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

A001263 Triangle of Narayana numbers T(n,k) = C(n-1,k-1)*C(n,k-1)/k with 1 <= k <= n, read by rows. Also called the Catalan triangle.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 1, 15, 50, 50, 15, 1, 1, 21, 105, 175, 105, 21, 1, 1, 28, 196, 490, 490, 196, 28, 1, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 1, 55, 825, 4950, 13860, 19404, 13860, 4950, 825
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of antichains (or order ideals) in the poset 2*(k-1)*(n-k) or plane partitions with rows <= k-1, columns <= n-k and entries <= 2. - Mitch Harris, Jul 15 2000
T(n,k) is the number of Dyck n-paths with exactly k peaks. a(n,k) = number of pairs (P,Q) of lattice paths from (0,0) to (k,n+1-k), each consisting of unit steps East or North, such that P lies strictly above Q except at the endpoints. - David Callan, Mar 23 2004
Number of permutations of [n] which avoid-132 and have k-1 descents. - Mike Zabrocki, Aug 26 2004
T(n,k) is the number of paths through n panes of glass, entering and leaving from one side, of length 2n with k reflections (where traversing one pane of glass is the unit length). - Mitch Harris, Jul 06 2006
Antidiagonal sums given by A004148 (without first term).
T(n,k) is the number of full binary trees with n internal nodes and k-1 jumps. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump. - Emeric Deutsch, Jan 18 2007
From Gary W. Adamson, Oct 22 2007: (Start)
The n-th row can be generated by the following operation using an ascending row of (n-1) triangular terms, (A) and a descending row, (B); e.g., row 6:
A: 1....3....6....10....15
B: 15...10....6.....3.....1
C: 1...15...50....50....15....1 = row 6.
Leftmost column of A,B -> first two terms of C; then followed by the operation B*C/A of current column = next term of row C, (e.g., 10*15/3 = 50). Continuing with the operation, we get row 6: (1, 15, 50, 50, 15, 1). (End)
The previous comment can be upgraded to: The ConvOffsStoT transform of the triangular series; and by rows, row 6 is the ConvOffs transform of (1, 3, 6, 10, 15). Refer to triangle A117401 as another example of the ConvOffsStoT transform, and OEIS under Maple Transforms. - Gary W. Adamson, Jul 09 2012
For a connection to Lagrange inversion, see A134264. - Tom Copeland, Aug 15 2008
T(n,k) is also the number of order-decreasing and order-preserving mappings (of an n-element set) of height k (height of a mapping is the cardinal of its image set). - Abdullahi Umar, Aug 21 2008
Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p.60]. See A033282 for the corresponding array of f-vectors for associahedra of type A_n. See A008459 and A145903 for the h-vectors for associahedra of type B and type D respectively. The Hilbert transform of this triangle (see A145905 for the definition of this transform) is A145904. - Peter Bala, Oct 27 2008
T(n,k) is also the number of noncrossing set partitions of [n] into k blocks. Given a partition P of the set {1,2,...,n}, a crossing in P are four integers [a, b, c, d] with 1 <= a < b < c < d <= n for which a, c are together in a block, and b, d are together in a different block. A noncrossing partition is a partition with no crossings. - Peter Luschny, Apr 29 2011
Noncrossing set partitions are also called genus 0 partitions. In terms of genus-dependent Stirling numbers of the second kind S2(n,k,g) that count partitions of genus g of an n-set into k nonempty subsets, one has T(n,k) = S2(n,k,0). - Robert Coquereaux, Feb 15 2024
Diagonals of A089732 are rows of A001263. - Tom Copeland, May 14 2012
From Peter Bala, Aug 07 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = 1/sqrt(y)*BesselI(1,2*sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang.
Generating function E(y)*E(x*y) = 1 + (1 + x)*y/(1!*2!) + (1 + 3*x + x^2)*y^2/(2!*3!) + (1 + 6*x + 6*x^2 + x^3)*y^3/(3!*4!) + .... Cf. A105278 with a generating function exp(y)*E(x*y).
The n-th power of this array has a generating function E(y)^n*E(x*y). In particular, the matrix inverse A103364 has a generating function E(x*y)/E(y). (End)
T(n,k) is the number of nonintersecting n arches above the x axis, starting and ending on vertices 1 to 2n, with k being the number of arches starting on an odd vertice and ending on a higher even vertice. Example: T(3,2)=3 [16,25,34] [14,23,56] [12,36,45]. - Roger Ford, Jun 14 2014
Fomin and Reading on p. 31 state that the rows of the Narayana matrix are the h-vectors of the associahedra as well as its dual. - Tom Copeland, Jun 27 2017
The row polynomials P(n, x) = Sum_{k=1..n} T(n, k)*x^(k-1), together with P(0, x) = 1, multiplied by (n+1) are the numerator polynomials of the o.g.f.s of the diagonal sequences of the triangle A103371: G(n, x) = (n+1)*P(n, x)/(1 - x)^{2*n+1}, for n >= 0. This is proved with Lagrange's theorem applied to the Riordan triangle A135278 = (1/(1 - x)^2, x/(1 - x)). See an example below. - Wolfdieter Lang, Jul 31 2017
T(n,k) is the number of Dyck paths of semilength n with k-1 uu-blocks (pairs of consecutive up-steps). - Alexander Burstein, Jun 22 2020
In case you were searching for Narayama numbers, the correct spelling is Narayana. - N. J. A. Sloane, Nov 11 2020
Named after the Canadian mathematician Tadepalli Venkata Narayana (1930-1987). They were also called "Runyon numbers" after John P. Runyon (1922-2013) of Bell Telephone Laboratories, who used them in a study of a telephone traffic system. - Amiram Eldar, Apr 15 2021 The Narayana numbers were first studied by Percy Alexander MacMahon (see reference, Article 495) as pointed out by Bóna and Sagan (see link). - Peter Luschny, Apr 28 2022
From Andrea Arlette España, Nov 14 2022: (Start)
T(n,k) is the degree distribution of the paths towards synchronization in the transition diagram associated with the Laplacian system over the complete graph K_n, corresponding to ordered initial conditions x_1 < x_2 < ... < x_n.
T(n,k) for n=2N+1 and k=N+1 is the number of states in the transition diagram associated with the Laplacian system over the complete bipartite graph K_{N,N}, corresponding to ordered (x_1 < x_2 < ... < x_N and x_{N+1} < x_{N+2} < ... < x_{2N}) and balanced (Sum_{i=1..N} x_i/N = Sum_{i=N+1..2N} x_i/N) initial conditions. (End)
From Gus Wiseman, Jan 23 2023: (Start)
Also the number of unlabeled ordered rooted trees with n nodes and k leaves. See the link by Marko Riedel. For example, row n = 5 counts the following trees:
((((o)))) (((o))o) ((o)oo) (oooo)
(((o)o)) ((oo)o)
(((oo))) ((ooo))
((o)(o)) (o(o)o)
((o(o))) (o(oo))
(o((o))) (oo(o))
The unordered version is A055277. Leaves in standard ordered trees are counted by A358371. (End)

