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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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

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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.

Links

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.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
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
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
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

Views

Author

N. J. A. Sloane

Keywords

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.

Links

Crossrefs

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
Cf. A000217, A001296, A001297, A001298, A007318, A028246, A039810-A039813, A048994, A087107-A087111, A087127, A094262, A127701.
See also A331155.
Cf. A038207, A086885, A132440, A138378, A173018, A238385.
Cf. A055154, A163353, A256894.
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)

A008290 Triangle T(n,k) of rencontres numbers (number of permutations of n elements with k fixed points).

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 9, 8, 6, 0, 1, 44, 45, 20, 10, 0, 1, 265, 264, 135, 40, 15, 0, 1, 1854, 1855, 924, 315, 70, 21, 0, 1, 14833, 14832, 7420, 2464, 630, 112, 28, 0, 1, 133496, 133497, 66744, 22260, 5544, 1134, 168, 36, 0, 1, 1334961, 1334960, 667485, 222480, 55650, 11088, 1890, 240, 45, 0, 1
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

This is a binomial convolution triangle (Sheffer triangle) of the Appell type: (exp(-x)/(1-x),x), i.e., the e.g.f. of column k is (exp(-x)/(1-x))*(x^k/k!). See the e.g.f. given by V. Jovovic below. - Wolfdieter Lang, Jan 21 2008
The formula T(n,k) = binomial(n,k)*A000166(n-k), with the derangements numbers (subfactorials) A000166 (see also the Charalambides reference) shows the Appell type of this triangle. - Wolfdieter Lang, Jan 21 2008
T(n,k) is the number of permutations of {1,2,...,n} having k pairs of consecutive right-to-left minima (0 is considered a right-to-left minimum for each permutation). Example: T(4,2)=6 because we have 1243, 1423, 4123, 1324, 3124 and 2134; for example, 1324 has right-to-left minima in positions 0-1,3-4 and 2134 has right-to-left minima in positions 0,2-3-4, the consecutive ones being joined by "-". - Emeric Deutsch, Mar 29 2008
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for the sequence aC(0,1). - Tom Copeland, Sep 10 2008
A refinement of this triangle is given by A036039. - Tom Copeland, Nov 06 2012
This triangle equals (A211229(2*n,2*k)) n,k >= 0. - Peter Bala, Dec 17 2014

Examples

			exp((y-1)*x)/(1-x) = 1 + y*x + (1/2!)*(1+y^2)*x^2 + (1/3!)*(2 + 3*y + y^3)*x^3 + (1/4!)*(9 + 8*y + 6*y^2 + y^4)*x^4 + (1/5!)*(44 + 45*y + 20*y^2 + 10*y^3 + y^5)*x^5 + ...
Triangle begins:
       1
       0      1
       1      0     1
       2      3     0     1
       9      8     6     0    1
      44     45    20    10    0    1
     265    264   135    40   15    0   1
    1854   1855   924   315   70   21   0  1
   14833  14832  7420  2464  630  112  28  0 1
  133496 133497 66744 22260 5544 1134 168 36 0 1
...
From _Peter Bala_, Feb 13 2017: (Start)
The infinitesimal generator has integer entries given by binomial(n,k)*(n-k-1)! for n >= 2 and 0 <= k <= n-2 and begins
   0
   0  0
   1  0  0
   2  3  0  0
   6  8  6  0 0
  24 30 20 10 0 0
...
It is essentially A238363 (unsigned and omitting the main diagonal), A211603 (with different offset) and appears to be A092271, again without the main diagonal. (End)

References

  • Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 173, Table 5.2 (without row n=0 and column k=0).
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 194.
  • Arnold Kaufmann, Introduction à la combinatorique en vue des applications, Dunod, Paris, 1968. See p. 92.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.

Links

Crossrefs

Mirror of triangle A098825.
Columns give A000166, A000240, A000387, A000449, A000475, A129135, A129136, A129149, A129153, A129217, A129218, A129238, A129255.
Cf. A055137, A008291, A098558.
Cf. A080955.
Cf. A000012, A000142 (row sums), A000354.
Cf. A170942. Sub-triangle of A211229.
Cf. A092271, A111492, A211603, A238363.
T(2n,n) gives A281262.
Cf. A000023, A133314, A238385, A320582, A335111.

Programs

  • Haskell
    a008290 n k = a008290_tabl !! n !! k
a008290_row n = a008290_tabl !! n
a008290_tabl = map reverse a098825_tabl
-- Reinhard Zumkeller, Dec 16 2013
  • Maple
    T:= proc(n,k) T(n, k):= `if`(k=0, `if`(n<2, 1-n, (n-1)*
      (T(n-1, 0)+T(n-2, 0))), binomial(n, k)*T(n-k, 0))
    end:
seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Mar 15 2013
  • Mathematica
    a[0] = 1; a[1] = 0; a[n_] := Round[n!/E] /; n >= 1 size = 8; Table[Binomial[n, k]a[n - k], {n, 0, size}, {k, 0, n}] // TableForm (* Harlan J. Brothers, Mar 19 2007 *)
T[n_, k_] := Subfactorial[n-k]*Binomial[n, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2017 *)
T[n_, k_] := If[n<1, Boole[n==0 && k==0], T[n, k] = T[n-1, k-1] + T[n-1, k]*(n-1-k) + T[n-1, k+1]*(k+1)]; (* Michael Somos, Sep 13 2024 *)
T[0, 0]:=1; T[n_, 0]:=T[n, 0]=n  T[n-1, 0]+(-1)^n; T[n_, k_]:=T[n, k]=n/k T[n-1, k-1];
Flatten@Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Oliver Seipel, Nov 26 2024 *)
  • PARI
    {T(n, k) = if(k<0 || k>n, 0, n!/k! * sum(i=0, n-k, (-1)^i/i!))}; /* Michael Somos, Apr 26 2000 */

Formula

T(n, k) = T(n-1, k)*n + binomial(n, k)*(-1)^(n-k) = T(n, k-1)/k + binomial(n, k)*(-1)^(n-k)/(n-k+1) = T(n-1, k-1)*n/k = T(n-k, 0)*binomial(n, k) = A000166(n-k)*binomial(n,k) [with T(0, 0) = 1]; so T(n, n) = 1, T(n, n-1) = 0, T(n, n-2) = n*(n-1)/2 for n >= 0.
Sum_{k=0..n} T(n, k) = Sum_{k=0..n} k * T(n, k) = n! for all n > 0, n, k integers. - Wouter Meeussen, May 29 2001
From Vladeta Jovovic, Aug 12 2002: (Start)
O.g.f. for k-th column: (1/k!)*Sum_{i>=k} i!*x^i/(1+x)^(i+1).
O.g.f. for k-th row: k!*Sum_{i=0..k} (-1)^i/i!*(1-x)^i. (End)
E.g.f.: exp((y-1)*x)/(1-x). - Vladeta Jovovic, Aug 18 2002
E.g.f. for number of permutations with exactly k fixed points is x^k/(k!*exp(x)*(1-x)). - Vladeta Jovovic, Aug 25 2002
Sum_{k=0..n} T(n, k)*x^k is the permanent of the n X n matrix with x's on the diagonal and 1's elsewhere; for x = 0, 1, 2, 3, 4, 5, 6 see A000166, A000142, A000522, A010842, A053486, A053487, A080954. - Philippe Deléham, Dec 12 2003; for x = 1+i see A009551 and A009102. - John M. Campbell, Oct 11 2011
T(n, k) = Sum_{j=0..n} A008290(n, j)*k^(n-j) is the permanent of the n X n matrix with 1's on the diagonal and k's elsewhere; for k = 0, 1, 2 see A000012, A000142, A000354. - Philippe Deléham, Dec 13 2003
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*binomial(j,k)*n!/j!. - Paul Barry, May 25 2006
T(n,k) = (n!/k!)*Sum_{j=0..n-k} ((-1)^j)/j!, 0 <= k <= n. From the Appell type of the triangle and the subfactorial formula.
T(n,0) = n*Sum_{j=0..n-1} (j/(j+1))*T(n-1,j), T(0,0)=1. From the z-sequence of this Sheffer triangle z(j)=j/(j+1) with e.g.f. (1-exp(x)*(1-x))/x. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
T(n,k) = (n/k)*T(n-1,k-1) for k >= 1. See above. From the a-sequence of this Sheffer triangle a(0)=1, a(n)=0, n >= 1 with e.g.f. 1. See the W. Lang link under A006232 for Sheffer a- and z-sequences. - Wolfdieter Lang, Jan 21 2008
From Henk P. J. van Wijk, Oct 29 2012: (Start)
T(n,k) = T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k=0 and
T(n,k) = T(n-1,k-1) + T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k>=1.
(End)
T(n,k) = A098825(n,n-k). - Reinhard Zumkeller, Dec 16 2013
Sum_{k=0..n} k^2 * T(n, k) = 2*n! if n > 1. - Michael Somos, Jun 06 2017
From Tom Copeland, Jul 26 2017: (Start)
The lowering and raising operators of this Appell sequence of polynomials P(n,x) are L = d/dx and R = x + d/dL log[exp(-L)/(1-L)] = x-1 + 1/(1-L) = x + L + L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L)^(-1) exp(-L) x^n = (1+L+L^2+...)(x-1)^n = n! Sum_{k=0..n} (x-1)^k / k!.
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
For more on the infinitesimal generator, noted by Bala below, see A238385. (End)
Sum_{k=0..n} k^m * T(n,k) = A000110(m)*n! if n >= m. - Zhujun Zhang, May 24 2019
Sum_{k=0..n} (k+1) * T(n,k) = A098558(n). - Alois P. Heinz, Mar 11 2022
From Alois P. Heinz, May 20 2023: (Start)
Sum_{k=0..n} (-1)^k * T(n,k) = A000023(n).
Sum_{k=0..n} (-1)^k * k * T(n,k) = A335111(n). (End)
T(n,k) = A145224(n,k)+A145225(n,k), refined by even and odd perms. - R. J. Mathar, Jul 06 2023

