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

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

Examples

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

References

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

Links

Crossrefs

Row sums of A271703 and for n >= 1 of A008297. Unsigned row sums of A111596.
A002868 is maximal element of the n-th row of A271703 and for n >= 1 of A008297.
Main diagonal of A257740 and of A319501.
Cf. A001263, A001700, A002378, A005408, A066668, A056953, A002720.
Cf. A000110, A052852, A082579, A132393, A255807, A255819, A318976.

Programs

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

Formula

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

A000579 Figurate numbers or binomial coefficients C(n,6).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 7, 28, 84, 210, 462, 924, 1716, 3003, 5005, 8008, 12376, 18564, 27132, 38760, 54264, 74613, 100947, 134596, 177100, 230230, 296010, 376740, 475020, 593775, 736281, 906192, 1107568, 1344904, 1623160, 1947792, 2324784, 2760681, 3262623
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of triangles (all of whose vertices lie inside the circle) formed when n points in general position on a circle are joined by straight lines - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), May 25 2000
Figurate numbers based on 6-dimensional regular simplex. According to Hyun Kwang Kim, it appears that every nonnegative integer can be represented as the sum of g = 13 of these numbers. - Jonathan Vos Post, Nov 28 2004
a(n) = A110555(n+1,6). - Reinhard Zumkeller, Jul 27 2005
a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7)^n. - Sergio Falcon, Feb 12 2007
Only prime in this sequence is 7. - Artur Jasinski, Dec 02 2007
6-dimensional triangular numbers, sixth partial sums of binomial transform of [1, 0, 0, 0, ...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009, R. J. Mathar, Jul 07 2009
The number of n-digit numbers the binary expansion of which contains 3 runs of 0's. Generally, the number of n-digit numbers with k runs of 0's is Sum_{i = k..n-k} binomial(i-1, k-1)*binomial(n-i, k) = C(n,2*k) = A034839(n,k) - Vladimir Shevelev, Jul 30 2010
The dimension of the space spanned by a 6-form that couples to M5-brane worldsheets wrapping 6-cycles inside tori (ref. Green,Miller,Vanhove eq. 3.10). - Stephen Crowley, Jan 09 2012
For a set of integers {1,2,...,n}, A253943(n) is the sum of the 2 smallest elements of each subset with 5 elements, which is 3*C(n+1,6) (for n>=5), hence A253943(n) = 3*a(n+1). - Serhat Bulut, Oktay Erkan Temizkan, Mar 13 2015
a(n) = fallfac(n, 6)/6! is also the number of independent components of an antisymmetric tensor of rank 6 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Number of orbits of Aut(Z^7) as function of the infinity norm n of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 645120. - Philippe A.J.G. Chevalier, Dec 28 2015
Coordination sequence for 6-dimensional cyclotomic lattice Z[zeta_7].

Examples

			a(9) = 84 = (1, 3, 3, 1) dot (1, 6, 15, 20) = (1 + 18 + 45 + 20). - _Gary W. Adamson_, Aug 02 2008
G.f. = x^6 + 7*x^7 + 28*x^8 + 84*x^9 + 210*x^10 + 462*x^11 + 924*x^12 + ...
For A = {1,2,3,4,5,6} subsets with 5 elements are {1,2,3,4,5}, {1,2,3,4,6}, {1,2,3,5,6}, {1,2,4,5,6}, {1,3,4,5,6}, {2,3,4,5,6}. Sum of 2 smallest elements of each subset: a(6) = (1+2) + (1+2) + (1+2) + (1+2) + (1+3) + (2+3) = 21 = 3*C(6+1,6) = 3*A000579(6+1). - _Serhat Bulut_, Oktay Erkan Temizkan, Mar 13 2015
a(7) = 7 from the seven independent components of an antisymmetric tensor A of rank 6 and dimension 7: A(1,2,3,4,5,6), A(1,2,3,4,5,7), A(1,2,3,4,6,7), A(1,2,3,5,6,7) A(1,2,4,5,6,7), A(1,2,3,5,6,7) and A(2,3,4,5,6,7). See a Dec 10 2015 comment. - _Wolfdieter Lang_, Dec 10 2015

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.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 7.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • 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).
  • Charles W. Trigg: Mathematical Quickies. New York: Dover Publications, Inc., 1985, p. 11, #32

Links

Crossrefs

Cf. A053135, A053128, A000580 (partial sums), A000581, A000582, A000217, A000292, A000332, A000389 (first differences), A104712 (fifth column, k=6).