Examples

			The initial rows of the triangle are:
  [1] 1
  [2] 1,  1
  [3] 1,  3,   1
  [4] 1,  6,   6,    1
  [5] 1, 10,  20,   10,    1
  [6] 1, 15,  50,   50,   15,    1
  [7] 1, 21, 105,  175,  105,   21,   1
  [8] 1, 28, 196,  490,  490,  196,  28,  1
  [9] 1, 36, 336, 1176, 1764, 1176, 336, 36, 1;
  ...
For all n, 12...n (1 block) and 1|2|3|...|n (n blocks) are noncrossing set partitions.
Example of umbral representation:
  A007318(5,k)=[1,5/1,5*4/(2*1),...,1]=(1,5,10,10,5,1),
  so A001263(5,k)={1,b(5)/b(1),b(5)*b(4)/[b(2)*b(1)],...,1}
  = [1,30/2,30*20/(6*2),...,1]=(1,15,50,50,15,1).
  First = last term = b.(5!)/[b.(0!)*b.(5!)]= 1. - _Tom Copeland_, Sep 21 2011
Row polynomials and diagonal sequences of A103371: n = 4,  P(4, x) = 1 + 6*x + 6*x^2 + x^3, and the o.g.f. of fifth diagonal is G(4, x) = 5* P(4, x)/(1 - x)^9, namely [5, 75, 525, ...]. See a comment above. - _Wolfdieter Lang_, Jul 31 2017

References

  • Berman and Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), pp. 103-124.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 196.
  • P. A. MacMahon, Combinatory Analysis, Vols. 1 and 2, Cambridge University Press, 1915, 1916; reprinted by Chelsea, 1960, Sect. 495.
  • T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, pp. 100-101.
  • A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.
  • T. K. Petersen, Eulerian Numbers, Birkhäuser, 2015, Chapter 2.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 17.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.36(a) and (b).

Links

Crossrefs

Other versions are in A090181 and A131198. - Philippe Deléham, Nov 18 2007
Cf. variants: A181143, A181144. - Paul D. Hanna, Oct 13 2010
Row sums give A000108 (Catalan numbers), n>0.
Columns give A000217, A002415, A006542, A006857, A108679. A084938.
Cf. A000372, A002083, A056932, A056939, A056940, A056941, A065329, A073345.
A145596, A145597, A145598, A145599. - Peter Bala, Oct 22 2008
A008459 (h-vectors type B associahedra), A033282 (f-vectors type A associahedra), A145903 (h-vectors type D associahedra), A145904 (Hilbert transform). - Peter Bala, Oct 27 2008
Cf. A016098 and A189232 for numbers of crossing set partitions.
Cf. A243752.
Cf. A089231, A103371, A135278.
Cf. A002378, A007318, A010790, A089732, A105278, A119900, A132710, A134264.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.
Cf. A000081, A005043, A032027, A055277, A358371.