Extensions

Comments and more terms from Michael Somos, Apr 26 2000 and Christian G. Bower, Apr 26 2000

A007840 Number of factorizations of permutations of n letters into ordered cycles.

Original entry on oeis.org

1, 1, 3, 14, 88, 694, 6578, 72792, 920904, 13109088, 207360912, 3608233056, 68495486640, 1408631978064, 31197601660080, 740303842925184, 18738231641600256, 503937595069600896, 14349899305396086912, 431322634732516137216, 13646841876634025159424
Offset: 0

Views

Author

Arnold Knopfmacher

Keywords

Comments

a(n) is the number of ways to seat n people at an unspecified number of circular tables and then linearly order the nonempty tables. - Geoffrey Critzer, Mar 18 2009
The terms of this sequence for n >= 1 are the row sums of A008275^2, the unsigned version of A039814. - Peter Bala, Jul 22 2014
Signed sequence is the base for an Appell sequence of polynomials with the e.g.f. e^(x*t)/[log(1+t) + 1] = exp(P(.,x),t) that is the umbral compositional inverse for A238385, reverse of A111492, i.e., umbrally evaluated UP(n,P(.,t))= x^n = P(n,UP(.,t)) where UP(n,t) are the polynomials of A238385. Umbrally evaluated means letting (A(.,t))^n = A(n,t) after substituting A for the independent variable of the polynomial. - Tom Copeland, Nov 15 2014
a(n) is the number of unimodal rooted forests on n labeled nodes (i.e., those forests that avoid the patterns 213 and 312). - Kassie Archer, Aug 30 2018
Number of permutations of [n] where fixed points at index j are j-colored and all other points are unicolored. - Alois P. Heinz, Apr 24 2020

Links

Crossrefs

Cf. A052860. Row sums of unsigned version of A039814.
Cf. A238385, A111492.

Programs

  • Maple
    a:= proc(n) a(n):= n!*`if`(n=0, 1, add(a(k)/(k!*(n-k)), k=0..n-1)) end:
seq(a(n), n=0..25);  # Alois P. Heinz, Nov 06 2012
  • Mathematica
    Table[Sum[Abs[StirlingS1[n, k]] k!, {k, 0, n}], {n, 0, 20}] (* Geoffrey Critzer, Mar 18 2009 *)
  • PARI
    a(n)=n!*polcoeff(1/(1+log(1-x +x*O(x^n))),n) /* Paul D. Hanna, Jul 19 2006 */
  • PARI
    {a(n)=local(CF=1+x*O(x)); for(k=0, n-1, CF=1/((n-k)-((n-k+1)\2)^2*x*CF)); n!*polcoeff(1/(1-x*CF), n)} /* Paul D. Hanna, Jul 19 2006 */
  • Sage
    def A007840_list(len):
    f, R, C = 1, [1], [1]+[0]*len
    for n in (1..len):
        f *= n
        for k in range(n, 0, -1):
            C[k] = -C[k-1]*((k-1)/k if k>1 else 1)
        C[0] = sum((-1)^k*C[k] for k in (1..n))
        R.append(C[0]*f)
    return R
print(A007840_list(20)) # Peter Luschny, Feb 21 2016

Formula

a(n) = Sum_{k=1..n} k! * s(n, k), s(n, k) = unsigned Stirling number of first kind; E.g.f. 1/(1+log(1-z)).
For n>0, a(n) is the permanent of the n X n matrix with entries a(i, i) = i and a(i, j) = 1 elsewhere. - Philippe Deléham, Dec 09 2003
a(n) = A052860(n)/n for n >= 1.
a(n) = n!*Sum_{k=0..n-1} a(k)/k!/(n-k) for n >= 1 with a(0)=1. - Paul D. Hanna, Jul 19 2006
E.g.f.: B(A(x)) where B(x) = 1/(1-x) and A(x) = log(1/(1-x)). - Geoffrey Critzer, Mar 18 2009
a(n) = D^n(1/(1-x)) evaluated at x = 0, where D is the operator exp(x)*d/dx. Cf. A006252. - Peter Bala, Nov 25 2011
E.g.f.: 1/(1+log(1-x)) = 1/(1 - x/(1 - x/(2 - x/(3 - 4*x/(4 - 4*x/(5 - 9*x/(6 - 9*x/(7 - 16*x/(8 - 16*x/(9 - ...)))))))))), a continued fraction. - Paul D. Hanna, Dec 31 2011
a(n) ~ n! * exp(n)/(exp(1)-1)^(n+1). - Vaclav Kotesovec, Jun 21 2013

Extensions

Extended June 1995

A038207 Triangle whose (i,j)-th entry is binomial(i,j)*2^(i-j).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 8, 12, 6, 1, 16, 32, 24, 8, 1, 32, 80, 80, 40, 10, 1, 64, 192, 240, 160, 60, 12, 1, 128, 448, 672, 560, 280, 84, 14, 1, 256, 1024, 1792, 1792, 1120, 448, 112, 16, 1, 512, 2304, 4608, 5376, 4032, 2016, 672, 144, 18, 1, 1024, 5120, 11520, 15360, 13440, 8064, 3360, 960, 180, 20, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