Programs

  • Magma
    [Binomial(n,6) : n in [0..50]]; // Wesley Ivan Hurt, Jul 13 2014
  • Maple
    A000579 := n->binomial(n,6);
ZL := [S, {S=Prod(B,B,B,B,B,B,B), B=Set(Z, 1 <= card)}, unlabeled]: seq(combstruct[count](ZL, size=n), n=7..40); # Zerinvary Lajos, Mar 13 2007
A000579:=-1/(z-1)**7; # Simon Plouffe in his 1992 dissertation, referring to offset 0.
seq(binomial(n,6),n=0..33); # Zerinvary Lajos, Jun 16 2008
G(x):=x^6*exp(x): f[0]:=G(x): for n from 1 to 39 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n]/6!,n=6..39); # Zerinvary Lajos, Apr 05 2009
  • Mathematica
    Table[Binomial[n, 6], {n, 6, 50}] (* Stefan Steinerberger, Apr 02 2006 *)
Table[n(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)/720, {n, 0, 100}] (* Artur Jasinski, Dec 02 2007 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,0,0,0,0,0,1},50] (* Harvey P. Dale, Dec 30 2012 *)
CoefficientList[ Series[ -7x^6/(x-1)^7,{x, 0, 35}], x]/7 (* Robert G. Wilson v, Jan 29 2015 *)
  • PARI
    a(n)=binomial(n,6) \\ Charles R Greathouse IV, Nov 20 2012
  • Python
    A000579_list, m = [], [1, -5, 10, -10, 5, -1, 0]
for _ in range(10**2):
    A000579_list.append(m[-1])
    for i in range(6):
        m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016

Formula

G.f.: x^6/(1-x)^7.
E.g.f.: exp(x)*x^6/720.
a(n) = (n^6 - 15*n^5 + 85*n^4 - 225*n^3 + 274*n^2 - 120*n)/720.
Conjecture: a(n+3) = Sum_{0 <= k, L, m <= n; k + L + m <= n} k*L*m. - Ralf Stephan, May 06 2005
Convolution of the nonnegative numbers (A001477) with the hexagonal numbers (A000389). Also convolution of the triangular numbers (A000217) with the tetrahedral numbers (A000292). - Sergio Falcon, Feb 12 2007
a(n) = n*(n - 1)*(n - 2)*(n - 3)*(n - 4)*(n - 5)/720. - Artur Jasinski, Dec 02 2007, R. J. Mathar, Jul 07 2009
Equals binomial transform of [1, 6, 15, 20, 15, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Aug 02 2008
a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 0, a(4) = 0, a(5) = 0, a(6) = 1, a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7). - Harvey P. Dale, Dec 30 2012
Sum_{n >= 0} a(n)/n! = e/720. Sum_{n >= 5} a(n)/(n-5)! = 4051*e/720. See A067653 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
Sum_{n >= 6} 1/a(n) = 6/5. - Hermann Stamm-Wilbrandt, Jul 13 2014
Sum_{n >= 6} (-1)^(n + 1)/a(n) = 192*log(2) - 661/5 = 0.8842586675... Also see A242023. - Richard R. Forberg, Aug 11 2014
a(n) = a(5-n) for all n in Z. - Michael Somos, Oct 07 2014
0 = a(n)*(+a(n+1) +5*a(n+2)) + a(n+1)*(-7*a(n+1) +a(n+2)) for all n in Z. - Michael Somos, Oct 07 2014
a(n) = 3*C(n+1,6) = 3*A000579(n+1). - Serhat Bulut, Oktay Erkan Temizkan, Mar 13 2015
a(n) = A000292(n-5)*A000292(n-2)/20. - R. J. Mathar, Nov 29 2015

Extensions

Some formulas that referred to other offsets corrected by R. J. Mathar, Jul 07 2009
I changed the offset to 0. This will require some further adjustments to the formulas. - N. J. A. Sloane, Aug 01 2010
Shevelev comment inserted and further adaptations to offset by R. J. Mathar, Aug 03 2010

A067764 Numerators of the coefficients in exp(x/(1-x)) power series.