Programs

  • GAP
    Flat(List([1..11],n->List([1..n],k->Binomial(n-1,k-1)*Binomial(n,k-1)/k))); # Muniru A Asiru, Jul 12 2018
  • Haskell
    a001263 n k = a001263_tabl !! (n-1) !! (k-1)
a001263_row n = a001263_tabl !! (n-1)
a001263_tabl = zipWith dt a007318_tabl (tail a007318_tabl) where
   dt us vs = zipWith (-) (zipWith (*) us (tail vs))
                          (zipWith (*) (tail us ++ [0]) (init vs))
-- Reinhard Zumkeller, Oct 10 2013
  • Magma
    /* triangle */ [[Binomial(n-1,k-1)*Binomial(n,k-1)/k : k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 19 2014
  • Maple
    A001263 := (n,k)->binomial(n-1,k-1)*binomial(n,k-1)/k;
a:=proc(n,k) option remember; local i; if k=1 or k=n then 1 else add(binomial(n+i-1, 2*k-2)*a(k-1,i),i=1..k-1); fi; end:
# Alternatively, as a (0,0)-based triangle:
R := n -> simplify(hypergeom([-n, -n-1], [2], x)): Trow := n -> seq(coeff(R(n,x),x,j), j=0..n): seq(Trow(n), n=0..9); # Peter Luschny, Mar 19 2018
  • Mathematica
    T[n_, k_] := If[k==0, 0, Binomial[n-1, k-1] Binomial[n, k-1] / k];
Flatten[Table[Binomial[n-1,k-1] Binomial[n,k-1]/k,{n,15},{k,n}]] (* Harvey P. Dale, Feb 29 2012 *)
TRow[n_] := CoefficientList[Hypergeometric2F1[1 - n, -n, 2, x], x];
Table[TRow[n], {n, 1, 11}] // Flatten (* Peter Luschny, Mar 19 2018 *)
aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
Table[Length[Select[aot[n],Length[Position[#,{}]]==k&]],{n,2,9},{k,1,n-1}] (* Gus Wiseman, Jan 23 2023 *)
T[1, 1] := 1; T[n_, k_]/;1<=k<=n := T[n, k] = (2n/k-1) T[n-1,k-1] + T[n-1, k]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Dec 31 2024 *)
  • PARI
    {a(n, k) = if(k==0, 0, binomial(n-1, k-1) * binomial(n, k-1) / k)};
  • PARI
    {T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^2*y^j)*x^m/m) +O(x^(n+1))),n,x),k,y)} \\ Paul D. Hanna, Oct 13 2010
  • Sage
    @CachedFunction
def T(n, k):
    if k == n or k == 1: return 1
    if k <= 0 or k > n: return 0
    return binomial(n, 2) * (T(n-1, k)/((n-k)*(n-k+1)) + T(n-1, k-1)/(k*(k-1)))
for n in (1..9): print([T(n, k) for k in (1..n)])  # Peter Luschny, Oct 28 2014