This infinite matrix is the square of the Pascal matrix (A007318) whose rows are [ 1,0,... ], [ 1,1,0,... ], [ 1,2,1,0,... ], ...
As an upper right triangle, table rows give number of points, edges, faces, cubes,
4D hypercubes etc. in hypercubes of increasing dimension by column. - Henry Bottomley, Apr 14 2000. More precisely, the (i,j)-th entry is the number of j-dimensional subspaces of an i-dimensional hypercube (see the Coxeter reference). - Christof Weber, May 08 2009
Number of different partial sums of 1+[1,1,2]+[2,2,3]+[3,3,4]+[4,4,5]+... with entries that are zero removed. - Jon Perry, Jan 01 2004
Row sums are powers of 3 (A000244), antidiagonal sums are Pell numbers (A000129). - Gerald McGarvey, May 17 2005
Riordan array (1/(1-2x), x/(1-2x)). - Paul Barry, Jul 28 2005
T(n,k) is the number of elements of the Coxeter group B_n with descent set contained in {s_k}, 0<=k<=n-1. For T(n,n), we interpret this as the number of elements of B_n with empty descent set (since s_n does not exist). - Elizabeth Morris (epmorris(AT)math.washington.edu), Mar 01 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then T(n,k) = the number of elements (x,y) of S for which y has exactly k more elements than x. - Ross La Haye, Oct 12 2007
T(n,k) is number of paths in the first quadrant going from (0,0) to (n,k) using only steps B=(1,0) colored blue, R=(1,0) colored red and U=(1,1). Example: T(3,2)=6 because we have BUU, RUU, UBU, URU, UUB and UUR. - Emeric Deutsch, Nov 04 2007
T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), and two kinds of step (1,0). - Joerg Arndt, Jul 01 2011
T(i,j) is the number of i-permutations of {1,2,3} containing j 1's. Example: T(2,1)=4 because we have 12, 13, 21 and 31; T(3,2)=6 because we have 112, 113, 121, 131, 211 and 311. - Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (2+x)^n. - N-E. Fahssi, Apr 13 2008
Sum of diagonals are Jacobsthal-numbers: A001045. - Mark Dols, Aug 31 2009
Triangle T(n,k), read by rows, given by [2,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 15 2009
Eigensequence of the triangle = A004211: (1, 3, 11, 49, 257, 1539, ...). - Gary W. Adamson, Feb 07 2010
f-vectors ("face"-vectors) for n-dimensional cubes [see e.g., Hoare]. (This is a restatement of Bottomley's above.) - Tom Copeland, Oct 19 2012
With P = Pascal matrix, the sequence of matrices I, A007318, A038207, A027465, A038231, A038243, A038255, A027466 ... = P^0, P^1, P^2, ... are related by Copeland's formula below to the evolution at integral time steps n= 0, 1, 2, ... of an exponential distribution exp(-x*z) governed by the Fokker-Planck equation as given in the Dattoli et al. ref. below. - Tom Copeland, Oct 26 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Unsigned diagonals of A133156 are rows of this array. - Tom Copeland, Oct 11 2014
Omitting the first row, this is the production matrix for A039683, where an equivalent differential operator can be found. - Tom Copeland, Oct 11 2016
T(n,k) is the number of functions f:[n]->[3] with exactly k elements mapped to 3. Note that there are C(n,k) ways to choose the k elements mapped to 3, and there are 2^(n-k) ways to map the other (n-k) elements to {1,2}. Hence, by summing T(n,k) as k runs from 0 to n, we obtain 3^n = Sum_{k=0..n} T(n,k). - Dennis P. Walsh, Sep 26 2017
Since this array is the square of the Pascal lower triangular matrix, the row polynomials of this array are obtained as the umbral composition of the row polynomials P_n(x) of the Pascal matrix with themselves. E.g., P_3(P.(x)) = 1 P_3(x) + 3 P_2(x) + 3 P_1(x) + 1 = (x^3 + 3 x^2 + 3 x + 1) + 3 (x^2 + 2 x + 1) + 3 (x + 1) + 1 = x^3 + 6 x^2 + 12 x + 8. - Tom Copeland, Nov 12 2018
T(n,k) is the number of 2-compositions of n+1 with some zeros allowed that have k zeros; see the Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020
Also the convolution triangle of A000079. - Peter Luschny, Oct 09 2022

Examples

			Triangle begins with T(0,0):
   1;
   2,  1;
   4,  4,  1;
   8, 12,  6,  1;
  16, 32, 24,  8,  1;
  32, 80, 80, 40, 10,  1;
  ... -  corrected by _Clark Kimberling_, Aug 05 2011
Seen as an array read by descending antidiagonals:
[0] 1, 2,  4,   8,    16,    32,    64,     128,     256, ...     [A000079]
[1] 1, 4,  12,  32,   80,    192,   448,    1024,    2304, ...    [A001787]
[2] 1, 6,  24,  80,   240,   672,   1792,   4608,    11520, ...   [A001788]
[3] 1, 8,  40,  160,  560,   1792,  5376,   15360,   42240, ...   [A001789]
[4] 1, 10, 60,  280,  1120,  4032,  13440,  42240,   126720, ...  [A003472]
[5] 1, 12, 84,  448,  2016,  8064,  29568,  101376,  329472, ...  [A054849]
[6] 1, 14, 112, 672,  3360,  14784, 59136,  219648,  768768, ...  [A002409]
[7] 1, 16, 144, 960,  5280,  25344, 109824, 439296,  1647360, ... [A054851]
[8] 1, 18, 180, 1320, 7920,  41184, 192192, 823680,  3294720, ... [A140325]
[9] 1, 20, 220, 1760, 11440, 64064, 320320, 1464320, 6223360, ... [A140354]

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 155.
  • H. S. M. Coxeter, Regular Polytopes, Dover Publications, New York (1973), p. 122.

Links

Crossrefs

See A065109 for a signed version.
Columns 0-10 are A000079, A001787, A001788(n-1), A001789, A003472, A054849, A002409(n-6), A054851, A140325(n-8), A140354(n-9), A172242.
T(2n,n) gives A059304.
Cf. A007318, A013609, A013610, A062715, A099089, A004211, A135387.
Cf. A001861, A008277, A027465, A027466, A038231, A038243, A038255, A039683, A112857, A118801, A132440, A133156, A133314, A167374, A218272, A238385, A323939.

Programs

  • GAP
    Flat(List([0..15], n->List([0..n], k->Binomial(n, k)*2^(n-k)))); # Stefano Spezia, Nov 21 2018
  • Haskell
    a038207 n = a038207_list !! n
a038207_list = concat $ iterate ([2,1] *) [1]
instance Num a => Num [a] where
   fromInteger k = [fromInteger k]
   (p:ps) + (q:qs) = p + q : ps + qs
   ps + qs         = ps ++ qs
   (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
    *                = []
-- Reinhard Zumkeller, Apr 02 2011
  • Haskell
    a038207' n k = a038207_tabl !! n !! k
a038207_row n = a038207_tabl !! n
a038207_tabl = iterate f [1] where
   f row = zipWith (+) ([0] ++ row) (map (* 2) row ++ [0])
-- Reinhard Zumkeller, Feb 27 2013
  • Magma
    /* As triangle */ [[(&+[Binomial(n,i)*Binomial(i,k): i in [k..n]]): k in [0..n]]: n in [0..15]]; // Vincenzo Librandi, Nov 16 2018
  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*2^(i-j), j = 0 .. i) end do; # yields sequence in triangular form - Emeric Deutsch, Nov 04 2007
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, n -> 2^(n-1)); # Peter Luschny, Oct 09 2022
  • Mathematica
    Table[CoefficientList[Expand[(y + x + x^2)^n], y] /. x -> 1, {n, 0,10}] // TableForm (* Geoffrey Critzer, Nov 20 2011 *)
Table[Binomial[n,k]2^(n-k),{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, May 22 2020 *)
  • PARI
    {T(n, k) = polcoeff((x+2)^n, k)}; /* Michael Somos, Apr 27 2000 */
  • Sage
    def A038207_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in range(dim): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1]+2*M[n-1,k]
    return M
A038207_triangle(9)  # Peter Luschny, Sep 20 2012

