 1) = A(-(k + 1),-z)^k and hence BINOMIAL(A(k,z)) = A(-(k + 1),-z). (End)
From Tom Copeland, Oct 19 2016: (Start)
Let a(1) = 1 + x + B(1) = x + 1/2 and a(n) = B(n) = (B.)^n, where B(n) are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted, signed row polynomials of this array: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... and p_n(x) = n * b(n-1), where b(n) are the partition polynomials of A133314 evaluated with these a(n).
Sum_{n > 0} R(n,-1/2) x^n/n! = 2 * tanh(x/2), where R(n,x) = Sum_{k = 1..n} T(n,k) x^(k-1) are the shifted row polynomials of this entry, so R(n,-1/2) = 4 * (2^(n+1)-1) B(n+1)/(n+1). (Cf. A000182.)
(End)
Also the Bernoulli numbers are given by B(n) = Sum_{k =1..n} (-1)^k T(n,k) / (k+1). - Tom Copeland, Nov 06 2016
G.f. for column k: k! x^k / Product_{i=1..k} (1-i*x). - Robert A. Russell, Sep 25 2018
a(j) <= A183109(j). - Manfred Boergens, Jul 25 2021
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