Formula

a(n, k) = C(n-1, k-1)*C(n, k-1)/k for k!=0; a(n, 0)=0.
Triangle equals [0, 1, 0, 1, 0, 1, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is Deléham's operator defined in A084938.
0Mike Zabrocki, Aug 26 2004
T(n, k) = C(n, k)*C(n-1, k-1) - C(n, k-1)*C(n-1, k) (determinant of a 2 X 2 subarray of Pascal's triangle A007318). - Gerald McGarvey, Feb 24 2005
T(n, k) = binomial(n-1, k-1)^2 - binomial(n-1, k)*binomial(n-1, k-2). - David Callan, Nov 02 2005
a(n,k) = C(n,2) (a(n-1,k)/((n-k)*(n-k+1)) + a(n-1,k-1)/(k*(k-1))) a(n,k) = C(n,k)*C(n,k-1)/n. - Mitch Harris, Jul 06 2006
Central column = A000891, (2n)!*(2n+1)! / (n!*(n+1)!)^2. - Zerinvary Lajos, Oct 29 2006
G.f.: (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) = Sum_{n>0, k>0} a(n, k)*x^n*y^k.
From Peter Bala, Oct 22 2008: (Start)
Relation with Jacobi polynomials of parameter (1,1):
Row n+1 generating polynomial equals 1/(n+1)*x*(1-x)^n*Jacobi_P(n,1,1,(1+x)/(1-x)). It follows that the zeros of the Narayana polynomials are all real and nonpositive, as noted above. O.g.f for column k+2: 1/(k+1) * y^(k+2)/(1-y)^(k+3) * Jacobi_P(k,1,1,(1+y)/(1-y)). Cf. A008459.
T(n+1,k) is the number of walks of n unit steps on the square lattice (i.e., each step in the direction either up (U), down (D), right (R) or left (L)) starting from the origin and finishing at lattice points on the x axis and which remain in the upper half-plane y >= 0 [Guy]. For example, T(4,3) = 6 counts the six walks RRL, LRR, RLR, UDL, URD and RUD, from the origin to the lattice point (1,0), each of 3 steps. Compare with tables A145596 - A145599.
Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
The o.g.f. for this array is I(1 + t*x + t*x^2 + t*x^3 + ...) = 1 + t*x + (t + t^2)*x^2 + (t + 3*t^2 + t^3)*x^3 + ... = 1/(1 - x*t/(1 - x/(1 - x*t/(1 - x/(1 - ...))))) (as a continued fraction). Cf. A108767, A132081 and A141618. (End)
G.f.: 1/(1-x-xy-x^2y/(1-x-xy-x^2y/(1-... (continued fraction). - Paul Barry, Sep 28 2010
E.g.f.: exp((1+y)x)*Bessel_I(1,2*sqrt(y)x)/(sqrt(y)*x). - Paul Barry, Sep 28 2010
G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n ). - Paul D. Hanna, Oct 13 2010
With F(x,t) = (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2))/(2*x) an o.g.f. in x for the Narayana polynomials in t, G(x,t) = x/(t+(1+t)*x+x^2) is the compositional inverse in x. Consequently, with H(x,t) = 1/ (dG(x,t)/dx) = (t+(1+t)*x+x^2)^2 / (t-x^2), the n-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*D_x)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*D_u)u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 04 2011
With offset 0, A001263 = Sum_{j>=0} A132710^j / A010790(j), a normalized Bessel fct. May be represented as the Pascal matrix A007318, n!/[(n-k)!*k!], umbralized with b(n)=A002378(n) for n>0 and b(0)=1: A001263(n,k)= b.(n!)/{b.[(n-k)!]*b.(k!)} where b.(n!) = b(n)*b(n-1)...*b(0), a generalized factorial (see example). - Tom Copeland, Sep 21 2011
With F(x,t) = {1-(1-t)*x-sqrt[1-2*(1+t)*x+[(t-1)*x]^2]}/2 a shifted o.g.f. in x for the Narayana polynomials in t, G(x,t)= x/[t-1+1/(1-x)] is the compositional inverse in x. Therefore, with H(x,t)=1/(dG(x,t)/dx)=[t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, (see A119900), the (n-1)-th Narayana polynomial in t is given by (1/n!)*((H(x,t)*d/dx)^n)x evaluated at x=0, i.e., F(x,t) = exp(x*H(u,t)*d/du) u, evaluated at u = 0. Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 30 2011
T(n,k) = binomial(n-1,k-1)*binomial(n+1,k)-binomial(n,k-1)*binomial(n,k). - Philippe Deléham, Nov 05 2011
A166360(n-k) = T(n,k) mod 2. - Reinhard Zumkeller, Oct 10 2013
Damped sum of a column, in leading order: lim_{d->0} d^(2k-1) Sum_{N>=k} T(N,k)(1-d)^N=Catalan(n). - Joachim Wuttke, Sep 11 2014
Multiplying the n-th column by n! generates the revert of the unsigned Lah numbers, A089231. - Tom Copeland, Jan 07 2016
Row polynomials: (x - 1)^(n+1)*(P(n+1,(1 + x)/(x - 1)) - P(n-1,(1 + x)/(x - 1)))/((4*n + 2)), n = 1,2,... and where P(n,x) denotes the n-th Legendre polynomial. - Peter Bala, Mar 03 2017
The coefficients of the row polynomials R(n, x) = hypergeom([-n,-n-1], [2], x) generate the triangle based in (0,0). - Peter Luschny, Mar 19 2018
Multiplying the n-th diagonal by n!, with the main diagonal n=1, generates the Lah matrix A105278. With G equal to the infinitesimal generator of A132710, the Narayana triangle equals Sum_{n >= 0} G^n/((n+1)!*n!) = (sqrt(G))^(-1) * I_1(2*sqrt(G)), where G^0 is the identity matrix and I_1(x) is the modified Bessel function of the first kind of order 1. (cf. Sep 21 2011 formula also.) - Tom Copeland, Sep 23 2020
T(n,k) = T(n,k-1)*C(n-k+2,2)/C(k,2). - Yuchun Ji, Dec 21 2020
From Sergii Voloshyn, Nov 25 2024: (Start)
G.f.: F(x,y) = (1-x*(1+y)-sqrt((1-x*(1+y))^2-4*y*x^2))/(2*x) is the solution of the differential equation x^3 * d^2(x*F(x,y))/dx^2 = y * d^2(x*F(x,y))/dy^2.
Let E be the operator x*D*D, where D denotes the derivative operator d/dx. Then (1/(n! (1 + n)!)) * E^n(x/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} C(n-1, k-1)*C(n, k-1)/k*x^k. For example, when n = 4 we have (1/4!/5!)*E^3(x/(1 - x)) = x (1 + 6 x + 6 x^2 + x^3)/(1 - x)^9. (End)