Formula

T(n, k) = Sum_{i=0..n} binomial(n,i)*binomial(i,k).
T(n, k) = (-1)^k*A065109(n,k).
G.f.: 1/(1-2*z-t*z). - Emeric Deutsch, Nov 04 2007
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
From the formalism of A133314, the e.g.f. for the row polynomials of A038207 is exp(x*t)*exp(2x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-2x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*2x). The results generalize for 2 replaced by any number. - Tom Copeland, Aug 18 2008
Sum_{k=0..n} T(n,k)*x^k = (2+x)^n. - Philippe Deléham, Dec 15 2009
n-th row is obtained by taking pairwise sums of triangle A112857 terms starting from the right. - Gary W. Adamson, Feb 06 2012
T(n,n) = 1 and T(n,k) = T(n-1,k-1) + 2*T(n-1,k) for kJon Perry, Oct 11 2012
The e.g.f. for the n-th row is given by umbral composition of the normalized Laguerre polynomials A021009 as p(n,x) = L(n, -L(.,-x))/n! = 2^n L(n, -x/2)/n!. E.g., L(2,x) = 2 -4*x +x^2, so p(2,x)= (1/2)*L(2, -L(.,-x)) = (1/2)*(2*L(0,-x) + 4*L(1,-x) + L(2,-x)) = (1/2)*(2 + 4*(1+x) + (2+4*x+x^2)) = 4 + 4*x + x^2/2. - Tom Copeland, Oct 20 2012
From Tom Copeland, Oct 26 2012: (Start)
From the formalism of A132440 and A218272:
Let P and P^T be the Pascal matrix and its transpose and H= P^2= A038207.
Then with D the derivative operator,
exp(x*z/(1-2*z))/(1-2*z)= exp(2*z D_z z) e^(x*z)= exp(2*D_x (x D_x)) e^(z*x)
= (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
= (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T
= Sum_{n>=0} z^n * 2^n Lag_n(-x/2)= exp[z*EF(.,x)], an o.g.f. for the f-vectors (rows) of A038207 where EF(n,x) is an e.g.f. for the n-th f-vector. (Lag_n(x) are the un-normalized Laguerre polynomials.)
Conversely,
exp(z*(2+x))= exp(2D_x) exp(x*z)= exp(2x) exp(x*z)
= (1 x x^2 x^3 ...) H^T (1 z z^2/2! z^3/3! ...)^T
= (1 z z^2/2! z^3/3! ...) H (1 x x^2 x^3 ...)^T
= exp(z*OF(.,x)), an e.g.f for the f-vectors of A038207 where
OF(n,x)= (2+x)^n is an o.g.f. for the n-th f-vector.
(End)
G.f.: R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ (1+y))*x/((2*k+2+ (1+y))*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
A038207 = exp[M*B(.,2)] where M = A238385-I and (B(.,x))^n = B(n,x) are the Bell polynomials (cf. A008277). B(n,2) = A001861(n). - Tom Copeland, Apr 17 2014
T = (A007318)^2 = A112857*|A167374| = |A118801|*|A167374| = |A118801*A167374| = |P*A167374*P^(-1)*A167374| = |P*NpdP*A167374|. Cf. A118801. - Tom Copeland, Nov 17 2016
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial 2^n*Sum_{k = 0..n} binomial(n,k)*x^k/k!. For example, the e.g.f. for the third subdiagonal is exp(x)*(8 + 24*x + 12*x^2 + 4*x^3/3) = 8 + 32*x + 80*x^2/2! + 160*x^3/3! + .... - Peter Bala, Mar 05 2017
T(3*k+2,k) = T(3*k+2,k+1), T(2*k+1,k) = 2*T(2*k+1,k+1). - Yuchun Ji, May 26 2020
From Robert A. Russell, Aug 05 2020: (Start)
G.f. for column k: x^k / (1-2*x)^(k+1).
E.g.f. for column k: exp(2*x) * x^k / k!. (End)
Also the array A(n, k) read by descending antidiagonals, where A(n, k) = (-1)^n*Sum_{j= 0..n+k} binomial(n + k, j)*hypergeom([-n, j+1], [1], 1). - Peter Luschny, Nov 09 2021

A001861 Expansion of e.g.f. exp(2*(exp(x) - 1)).

Original entry on oeis.org

1, 2, 6, 22, 94, 454, 2430, 14214, 89918, 610182, 4412798, 33827974, 273646526, 2326980998, 20732504062, 192982729350, 1871953992254, 18880288847750, 197601208474238, 2142184050841734, 24016181943732414, 278028611833689478, 3319156078802044158, 40811417293301014150
Offset: 0

Views

Author

N. J. A. Sloane, Simon Plouffe

Keywords

Comments

Values of Bell polynomials: ways of placing n labeled balls into n unlabeled (but 2-colored) boxes.
First column of the square of the matrix exp(P)/exp(1) given in A011971. - Gottfried Helms, Mar 30 2007
Base matrix in A011971, second power in A078937, third power in A078938, fourth power in A078939. - Gottfried Helms, Apr 08 2007
Equals row sums of triangle A144061. - Gary W. Adamson, Sep 09 2008
Equals eigensequence of triangle A109128. - Gary W. Adamson, Apr 17 2009
Hankel transform is A108400. - Paul Barry, Apr 29 2009
The number of ways of putting n labeled balls into a set of bags and then putting the bags into 2 labeled boxes. An example is given below. - Peter Bala, Mar 23 2013
The f-vectors of n-dimensional hypercube are given by A038207 = exp[M*B(.,2)] = exp[M*A001861(.)] where M = A238385-I and (B(.,x))^n = B(n,x) are the Bell polynomials (cf. A008277). - Tom Copeland, Apr 17 2014
Moments of the Poisson distribution with mean 2. - Vladimir Reshetnikov, May 17 2016
Exponential self-convolution of Bell numbers (A000110). - Vladimir Reshetnikov, Oct 06 2016

Examples

			a(2) = 6: The six ways of putting 2 balls into bags (denoted by { }) and then into 2 labeled boxes (denoted by [ ]) are
01: [{1,2}] [ ];
02: [ ] [{1,2}];
03: [{1}] [{2}];
04: [{2}] [{1}];
05: [{1} {2}] [ ];
06: [ ] [{1} {2}].
- _Peter Bala_, Mar 23 2013

References

  • 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).

Links

Crossrefs

For boxes of 1 color, see A000110, for 3 colors see A027710, for 4 colors see A078944, for 5 colors see A144180, for 6 colors see A144223, for 7 colors see A144263, for 8 colors see A221159.
First column of A078937.
Equals 2*A035009(n), n>0.
Row sums of A033306, A036073, A049020, and A144061.
Cf. A000110, A000587, A002871, A027710, A056857, A068199, A068200, A068201, A078937, A078938, A078944, A078945, A109128, A129323, A129324, A129325, A129327, A129328, A129329, A129331, A129332, A129333, A144180, A144223, A144263, A189233, A213170, A221159, A221176.

Programs

  • Magma
    [&+[2^k*StirlingSecond(n, k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, May 18 2019
  • Maple
    A001861:=n->add(Stirling2(n,k)*2^k, k=0..n); seq(A001861(n), n=0..20); # Wesley Ivan Hurt, Apr 18 2014
# second Maple program:
b:= proc(n, m) option remember;
     `if`(n=0, 2^m, m*b(n-1, m)+b(n-1, m+1))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..25);  # Alois P. Heinz, Aug 04 2021
  • Mathematica
    Table[Sum[StirlingS2[n, k]*2^k, {k, 0, n}], {n, 0, 21}] (* Geoffrey Critzer, Oct 06 2009 *)
mx = 16; p = 1; Range[0, mx]! CoefficientList[ Series[ Exp[ (Exp[p*x] - p - 1)/p + Exp[x]], {x, 0, mx}], x] (* Robert G. Wilson v, Dec 12 2012 *)
Table[BellB[n, 2], {n, 0, 20}] (* Vaclav Kotesovec, Jan 06 2013 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(exp(2*(exp(x+x*O(x^n))-1)),n))
  • PARI
    {a(n)=polcoeff(sum(m=0, n, 2^m*x^m/prod(k=1,m,1-k*x +x*O(x^n))), n)} /* Paul D. Hanna, Feb 15 2012 */
  • PARI
    {a(n) = sum(k=0, n, 2^k*stirling(n, k, 2))} \\ Seiichi Manyama, Jul 28 2019
  • Sage
    expnums(30, 2) # Zerinvary Lajos, Jun 26 2008