Original entry on oeis.org

1, 1, 3, 13, 73, 167, 4051, 37633, 43817, 4596553, 58941091, 274691047, 12470162233, 202976401213, 1178339174801, 65573803186921, 99264170666917, 994319127823939, 588633468315403843, 13564373693588558173, 109232642628695218147, 752832094524169066031
Offset: 0

Views

Author

Benoit Cloitre, Feb 03 2002

Keywords

Comments

Define c(n) = a(n)/A067653(n). For a given sequence s(n) consider P[s(n)](z) := e^(-z/(1-z))*Sum_{k>=0} s(k)c(k)z^k. Regarding complex-valued abelian limitation the following holds true: if s(n) is convergent (to the limit s) then lim_{z->+1} P[s(n)](z)=s in a certain subdomain D of the unit circle. There are two constraints: (1) D contains the line [0,1[. (2) There is a d > 0 such that the intersection of {w|Re(w) > 1-d} and D is a nonempty subset of a generalized Stolz set defined by {w||Im(w)| <= t*(1 - Re(w))^(3/2)}, t < 1. If z tends to +1 from outside such a domain, that limit doesn't exist in general. - Hieronymus Fischer, Oct 20 2010
The ratio sequence given by c(n) = a(n)/A067653(n) also occurs in certain row and column sums related to Pascal's triangle, as in the two formulas given below. - Richard R. Forberg, Dec 26 2013

Examples

			Example for first formula. 1/1! + 3/2! + 3/3! + 1/4! = 73/24.
Example for 2nd formula. A000332 = 0, 0, 0, 0, 1, 5, 15, 35, 70, 126, ...; a(4) = 0/0! + 1/1! + 5/2! + 15/3! + 35/4! + 70/5! + 126/6! + ... = 73*e/24.
exp(x/(1-x)) = 1 + x + 3/2*x^2 + 13/6*x^3 + 73/24*x^4 + 167/40*x^5 + 4051/720*x^6 + 37633/5040*x^7 + 43817/4480*x^8 + 4596553/362880*x^9 + ... .

References

  • O. Perron, Über das infinitäre Verhalten der Koeffizienten einer gewissen Potenzreihe, Archiv d. Math. u. Phys. (3), Vol. 22, pp. 329-340, 1914.
  • H. Fischer, Eine Theorie komplexwertiger Abelscher Limitierungsmethoden (A theory of complex valued abelian limitation methods), Dissertation (1987), pp. 29-32.
  • K. Zeller, W. Beekmann, Theorie der Limitierungsverfahren, Springer-Verlag, Berlin (1970).

Links

Crossrefs

Cf. A067653.

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(x/(1-x)))); [Numerator(b[n]): n in [1..m]]; // G. C. Greubel, Dec 04 2018
  • Maple
    b:= proc(n) option remember; `if`(n=0, 1,
      add((n-k)*b(k), k=0..n-1)/n)
    end:
a:= n-> numer(b(n)):
seq(a(n), n=0..25);  # Alois P. Heinz, May 12 2016
  • Mathematica
    Table[Numerator@ SeriesCoefficient[Exp[x/(1 - x)], {x, 0, n}], {n, 19}] (* Michael De Vlieger, Dec 14 2015 *)
r[n_] := If[n == 0, 1, Hypergeometric1F1[1 - n, 2, -1]]; Table[Numerator@ r[n], {n, 0, 21}] (* Peter Luschny, Feb 02 2019 *)
  • PARI
    a(n) = numerator(sum(k=1, n, binomial(n-1, k-1)/k!)); \\ Altug Alkan, Dec 14 2015
  • SageMath
    [1] + [numerator(sum(binomial(n-1, j-1)/factorial(j) for j in (1..n))) for n in (1..30)] # G. C. Greubel, Dec 04 2018

Formula

a(n) is the numerator of Sum_{i=1..n} binomial(n-1, i-1)/i!.
a(n) is also the numerator of (Sum_{m>=0} binomial(n+m-1,n)/m!)/e, with A067653(n) as the denominator. See as example A000332 = binomial(n,4) below. - Richard R. Forberg, Dec 26 2013
a(n) = numerator(hypergeom([1 - n], [2], -1)) for n > 0. - Peter Luschny, Feb 02 2019

Extensions

a(0)=1 prepended by Alois P. Heinz, May 12 2016

A321939 Numerators in the asymptotic expansion of the Maclaurin coefficients of exp(x/(1-x)).