Extensions

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A056939 Array read by antidiagonals: number of antichains (or order ideals) in the poset 3*m*n or plane partitions with rows <= m, columns <= n and entries <= 3.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 10, 10, 1, 1, 20, 50, 20, 1, 1, 35, 175, 175, 35, 1, 1, 56, 490, 980, 490, 56, 1, 1, 84, 1176, 4116, 4116, 1176, 84, 1, 1, 120, 2520, 14112, 24696, 14112, 2520, 120, 1, 1, 165, 4950, 41580, 116424, 116424, 41580, 4950, 165, 1
Offset: 0

Views

Author

Mitch Harris

Keywords

Comments

Triangle of generalized binomial coefficients (n,k)A342889.%20This%20array%20is%20the%20main%20subject%20of%20the%20long%20article%20by%20Felsner%20et%20al.%20(2011).%20-%20_N.%20J.%20A.%20Sloane">3; cf. A342889. This array is the main subject of the long article by Felsner et al. (2011). - _N. J. A. Sloane, Apr 03 2021
This triangle is mentioned by Hoggatt (1977). - N. J. A. Sloane, Mar 27 2021
Determinants of 3 X 3 subarrays of Pascal's triangle A007318 (a matrix entry being set to 0 when not present). - Gerald McGarvey, Feb 24 2005
Also determinants of 3 X 3 arrays whose entries come from a single row: T(n,k) = det [C(n,k),C(n,k-1),C(n,k-2); C(n,k+1),C(n,k),C(n,k-1); C(n,k+2),C(n,k+1),C(n,k)]. - Peter Bala, May 10 2012
From Gary W. Adamson, Jul 10 2012: (Start)
The triangular view of this triangle is
1;
1, 1;
1, 4, 1;
1, 10, 10, 1;
1, 20, 50, 20, 1;
The n-th row of this triangle is generated by applying the ConvOffs transform to the first n terms of 1, 4, 10, 20, ... (A000292 without leading zero). See A214281 for a procedural definition of the transformation and search "ConvOffs" for more examples. (End)
Define polynomials p(n, x) = hypergeom([-1 - n, -n, 1 - n], [2, 3], -x). If the triangle is extended by the diagonal 1, 0, 0,... on the right side the resulting (0, 0)-based triangle is T*(n, k) = [x^k] p(n, x). The polynomials evaluated at x = 1 gives the number of Baxter permutations of length n (see the formula given by Richard L. Ollerton in A001181). - Peter Luschny, Dec 28 2022

Examples

			The initial rows of the array are:
     1      1      1      1      1      1 ...
     1      4     10     20     35     56 ...
     1     10     50    175    490   1176 ...
     1     20    175    980   4116  14112 ...
     1     35    490   4116  24696 116424 ...
     1     56   1176  14112 116424 731808 ...
     ...
Considered as a triangle, the initial rows are:
  [1],
  [1, 1],
  [1, 4, 1],
  [1, 10, 10, 1],
  [1, 20, 50, 20, 1],
  [1, 35, 175, 175, 35, 1],
  [1, 56, 490, 980, 490, 56, 1],
  [1, 84, 1176, 4116, 4116, 1176, 84, 1],
  [1, 120, 2520, 14112, 24696, 14112, 2520, 120, 1],
  [1, 165, 4950, 41580, 116424, 116424, 41580, 4950, 165, 1],
  [1, 220, 9075, 108900, 457380, 731808, 457380, 108900, 9075, 220, 1]
  ...

References

  • Berman and Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), p. 103-124
  • R. P. Stanley, Theory and application of plane partitions. II. Studies in Appl. Math. 50 (1971), p. 259-279. Thm. 18.1

Links

Crossrefs

Cf. A000372, A056932, A001263, A056940, A056941.
Antidiagonals sum to A001181 (Baxter permutations). Cf. A197208.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1..12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Maple
    # To get initial terms of the array - N. J. A. Sloane, Apr 20 2021
bb := (k,l) -> binomial(k+l,k)*binomial(k+l+1,k)*binomial(k+l+2,k)*2/((k+1)^2*(k+2));
for k from 0 to 8 do
lprint([seq(bb(k,l),l=0..8)]);
od:
  • Mathematica
    t[n_, m_] = 2*Binomial[n, m]*Binomial[n + 1, m + 1]* Binomial[n + 2, m + 2]/((n - m + 1)^2*(n - m + 2)); Flatten[Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]] (* Roger L. Bagula, Jan 28 2009 *)
  • PARI
    \\ cf. A359363
C=binomial;
T(n,k)=if(n==0&&k==0,1,(C(n+1,k-1)*C(n+1,k)*C(n+1,k+1))/(C(n+1,1)*C(n+1,2)));
for(n=1,10,for(k=1,n,print1(T(n,k),", "));print()); \\ Joerg Arndt, Jan 04 2024