Formula

a(n) = Sum_{k=0..n} 2^k*Stirling2(n, k). - Emeric Deutsch, Oct 20 2001
a(n) = exp(-2)*Sum_{k>=1} 2^k*k^n/k!. - Benoit Cloitre, Sep 25 2003
G.f. satisfies 2*(x/(1-x))*A(x/(1-x)) = A(x) - 1; twice the binomial transform equals the sequence shifted one place left. - Paul D. Hanna, Dec 08 2003
PE = exp(matpascal(5)-matid(6)); A = PE^2; a(n)=A[n,1]. - Gottfried Helms, Apr 08 2007
G.f.: 1/(1-2x-2x^2/(1-3x-4x^2/(1-4x-6x^2/(1-5x-8x^2/(1-6x-10x^2/(1-... (continued fraction). - Paul Barry, Apr 29 2009
O.g.f.: Sum_{n>=0} 2^n*x^n / Product_{k=1..n} (1-k*x). - Paul D. Hanna, Feb 15 2012
a(n) ~ exp(-2-n+n/LambertW(n/2))*n^n/LambertW(n/2)^(n+1/2). - Vaclav Kotesovec, Jan 06 2013
G.f.: (G(0) - 1)/(x-1)/2 where G(k) = 1 - 2/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 16 2013
G.f.: 1/Q(0) where Q(k) = 1 + x*k - x - x/(1 - 2*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 07 2013
G.f.: ((1+x)/Q(0)-1)/(2*x), where Q(k) = 1 - (k+1)*x - 2*(k+1)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: T(0)/(1-2*x), where T(k) = 1 - 2*x^2*(k+1)/( 2*x^2*(k+1) - (1-2*x-x*k)*(1-3*x-x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 24 2013
a(n) = Sum_{k=0..n} A033306(n,k) = Sum_{k=0..n} binomial(n,k)*Bell(k)*Bell(n-k), where Bell = A000110 (see Motzkin, p. 170). - Danny Rorabaugh, Oct 18 2015
a(0) = 1 and a(n) = 2 * Sum_{k=0..n-1} binomial(n-1,k)*a(k) for n > 0. - Seiichi Manyama, Sep 25 2017 [corrected by Ilya Gutkovskiy, Jul 12 2020]

A094587 Triangle of permutation coefficients arranged with 1's on the diagonal. Also, triangle of permutations on n letters with exactly k+1 cycles and with the first k+1 letters in separate cycles.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24, 12, 4, 1, 120, 120, 60, 20, 5, 1, 720, 720, 360, 120, 30, 6, 1, 5040, 5040, 2520, 840, 210, 42, 7, 1, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 3628800
Offset: 0

Views

Author

Paul Barry, May 13 2004

Keywords

Comments

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016
Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007
From Tom Copeland, Nov 01 2007: (Start)
T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = Sum_{j=0..n} (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x), formally
5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)
6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)
7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
From Peter Bala, Jul 10 2008: (Start)
This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0.....................1...............2.......3......4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.
Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.
When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).
We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).
An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).
Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)
(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009
Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019
The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009
T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011
T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011
T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012
The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013
For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016
See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016
Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016
The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017
Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017
If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

Examples

			Rows begin {1}, {1,1}, {2,2,1}, {6,6,3,1}, ...
For n=3 and k=1, T(3,1)=6 since there are exactly 6 permutations of {1,2,3,4} with exactly 2 cycles and with 1 and 2 in separate cycles. The permutations are (1)(2 3 4), (1)(2 4 3), (1 3)(2 4), (1 4)(2 3), (1 3 4)(2), and (1 4 3)(2). - _Dennis P. Walsh_, Jan 24 2011
Triangle begins:
     1,
     1,    1,
     2,    2,    1,
     6,    6,    3,    1,
    24,   24,   12,    4,    1,
   120,  120,   60,   20,    5,    1,
   720,  720,  360,  120,   30,    6,    1,
  5040, 5040, 2520,  840,  210,   42,    7,    1
The production matrix is:
      1,     1,
      1,     1,     1,
      2,     2,     1,    1,
      6,     6,     3,    1,    1,
     24,    24,    12,    4,    1,   1,
    120,   120,    60,   20,    5,   1,   1,
    720,   720,   360,  120,   30,   6,   1,   1,
   5040,  5040,  2520,  840,  210,  42,   7,   1,   1,
  40320, 40320, 20160, 6720, 1680, 336,  56,   8,   1,   1
which is the exponential Riordan array A094587, or [1/(1-x),x], with an extra superdiagonal of 1's.
Inverse begins:
   1,
  -1,  1,
   0, -2,  1,
   0,  0, -3,  1,
   0,  0,  0, -4,  1,
   0,  0,  0,  0, -5,  1,
   0,  0,  0,  0,  0, -6,  1,
   0,  0,  0,  0,  0,  0, -7,  1

Links

Crossrefs

Cf. A000166 (alt. row sums), A000522 (row sums).
Cf. A068424, A036039, A173333, A213936, A263916.
Cf. A000670, A008279, A021009, A048994, A132013, A132014, A132159, A132440, A133314, A218234, A235706, A238385.

Programs

  • Haskell
    a094587 n k = a094587_tabl !! n !! k
a094587_row n = a094587_tabl !! n
a094587_tabl = map fst $ iterate f ([1], 1)
   where f (row, i) = (map (* i) row ++ [1], i + 1)
-- Reinhard Zumkeller, Jul 04 2012
  • Maple
    T := proc(n, m): n!/m! end: seq(seq(T(n, m), m=0..n), n=0..9);  # Johannes W. Meijer, Oct 07 2009, revised Nov 25 2012
# Alternative: Note that if you leave out 'abs' you get A021009.
T := proc(n, k) option remember; if n = 0 and k = 0 then 1 elif k < 0 or k > n then 0 else abs((n + k)*T(n-1, k) - T(n-1, k-1)) fi end: #  Peter Luschny, Dec 30 2021
  • Mathematica
    Flatten[Table[Table[n!/k!, {k,0,n}], {n,0,10}]] (* Geoffrey Critzer, Dec 11 2011 *)
  • Sage
    def A094587_row(n): return (factorial(n)*exp(x).taylor(x,0,n)).list()
for n in (0..7): print(A094587_row(n)) # Peter Luschny, Sep 28 2017