Original entry on oeis.org

1, -5, -479, -15313, 710401, -3532731539, -1439747442109, -34886932972781, -171887027703456763, -6317295244143234168127, -2059266220658860906379923, -16155159358654324183625719723, -125609753430605939189919003924509
Offset: 0

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Author

Richard P. Brent, Dec 05 2018

Keywords

Comments

If r(n) = A067764(n)/A067653(n) then r(n)/(exp(2*sqrt(n))/(2*n^(3/4)*sqrt(Pi*e))) has an asymptotic expansion in ascending powers of 1/sqrt(n) whose coefficients are rational numbers 1, -5/48, etc. The sequence gives the numerators of these rational numbers.
Another expression for r(n), n > 0, is r(n) = M(n+1,2,1)/e, where M(a,b,z) = 1F1(a;b;z) is a confluent hypergeometric function (Kummer function).
The same rational numbers, except for signs, occur in the asymptotic expansion of the Maclaurin coefficients of exp(1/(1-x))*E1(1/(1-x)), where E1(x) is an exponential integral. See Lemmas 1-2 and Theorem 5 of the preprint by Brent et al. (2018).

Examples

			The asymptotic expansion is 1 - 5*h/48 - 479*h^2/4608 - 15313*h^3/3317760 + ..., where h = 1/sqrt(n).

References

  • L. J. Slater, Confluent Hypergeometric Functions, Cambridge University Press, 1960.

Links

Crossrefs

The denominators are A321940. The formula for A321939(n)/A321940(n) in Theorem 5 of Brent et al. (2018) uses A321937(n)/A321938(n). The sequence A321941 can be defined using A321939 and A321940.

Formula

A formula is given in Theorem 5, and a recurrence in Lemma 7, of Brent et al. (2018).

A321940 Denominators in the asymptotic expansion of the Maclaurin coefficients of exp(x/(1-x)).

Original entry on oeis.org

1, 48, 4608, 3317760, 127401984, 214035333120, 308210879692800, 2958824445050880, 5680942934497689600, 134979204123665104896000, 18141205034220590098022400, 56600559706768241105829888000
Offset: 0

Views

Author

Richard P. Brent, Dec 08 2018

Keywords

Comments

If r(n) = A067764(n)/A067653(n) then r(n)/(exp(2*sqrt(n))/(2*n^(3/4)*sqrt(Pi*e))) has an asymptotic expansion in ascending powers of 1/sqrt(n) whose coefficients are rational numbers 1, -5/48, etc. The sequence gives the denominators of these rational numbers.
Another expression for r(n), n > 0, is r(n) = M(n+1,2,1)/e, where M(a,b,z) = 1F1(a;b;z) is a confluent hypergeometric function (Kummer function).
The same rational numbers, except for signs, occur in the asymptotic expansion of the Maclaurin coefficients of exp(1/(1-x))*E1(1/(1-x)), where E1(x) is an exponential integral. See Lemmas 1-2 and Theorem 5 of the preprint by Brent et al. (2018).

Examples

			The asymptotic expansion is 1 - 5*h/48 - 479*h^2/4608 - 15313*h^3/3317760 + ..., where h = 1/sqrt(n).

References

  • L. J. Slater, Confluent Hypergeometric Functions, Cambridge University Press, 1960.

Links

Crossrefs

The numerators are A321939. The formula in Theorem 5 of Brent et al. (2018) uses A321937(n)/A321938(n).

Formula

A formula is given in Theorem 5, and a recurrence in Lemma 7, of Brent et al. (2018).

A380271 Denominators of coefficients in expansion of exp(-1 + 1 / Product_{k>=1} (1 - x^k)).

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 720, 1008, 40320, 72576, 3628800, 39916800, 95800320, 6227020800, 3487131648, 1307674368000, 20922789888000, 2845499424768, 6402373705728000, 24329020081766400, 187146308321280000, 51090942171709440000, 224800145555521536000, 25852016738884976640000
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 18 2025

Keywords

Examples

			1, 1, 5/2, 31/6, 265/24, 2621/120, 31621/720, 85319/1008, 6574961/40320, ...

Crossrefs

Cf. A000041, A017666, A058892, A066186, A067653, A098988, A380171 (numerators).

Programs

  • Mathematica
    nmax = 23; CoefficientList[Series[Exp[-1 + 1/Product[1 - x^k, {k, 1, nmax}]], {x, 0, nmax}], x] // Denominator
b[0] = 1; b[n_] := b[n] = (1/n) Sum[k PartitionsP[k] b[n - k], {k, 1, n}]; a[n_] := Denominator[b[n]]; Table[a[n], {n, 0, 23}]

Formula

b(0) = 1, b(n) = (1/n) * Sum_{k=1..n} k * A000041(k) * b(n-k), a(n) = denominator of b(n).
Showing 1-7 of 7 results.

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