Formula

Product_{k=0..2} binomial(n+m+k, m+k)/binomial(n+k, k) gives the array as a square.
T(n,m) = 2*binomial(n, m)*binomial(n+1, m+1)*binomial(n+2, m+2)/((n-m+1)^2*(n-m+2)). - Roger L. Bagula, Jan 28 2009
From Peter Bala, Oct 13 2011: (Start)
T(n,k) = 2/((n+1)*(n+2)*(n+3))*C(n+1,k)*C(n+2,k+2)*C(n+3,k+1);
T(n,k) = 2/((n+1)*(n+2)*(n+3))*C(n+1,k+1)*C(n+2,k)*C(n+3,k+2). Cf. A197208.
T(n-1,k-1)*T(n,k+1)*T(n+1,k) = T(n-1,k)*T(n,k-1)*T(n+1,k+1).
Define a(r,n) = n!*(n+1)!*...*(n+r)!. The triangle whose (n,k)-th entry is a(r,0)*a(r,n)/(a(r,k)*a(r,n-k)) is A007318 (r = 0), A001263 (r = 1), A056939 (r = 2), A056940 (r = 3) and A056941 (r = 4). (End)
The column generating functions of the square array (starting at column 1) are 1/(1 - x)^4, (1 + 3*x + x^2)/(1 - x)^7, (1 + 10*x + 20*x^2 + 10*x^3 + x^4)/(1 - x)^10, ..., where the numerator polynomials are the row polynomials of A087647. See Barry p. 31. - Peter Bala, Oct 18 2023

A056941 Number of antichains (or order ideals) in the poset 5*m*n or plane partitions with not more than m rows, n columns and entries <= 5.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 21, 21, 1, 1, 56, 196, 56, 1, 1, 126, 1176, 1176, 126, 1, 1, 252, 5292, 14112, 5292, 252, 1, 1, 462, 19404, 116424, 116424, 19404, 462, 1, 1, 792, 60984, 731808, 1646568, 731808, 60984, 792, 1, 1, 1287, 169884, 3737448, 16818516, 16818516, 3737448, 169884, 1287, 1
Offset: 0

Views

Author

Mitch Harris

Keywords

Comments

Triangle of generalized binomial coefficients (n,k)A342889.%20-%20_N.%20J.%20A.%20Sloane">5; cf. A342889. - _N. J. A. Sloane, Apr 03 2021

Examples

			The array starts:
  [1    1      1        1          1           1            1 ...]
  [1    6     21       56        126         252          462 ...]
  [1   21    196     1176       5292       19404        60984 ...]
  [1   56   1176    14112     116424      731808      3737448 ...]
  [1  126   5292   116424    1646568    16818516    133613766 ...]
  [1  252  19404   731808   16818516   267227532   3184461423 ...]
  [1  462  60984  3737448  133613766  3184461423  55197331332 ...]
  [...]
Considered as a triangle, the initial rows are:
   1;
   1,   1;
   1,   6,     1;
   1,  21,    21,      1;
   1,  56,   196,     56,       1;
   1, 126,  1176,   1176,     126,      1;
   1, 252,  5292,  14112,    5292,    252,     1;
   1, 462, 19404, 116424,  116424,  19404,   462,   1;
   1, 792, 60984, 731808, 1646568, 731808, 60984, 792, 1; ...

References

  • Berman and Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), p. 103-124
  • P. A. MacMahon, Combinatory Analysis, Section 495, 1916.
  • R. P. Stanley, Theory and application of plane partitions. II. Studies in Appl. Math. 50 (1971), p. 259-279. Thm. 18.1

Links

Crossrefs

Cf. A000372, A001263, A005363 (row sums), A056932, A056939, A056940, A142465, A142467, A142468.
Antidiagonals sum to A005363 (Hoggatt sequence).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Magma
    A056941:= func< n,k | (&*[Binomial(n+j,k)/Binomial(k+j,k): j in [0..4]]) >;
[A056941(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 14 2022
  • Mathematica
    T[n_, k_] := Product[Binomial[n+j, k]/Binomial[k+j, k], {j,0,4}];
Table[T[n, k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 14 2022 *)
  • PARI
    A056941(n,m)=prod(k=0,4,binomial(n+m+k,m+k)/binomial(n+k,k)) \\ as an array \\ M. F. Hasler, Sep 26 2018
  • SageMath
    def A056941(n,k): return product(binomial(n+j,k)/binomial(k+j,k) for j in (0..4))
flatten([[A056941(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Nov 14 2022