Formula

T(n, k) = n!/k! if n >= k >= 0, otherwise 0.
T(n, k) = Sum_{i=k..n} |S1(n+1, i+1)*S2(i, k)| * (-1)^i, with S1, S2 the Stirling numbers.
T(n,k) = (n-k)*T(n-1,k) + T(n-1,k-1). E.g.f.: exp(x*y)/(1-y) = 1 + (1+x)*y + (2+2*x+x^2)*y^2/2! + (6+6*x+3*x^2+x^3)*y^3/3!+ ... . - Peter Bala, Jul 10 2008
A094587 = 1 / ((-1)*A129184 * A127648 + I), I = Identity matrix. - Gary W. Adamson, May 03 2009
From Johannes W. Meijer, Oct 07 2009: (Start)
The o.g.f. of right hand column k is Gf(z;k) = (k-1)!/(1-z)^k, k => 1.
The recurrence relations of the right hand columns lead to Pascal's triangle A007318. (End)
Let f(x) = (1/x)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+1)*R(n,x)-x*R'(n,x). Cf. A132159. - Peter Bala, Oct 28 2011
A padded shifted version of this lower triangular matrix with zeros in the first column and row except for a one in the diagonal position is given by integral(t=0 to t=infinity) exp[-t(I-P)] = 1/(I-P) = I + P^2 + P^3 + ... where P is the infinitesimal generator matrix A218234 and I the identity matrix. The non-padded version is given by P replaced by A132440. - Tom Copeland, Oct 25 2012
From Peter Bala, Aug 28 2013: (Start)
The row polynomials R(n,x) form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x).
Let P(n,x) = Product_{k=0..n-1} (x + k) denote the rising factorial polynomial sequence with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (6, 6, 3, 1) so P(3,x + 1) = (x + 1)*(x + 2)*(x + 3) = 6 + 6*x + 3*x*(x + 1) + x*(x + 1)*(x + 2). (End)
From Tom Copeland, Apr 21 & 26, and Aug 13 2014: (Start)
T-I = M = -A021009*A132440*A021009 with e.g.f. y*exp(x*y)/(1-y). Cf. A132440. Dividing the n-th row of M by n generates the (n-1)th row of T.
T = 1/(I - A132440) = {2*I - exp[(A238385-I)]}^(-1) = unsigned exp[(I-A238385)] = exp[A000670(.)*(A238385-I)] = , umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)/(1-y), so the row polynomials form an Appell sequence with lowering operator d/dx and raising operator x + 1/(1-D).
With L(n,m,x)= Laguerre polynomials of order m, the row polynomials are (-1)^n*n!*L(n,-1-n,x) = (-1)^n*(-1!/(-1-n)!)*K(-n,-1-n+1,x) = n!* K(-n,-n,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-1-n,-:xD:) = (-1)^n*x^(n+1)*:Dx:^n*x^(-1-n) = (-1)^n*x*:xD:^n*x^(-1) = (-1)^n*n!*binomial(xD-1,n) = n!*K(-n,-n,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706 and A132159.
The n-th row of signed M has the coefficients of d[(-:xD:)^n]/d(:Dx:)= f[d/d(-:xD:)](-:xD:)^n with f(y)=y/(y-1), :Dx:^n= n!L(n,0,-:xD:), and (-:xD:)^n = n!L(n,0,:Dx:). M has the coefficients of [D/(1-D)]x^n. (End)
From Tom Copeland, Nov 18 2015: (Start)
Coefficients of the row polynomials of the e.g.f. Sum_{n>=0} P_n(b1,b2,..,bn;t) x^n/n! = e^(P.(..;t) x) = e^(xt) / (1-b.x) = (1 + b1 x + b2 x^2 + b3 x^3 + ...) e^(xt) = 1 + (b1 + t) x + (2 b2 + 2 b1 t + t^2) x^2/2! + (6 b3 + 6 b2 t + 3 b1 t^2 + t^3) x^3/3! + ... , with lowering operator L = d/dt, i.e., L P_n(..;t) = n * P_(n-1)(..;t), and raising operator R = t + d[log(1 + b1 D + b2 D^2 + ...)]/dD = t - Sum_{n>=1} F(n,b1,..,bn) D^(n-1), i.e., R P_n(..,;t) = P_(n+1)(..;t), where D = d/dt and F(n,b1,..,bn) are the Faber polynomials of A263916.
Also P_n(b1,..,bn;t) = CIP_n(t-F(1,b1),-F(2,b1,b2),..,-F(n,b1,..,bn)), the cycle index polynomials A036039.
(End)
The raising operator R = x + 1/(1-D) = x + 1 + D + D^2 + ... in matrix form acting on an o.g.f. (formal power series) is the transpose of the production matrix M below. The linear term x is the diagonal of ones after transposition. The other transposed diagonals come from D^m x^n = n! / (n-m)! x^(n-m). Then P(n,x) = (1,x,x^2,..) M^n (1,0,0,..)^T is a matrix representation of R P(n-1,x) = P(n,x). - Tom Copeland, Aug 17 2016
The row polynomials have e.g.f. e^(xt)/(1-t) = exp(t*q.(x)), umbrally. With p_n(x) the row polynomials of A132013, q_n(x) = v_n(p.(u.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form [T] = [q] = [v]*[p]*[u]. Conversely, p_n(x) = u_n (q.(v.(x))). - Tom Copeland, Nov 10 2016
From the Appell sequence formalism, 1/(1-b.D) t^n = P_n(b1,b2,..,bn;t), the generalized row polynomials noted in the Nov 18 2015 formulas, consistent with the 2007 comments. - Tom Copeland, Nov 22 2016
From Peter Bala, Feb 18 2017: (Start)
G.f.: Sum_{n >= 1} (n*x)^(n-1)/(1 + (n - t)*x)^n = 1 + (1 + t)*x + (2 + 2*t + t^2)*x^2 + ....
n-th row polynomial R(n,t) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^k*(x + k - t)^(n-k) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^(n-k)*(x + k + t)^k, for arbitrary x. The particular case of the latter sum when x = 0 and t = 1 is identity 10.35 in Gould, Vol.4. (End)
Rodrigues-type formula for the row polynomials: R(n, x) = -exp(x)*Int(exp(-x)* x^n, x), for n >= 0. Recurrence: R(n, x) = x^n + n*R(n-1, x), for n >= 1, and R(0, x) = 1. d/dx(R(n, x)) = R(n, x) - x^n, for n >= 0 (compare with the formula from Peter Bala, Aug 28 2013). - Wolfdieter Lang, Dec 23 2019
T(n, k) = Sum_{i=0..n-k} A048994(n-k, i) * n^i for 0 <= k <= n. - Werner Schulte, Jul 26 2022

Extensions

Edited by Johannes W. Meijer, Oct 07 2009
New description from Dennis P. Walsh, Jan 24 2011

A132440 Infinitesimal Pascal matrix: generator (lower triangular matrix representation) of the Pascal matrix, the classical operator xDx, iterated Laguerre transforms, associated matrices of the list partition transform and general Euler transformation for sequences.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0
Offset: 0

Views

Author

Tom Copeland, Nov 13 2007, Nov 15 2007, Nov 22 2007, Dec 02 2007

Keywords

Comments

Let M(t) = exp(t*T) = lim_{n->oo} (1 + t*T/n)^n.
Pascal matrix = [ binomial(n,k) ] = M(1) = exp(T), truncating the series gives the n X n submatrices.
Inverse Pascal matrix = M(-1) = exp(-T) = matrix for inverse binomial transform.
A(j) = T^j / j! equals the matrix [binomial(n,k) * delta(n-k-j)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e., A(j) is a matrix with all the terms 0 except for the j-th lower (or main for j=0) diagonal, which equals that of the Pascal triangle. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n,k) * d(n-k)] which include as a subset the invertible associated matrices of the list partition transform (LPT) of A133314.
For sequences with b(0) = 1, umbrally,
M[b(.)] = exp(b(.)*T) = [ binomial(n,k) * b(n-k) ] = matrices associated to b by LPT.
[M[b(.)]]^(-1) = exp(c(.)*T) = [ binomial(n,k) * c(n-k) ] = matrices associated to c, where c = LPT(b) . Or,
[M[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[M(b(.))] = M[LPT(b(.))]= M[c(.)].
This is related to xDx, the iterated Laguerre transform and the general Euler transformation of a sequence through the comments in A132013 and A132014 and the relation [Sum_{k=0..n} binomial(n,k) * b(n-k) * d(k)] = M(b)*d, (n-th term). See also A132382.
If b(n,x) is a binomial type Sheffer sequence, then M[b(.,x)]*s(y) = s(x+y) when s(y) = (s(0,y),s(1,y),s(2,y),...) is an array for a Sheffer sequence with the same delta operator as b(n,x) and [M[b(.,x)]]^(-1) is given by the formulas above with b(n) replaced by b(n,x) as b(0,x)=1 for a binomial-type Sheffer sequence.
T = I - A132013 and conversely A132013 = I - T, which is the matrix representation for the iterated mixed order Laguerre transform characterized in A132013 (and A132014).
(I-T)^m generates the group [A132013]^m for m = 0,1,2,... discussed in A132014.
The inverse is 1/(I-T) = I + T + T^2 + T^3 + ... = [A132013]^(-1) = A094587 with the associated sequence (0!,1!,2!,3!,...) under the LPT.
And 1/(I-T)^2 = I + 2*T + 3*T^2 + 4*T^3 + ... = [A132013]^(-2) = A132159 with the associated sequence (1!,2!,3!,4!,...) under the LPT.
The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or e.g.f.'s EA(x) and EB(x).
1) b(0) = 0, b(n) = n * a(n-1),
2) B(x) = xDx A(x)
3) B(x) = x * Lag(1,-:xD:) A(x)
4) EB(x) = x * EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x) is the Laguerre polynomial.
So the exponentiated operator can be characterized as
5) exp(t*T) A(x) = exp(t*xDx) A(x) = [Sum_{n=0,1,...} (t*x)^n * Lag(n,-:xD:)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x), a generalized Euler transformation for an o.g.f.,
6) exp(t*T) EA(x) = exp(t*x)*EA(x) = exp[(t+a(.))*x], gen. Euler trf. for an e.g.f.
7) exp(t*T) * a = M(t) * a = [Sum_{k=0..n} binomial(n,k) * t^(n-k) * a(k)].
The umbral extension of formulas 5, 6 and 7 gives formally
8) exp[c(.)*T] A(x) = exp(c(.)*xDx) A(x) = [Sum_{n>=0} (c(.)*x)^n * Lag(n,-:xD:)] A(x) = [exp{[c(.)*u/(1-c(.)*u)]*:xD:} / (1-c(.)*u) ] A(x) (eval. at u=x) = A[x/(1-c(.)*x)]/(1-c(.)*x), where the umbral evaluation should be applied only after a power series in c is obtained,
9) exp[c(.)*T] EA(x) = exp(c(.)*x)*EA(x) = exp[(c(.)+a(.))*x]
10) exp[c(.)*T] * a = M[c(.)] * a = [Sum_{k=0..n} binomial(n,k) * c(n-k) * a(k)] .
The n X n principal submatrix of T is nilpotent, in particular, [Tsub_n]^(n+1) = 0, n=0,1,2,3,....
Note (xDx)^n = x^n D^n x^n = x^n n! (:Dx:)^n/n! = x^n n! Lag(n,-:xD:).
The operator xDx is an important, classical operator explored by among others Dattoli, Al-Salam, Carlitz and Stokes and even earlier investigators.
For a recent treatment of xDx, DxD and more general operators see the paper "Laguerre-type derivatives: Dobinski relations and combinatorial identities". - Karol A. Penson, Sep 15 2009
See Copeland's link for generalized Laguerre functions and connection to fractional differ-integrals in exercises through (:Dx:)^a/a!=(D^a x^a)/a!. - Tom Copeland, Nov 17 2011
From Tom Copeland, Apr 25 2014: (Start)
Conjugation or "similarity" transformations of [dP]=A132440 have an operator interpretation (cf. A074909 and A238363):
In general, select two operators A and B such that A^n = F1(n,B) and B^n = F2(n,A); then A^n = F1(n,F2(.,A)) and B^n = F2(n,F1(.,B)), evaluated umbrally, i.e., F1(n,F2(.,x))=F2(n,F1(.,x))=x^n, implying the polynomials F1 and F2 are an umbral compositional inverse pair.
One such pair are the Bell polynomials Bell(n,x) and falling factorials (x)_n with Bell(n,:xD:)=(xD)^n and (xD)_n=:xD:^n (cf. A074909). Another are the Laguerre polynomials LN(n,x)= n!*Lag(n,x) (A021009), which are umbrally self-inverse, with LN(n,-:xD:)=:Dx:^n and LN(n,:Dx:)= (-:xD:)^n with :Dx:^n=D^n*x^n.
Evaluating, for n>=0, the operator derivative d(B^n)/dA = d(F2(n,A))/dA in the basis B^n, i.e., with A^n finally replaced by F1(n,B), or A^n=F1(.,B)^n=F1(n,B), is equivalent to the matrix conjugation
A) [F2]*[dP]*[F1]
B) = [F2]*[dP]*[F2]^(-1)
C) = [F1]^(-1)*[dP]*[F1],
where [F1] is the lower triangular matrix with the n-th row the coefficients of F1(n,x) and analogously for [F2].
So, given the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose, form the power series V(x)=Rv*Cv(x).
D) dV(B)/dA = Rv * [F2]*[dP]*[F1] * Cv(B).
E) With A=D and B=D, F1(n,x)=F2(n,x)=x^n and [F1]=[F2]=I. Then d(B^n)/dA = d(D^n)/dD = n * D^(n-1); therefore, consistently [F2]*[dP]*[F1] = [dP] and dV(D)/dD = Rv * [dP] * Cv(D). (End)

Examples

			Matrix T begins
  0;
  1,0;
  0,2,0;
  0,0,3,0;
  0,0,0,4,0;
  ...

References

  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015, (x^n D^n x^n on p. 187).

Links

Programs

Formula

T = log(P) with the Pascal matrix P:=A007318. This should be read as T_N = log(P_N) with P_N the N X N matrix P, N>=2. Because P_N is lower triangular with all diagonal elements 1, the series log(1_N-(1_N-P_N)) stops after N-1 terms because (1_N-P_N)^N is the 0_N-matrix. - Wolfdieter Lang, Oct 14 2010
Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L p_n(x) = n * p_(n-1)(x) and R p_n(x) = p_(n+1)(x), the matrix T represents the action of R*L*R in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R = D^(-1). - Tom Copeland, Oct 25 2012
From Tom Copeland, Apr 26 2014: (Start)
A) T = exp(A238385-I) - I
B) = [St1]*P*[St2] - I
C) = [St1]*P*[St1]^(-1) - I
D) = [St2]^(-1)*P*[St2] - I
E) = [St2]^(-1)*P*[St1]^(-1) - I
where P=A007318, [St1]=padded A008275 just as [St2]=A048993=padded A008277, and I=identity matrix. (End)
From Robert Israel, Oct 02 2015: (Start)
G.f. Sum_{k >= 1} k x^((k+3/2)^2/2 - 17/8) is related to Jacobi theta functions.
If 8*n+17 = y^2 is a square, then a(n) = (y-3)/2, otherwise a(n) = 0. (End)

Extensions

Missing zero added in table by Tom Copeland, Feb 25 2014

A003713 Expansion of e.g.f. log(1/(1+log(1-x))).

Original entry on oeis.org

0, 1, 2, 7, 35, 228, 1834, 17582, 195866, 2487832, 35499576, 562356672, 9794156448, 186025364016, 3826961710272, 84775065603888, 2011929826983504, 50929108873336320, 1369732445916318336, 39005083331889816960, 1172419218038422659456, 37095226237402478348544
Offset: 0

Views

Author

N. J. A. Sloane, R. H. Hardin, Simon Plouffe

Keywords

Comments

a(n+1) is the permanent of the n X n matrix M with M(i,i) = i+1, other entries 1. - Philippe Deléham, Nov 03 2005
Supernecklaces of type III (cycles of cycles). - Ricardo Bittencourt, May 05 2013
Unsigned coefficients for the raising / creation operator R for the Appell sequence of polynomials A238385: R = x + 1 - 2 D + 7 D^2/2! - 35 D^3/3! + ... . - Tom Copeland, May 09 2016

References

  • J. Ginsburg, Iterated exponentials, Scripta Math., 11 (1945), 340-353.
  • 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).

Links

Crossrefs

a(n)=|A039814(n, 1)| (first column of triangle). Cf. A000268, A000310, A000359, A000406, A001765.
Cf. A007840, A089064.
Cf. A238385.

Programs

  • Maple
    series(ln(1/(1+ln(1-x))),x,17);
with (combstruct): M[ 1798 ] := [ A,{A=Cycle(Cycle(Z))},labeled ]:
  • Mathematica
    With[{nn=20},CoefficientList[Series[Log[1/(1+Log[1-x])],{x,0,nn}],x]Range[0,nn]!] (* Harvey P. Dale, Dec 15 2012 *)
Table[Sum[(-1)^(n-k) * (k-1)! * StirlingS1[n, k], {k, 1, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 19 2024 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(-log(1+log(1-x+x*O(x^n))),n))

Formula

Sum_{k=1..n} (k-1)!*|Stirling1(n, k)|. - Vladeta Jovovic, Sep 14 2003
a(n+1) = n! * Sum_{k=0..n} A007840(k)/k!. E.g., a(4) = 228 = 24*(1/1 + 1/1 + 3/2 + 14/6 + 88/24) = 24 + 24 + 36 + 56 + 88. - Philippe Deléham, Dec 10 2003
a(n) ~ (n-1)! * (exp(1)/(exp(1)-1))^n. - Vaclav Kotesovec, Jun 21 2013
a(0) = 0; a(n) = (n-1)! + Sum_{k=1..n-1} binomial(n-1,k) * (k-1)! * a(n-k). - Ilya Gutkovskiy, Jul 18 2020

Extensions

Thanks to Paul Zimmermann for comments.

A238363 Coefficients for the commutator for the logarithm of the derivative operator [log(D),x^n D^n]=d[(xD)!/(xD-n)!]/d(xD) expanded in the operators :xD:^k.