Formula

From Peter Bala, Oct 13 2011: (Start)
A(n, k) = Product_{j=0..4} C(n+k+j, k+j)/C(n+j, j) gives the array as a square.
g(n-1, k-1)*g(n, k+1)*g(n+1, k) = g(n-1, k)*g(n, k-1)*g(n+1, k+1) where g(n, k) is the array A(n, k) and triangle T(n, k).
Define f(r,n) = n!*(n+1)!*...*(n+r)!. The triangle whose (n,k)-th entry is f(r,0)*f(r,n)/(f(r,k)*f(r,n-k)) is A007318 (r = 0), A001263 (r = 1), A056939 (r = 2), A056940 (r = 3) and A056941 (r = 4). (End)
From Peter Bala, May 10 2012: (Start)
Determinants of 5 X 5 subarrays of Pascal's triangle A007318 (a matrix entry being set to 0 when not present).
Also determinants of 5 X 5 arrays whose entries come from a single row:
det [C(n,k), C(n,k-1), C(n,k-2), C(n,k-3), C(n,k-4); C(n,k+1), C(n,k), C(n,k-1), C(n,k-2), C(n,k-3); C(n,k+2), C(n,k+1), C(n,k), C(n,k-1), C(n,k-2); C(n,k+3), C(n,k+2), C(n,k+1), C(n,k), C(n,k-1); C(n,k+4), C(n,k+3), C(n,k+2), C(n,k+1), C(n,k)]. (End)
From G. C. Greubel, Nov 14 2022: (Start)
T(n, k) = Product_{j=0..4} binomial(n+j, k)/binomial(k+j, k) (gives the triangle).
Sum_{k=0..n} T(n, k) = A005363(n). (End)

Extensions

Edited by M. F. Hasler, Sep 26 2018

A056940 Number of antichains (or order ideals) in the poset 4*m*n or plane partitions with at most m rows and n columns and entries <= 4.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 15, 15, 1, 1, 35, 105, 35, 1, 1, 70, 490, 490, 70, 1, 1, 126, 1764, 4116, 1764, 126, 1, 1, 210, 5292, 24696, 24696, 5292, 210, 1, 1, 330, 13860, 116424, 232848, 116424, 13860, 330, 1, 1, 495, 32670, 457380, 1646568, 1646568, 457380, 32670, 495, 1
Offset: 0

Views

Author

Mitch Harris

Keywords

Comments

Triangle of generalized binomial coefficients (n,k)A342889.%20-%20_N.%20J.%20A.%20Sloane">4; cf. A342889. - _N. J. A. Sloane, Apr 03 2021
Determinants of 4 X 4 subarrays of Pascal's triangle A007318 (a matrix entry being set to 0 when not present). - Gerald McGarvey, Feb 24 2005
Row sums are: {1, 2, 7, 32, 177, 1122, 7898, 60398, 494078, 4274228, 38763298, ...}. - Roger L. Bagula, Mar 08 2010
Also determinants of 4x4 arrays whose entries come from a single row: T(n,k) = det [C(n,k), C(n,k-1), C(n,k-2), C(n,k-3); C(n,k+1), C(n,k), C(n,k-1), C(n,k-2); C(n,k+2), C(n,k+1), C(n,k), C(n,k-1); C(n,k+3), C(n,k+2), C(n,k+1), C(n,k)]. - Peter Bala, May 10 2012

Examples

			Triangle begins as:
  1.
  1,   1.
  1,   5,     1.
  1,  15,    15,      1.
  1,  35,   105,     35,      1.
  1,  70,   490,    490,     70,      1.
  1, 126,  1764,   4116,   1764,    126,     1.
  1, 210,  5292,  24696,  24696,   5292,   210,   1.
  1, 330, 13860, 116424, 232848, 116424, 13860, 330, 1. - _Roger L. Bagula_, Mar 08 2010

Links

Crossrefs

Cf. A000372, A056932, A001263.
Antidiagonals sum to A005362 (Hoggatt sequence).
Cf. A056939 (q=2), A056940 (q=3), A056941 (q=4), A142465 (q=5), A142467 (q=6), A142468 (q=7), A174109 (q=8).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Mathematica
    c[n_, q_] = Product[i + j, {j, 0, q}, {i, 1, n}];
T[n_, m_, q_] = c[n, q]/(c[m, q]*c[n - m, q]);
Table[T[n, k, 3], {n, 0, 10}, {k, 0, n}]//Flatten (* Roger L. Bagula, Mar 08 2010 *)(* modified by G. C. Greubel, Apr 13 2019 *)
  • PARI
    A056940(n,m)=prod(k=0,3,binomial(n+m+k,m+k)/binomial(n+k,k)) \\ M. F. Hasler, Sep 26 2018

Formula

Product_{k=0..3} C(n+m+k, m+k)/C(n+k, k) gives the array as a square.
T(n,m,q) = c(n,q)/(c(m,q)*c(n-m,q)) with c(n,q) = Product_{i=1..n, j=0..q} (i + j), q = 3. - Roger L. Bagula, Mar 08 2010
From Peter Bala, Oct 13 2011: (Start)
T(n-1,k-1)*T(n,k+1)*T(n+1,k) = T(n-1,k)*T(n,k-1)*T(n+1,k+1).
Define f(r,n) = n!*(n+1)!*...*(n+r)!. The triangle whose (n,k)-th entry is f(r,0)*f(r,n)/(f(r,k)*f(r,n-k)) is A007318 (r = 0), A001263 (r = 1), A056939 (r = 2), A056940 (r = 3) and A056941 (r = 4). (End)

Extensions

Edited by M. F. Hasler, Sep 26 2018

A142467 Triangle T(n,m) read by rows: T(n,m) = Product_{i=0..6} binomial(n+i,m)/binomial(m+i,m).