Original entry on oeis.org

1, -1, 2, 2, -3, 3, -6, 8, -6, 4, 24, -30, 20, -10, 5, -120, 144, -90, 40, -15, 6, 720, -840, 504, -210, 70, -21, 7, -5040, 5760, -3360, 1344, -420, 112, -28, 8, 40320, -45360, 25920, -10080, 3024, -756, 168, -36, 9, -362880, 403200, -226800, 86400, -25200, 6048, -1260, 240, -45, 10
Offset: 1

Views

Author

Tom Copeland, Feb 25 2014

Keywords

Comments

Let D=d/dx and [A,B]=A·B-B·A. Then each row corresponds to the coefficients of the operators :xD:^k = x^k D^k in the expansion of the commutator [log(D),:xD:^n]=[-log(x),:xD:^n]=sum(k=0 to n-1, a(n,k) :xD:^k). The e.g.f. is derived from [log(D), exp(t:xD:)]=[-log(x), exp(t:xD:)]= log(1+t)exp(t:xD:), using the shift property exp(t:xD:)f(x)=f((1+t)x).
The reversed unsigned array is A111492.
See the mathoverflow link and link therein to an associated mathstackexchange question for other formulas for log(D). In addition, R_x = log(D) = -log(x) + c - sum[n=1 to infnty, (-1)^n 1/n :xD:^n/n!]=
-log(x) + Psi(1+xD) = -log(x) + c + Ein(:xD:), where c is the Euler-Mascheroni constant, Psi(x), the digamma function, and Ein(x), a breed of the exponential integrals (cf. Wikipedia). The :xD:^k ops. commute; therefore, the commutator reduces to the -log(x) term.
Also the n-th row corresponds to the expansion of d[(xD)!/(xD-n)!]/d(xD) = d[:xD:^n]/d(xD) in the operators :xD:^k, or, equivalently, the coefficients of x in d[z!/(z-n)!]/dz=d[St1(n,z)]]/dz evaluated umbrally with z=St2(.,x), i.e., z^n replaced by St2(n,x), where St1(n,x) and St2(n,x) are the signed and unsigned Stirling polynomials of the first (A008275) and second (A008277) kinds. The derivatives of the unsigned St1 are A028421. See examples. This formalism follows from the relations between the raising and lowering operators presented in the MathOverflow link and the Pincherle derivative. The results can be generalized through the operator relations in A094638, which are related to the celebrated Witt Lie algebra and pseudodifferential operators / symbols, to encompass other integral arrays.
A002741(n)*(-1)^(n+1) (row sums), A002104(n)*(-1)^(n+1) (alternating row sums). Column sequences: A133942(n-1), A001048(n-1), A238474, ... - Wolfdieter Lang, Mar 01 2014
Add an additional head row of zeros to the lower triangular array and denote it as T (with initial indexing in columns and rows being 0). Let dP = A132440, the infinitesimal generator for the Pascal matrix, and I, the identity matrix, then exp(T)=I+dP, i.e., T=log(I+dP). Also, (T_n)^n=0, where T_n denotes the n X n submatrix, i.e., T_n is nilpotent of order n. - Tom Copeland, Mar 01 2014
Any pair of lowering and raising ops. L p(n,x) = n·p(n-1,x) and R p(n,x) = p(n+1,x) satisfy [L,R]=1 which implies (RL)^n = St2(n,:RL:), and since (St2(·,u))!/(St2(·,u)-n)!= u^n, when evaluated umbrally, d[(RL)!/(RL-n)!]/d(RL) = d[:RL:^n]/d(RL) is well-defined and gives A238363 when the LHS is reduced to a sum of :RL:^k terms, exactly as for L=d/dx and R=x above. (Note that R_x above is a raising op. different from x, with associated L_x=-xD.) - Tom Copeland, Mar 02 2014
For relations to colored forests, disposition of flags on flagpoles, and the colorings of the vertices of the complete graphs K_n, encoded in their chromatic polynomials, see A130534. - Tom Copeland, Apr 05 2014
The unsigned triangle, omitting the main diagonal, gives A211603. See also A092271. Related to the infinitesimal generator of A008290. - Peter Bala, Feb 13 2017

Examples

			The first few row polynomials are
p(1,x)=  1
p(2,x)= -1 + 2x
p(3,x)=  2 - 3x + 3x^2
p(4,x)= -6 + 8x - 6x^2 + 4x^3
p(5,x)= 24 -30x +20x^2 -10x^3 + 5x^4
...........
For n=3: z!/(z-3)!=z^3-3z^2+2z=St1(3,z) with derivative 3z^2-6z+2, and
3·St2(2,x)-6·St2(1,x)+2=3(x^2+x)-6x+2=3x^2-3x+2=p(3,x). To see the relation to the operator formalism, note that (xD)^k=St2(k,:xD:) and (xD)!/(xD-k)!=[St2(·,:xD:)]!/[St2(·,:xD:)-k]!= :xD:^k.
The triangle a(n,k) begins:
n\k       0       1       2      3      4     5      6    7   8   9 ...
1:        1
2:       -1       2
3:        2      -3       3
4:       -6       8      -6      4
5:       24     -30      20    -10      5
6:     -120     144     -90     40    -15     6
7:      720    -840     504   -210     70   -21      7
8:    -5040    5760   -3360   1344   -420   112    -28    8
9:    40320  -45360   25920 -10080   3024  -756    168  -36   9
10: -362880  403200 -226800  86400 -25200  6048  -1260  240 -45  10
... formatted by _Wolfdieter Lang_, Mar 01 2014
-----------------------------------------------------------------------

Links

Crossrefs

Cf. A094587, A132013, A132014, A132159, A235706, A238385.
Cf. A008290, A092271, A211603.

Programs

  • Mathematica
    a[n_, k_] := (-1)^(n-k-1)*n!/((n-k)*k!); Table[a[n, k], {n, 1, 10}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jul 09 2015 *)

Formula

a(n,k) = (-1)^(n-k-1)*n!/((n-k)*k!) for k=0 to (n-1).
E.g.f.: log(1+t)*exp(x*t).
E.g.f.for unsigned array: -log(1-t)*exp(x*t).
The lowering op. for the row polynomials is L=d/dx, i.e., L p(n,x) = n*p(n-1,x).
An e.g.f. for an unsigned related version is -log(1+t)*exp(x*t)/t= exp(t*s(·,x)) with s(n,x)=(-1)^n * p(n+1,-x)/(n+1). Let L=d/dx and R= x-(1/((1-D)log(1-D))+1/D),then R s(n,x)= s(n+1,x) and L s(n,x)= n*s(n-1,x), defining a special Sheffer sequence of polynomials, an Appell sequence. So, R (-1)^(n-1) p(n,-x)/n = (-1)^n p(n+1,-x)/(n+1).
From Tom Copeland, Apr 17 2014: (Start)
Dividing each diagonal by its first element (-1)^(n-1)*(n-1)! yields the reverse of A104712.
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x). Then with dP = A132440, M = padded A238363 = A238385-I, I = identity matrix, and (B(.,x))^n = B(n,x) = the n-th Bell polynomial Bell(n,x) of A008277,
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x), and
B) P(:xD:)=exp(dP:xD:)=exp[(e^M-I):xD:]=exp[M*B(.,:xD:)]=exp[M*xD]=
(1+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x].
C) P(x)^m = P(m*x). P(2x) = A038207(x) = exp[M*B(.,2x)], face vectors of n-D hypercubes. (End)
From Tom Copeland, Apr 26 2014: (Start)
M = padded A238363 = A238385-I
A) = [St1]*[dP]*[St2] = [padded A008275]*A132440*A048993
B) = [St1]*[dP]*[St1]^(-1)
C) = [St2]^(-1)*[dP]*[St2]
D) = [St2]^(-1)*[dP]*[St1]^(-1),
where [St1]=padded A008275 just as [St2]=A048993=padded A008277.
E) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I + dP)^x*[St1].
F) exp(x*M) = [St1]*P(x)*[St2] = (I + dP)^x,
where (I + dP)^x = sum(k>=0, C(x,k)*dP^k).
Let the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose. Form the power series V(x)= Rv * Cv(x) and W(y) := V(x.) evaluated umbrally with (x.)^n = x_n = (y)_n = y!/(y-n)!. Then
G) U(:xD:) = dV(:xD:)/d(xD) = dW(xD)/d(xD) evaluated with (xD)^n = Bell(n,:xD:),
H) U(x) = dV(x.)/dy := dW(y)/dy evaluated with y^n=y_n=Bell(n,x), and
I) U(x) = Rv * M * Cv(x). (Cf. A132440, A074909.) (End)
The Bernoulli polynomials Ber_n(x) are related to the polynomials q_n(x) = p(n+1,x) / (n+1) with the e.g.f. [log(1+t)/t] e^(xt) (cf. s_n (x) above) as Ber_n(x) = St2_n[q.(St1.(x))], umbrally, or [St2]*[q]*[St1], in matrix form. Since q_n(x) is an Appell sequence of polynomials, q_n(x) = [log(1+D_x)/D_x]x^n. - Tom Copeland, Nov 06 2016

Extensions

Pincherle formalism added by Tom Copeland, Feb 27 2014
Showing 1-10 of 16 results. Next

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