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 36, 36, 1, 1, 120, 540, 120, 1, 1, 330, 4950, 4950, 330, 1, 1, 792, 32670, 108900, 32670, 792, 1, 1, 1716, 169884, 1557270, 1557270, 169884, 1716, 1, 1, 3432, 736164, 16195608, 44537922, 16195608, 736164, 3432, 1
Offset: 0

Views

Author

Roger L. Bagula, Sep 20 2008

Keywords

Comments

Triangle of generalized binomial coefficients (n,k)A342889.%20-%20_N.%20J.%20A.%20Sloane">7; cf. A342889. - _N. J. A. Sloane, Apr 03 2021

Examples

			Triangle begins as:
  1;
  1,     1;
  1,     8,       1;
  1,    36,      36,         1;
  1,   120,     540,       120,         1;
  1,   330,    4950,      4950,       330,         1;
  1,   792,   32670,    108900,     32670,       792,         1;
  1,  1716,  169884,   1557270,   1557270,    169884,      1716,       1;
  1,  3432,  736164,  16195608,  44537922,  16195608,    736164,    3432,    1;
  1,  6435, 2760615, 131589315, 868489479, 868489479, 131589315, 2760615, 6435, 1;

Links

Crossrefs

Cf. A001263, A005365 (row sums), A056939, A056940, A056941.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Magma
    [(&*[Binomial(n+j,k)/Binomial(k+j,k): j in [0..6]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 13 2022
  • Mathematica
    T[n_, k_]:= Product[Binomial[n+j, k]/Binomial[k+j, k], {j,0,6}];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Nov 13 2022 *)
  • PARI
    T(n, k) = prod(j=0, 6, binomial(n+j, k)/binomial(k+j, k)); \\ Seiichi Manyama, Apr 01 2021
  • SageMath
    def A142467(n,k): return product(binomial(n+j,k)/binomial(k+j,k) for j in (0..6))
flatten([[A142467(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Nov 13 2022

Formula

T(n,m) = A142465(n,m)*binomial(n+6,m)/binomial(m+6,m).
Sum_{k=0..n} T(n, k) = A005365(n).

Extensions

Edited by the Associate Editors of the OEIS, May 17 2009

A142468 An eight-products triangle sequence of coefficients: T(n,k) = binomial(n,k) * Product_{j=1..7} j!*(n+j)!/((k+j)!*(n-k+j)!).

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 45, 45, 1, 1, 165, 825, 165, 1, 1, 495, 9075, 9075, 495, 1, 1, 1287, 70785, 259545, 70785, 1287, 1, 1, 3003, 429429, 4723719, 4723719, 429429, 3003, 1, 1, 6435, 2147145, 61408347, 184225041, 61408347, 2147145, 6435, 1
Offset: 0

Views

Author

Roger L. Bagula, Sep 20 2008

Keywords

Comments

Triangle of generalized binomial coefficients (n,k)A342889.%20-%20_N.%20J.%20A.%20Sloane">8; cf. A342889. - _N. J. A. Sloane, Apr 03 2021
Row sums are {1, 2, 11, 92, 1157, 19142, 403691, 10312304, 311348897, 10826298914, 426196716090, ...}.
The general function is T(n,m)L = binomial(n,m)*Product{k=1..L} k!*(n + k)!/((m + k)!*(n - m + k)!) to give the quadratic row {1, L+2, 1}.

Examples

			Triangle begins as:
  1;
  1,    1;
  1,    9,       1;
  1,   45,      45,        1;
  1,  165,     825,      165,         1;
  1,  495,    9075,     9075,       495,        1;
  1, 1287,   70785,   259545,     70785,     1287,       1;
  1, 3003,  429429,  4723719,   4723719,   429429,    3003,    1;
  1, 6435, 2147145, 61408347, 184225041, 61408347, 2147145, 6435, 1;

Links

Crossrefs

Cf. A001263, A056939, A056940, A056941.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Magma
    B:=Binomial; [B(n,k)*(&*[B(n+2*j, k+j)/B(n+2*j, j): j in [1..7]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Nov 14 2019, Mar 03 2021
  • Maple
    b:= binomial;
T:= (n, k) -> b(n, k)*mul(b(n+2*j, k+j)/b(n+2*j, j), j = 1..7);
seq(seq(T(n, k), k = 0..n), n = 0..10); # G. C. Greubel, Nov 14 2019, Mar 03 2021
  • Mathematica
    T[n_, k_]:= T[n,k]= With[{B=Binomial}, B[n,k]* Product[B[n+2*j,k+j]/B[n+2*j,j], {j, 7}] ];
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Nov 14 2019, Mar 03 2021 *)
  • PARI
    T(n,k) = b=binomial; b(n,k)*prod(j=1,7, b(n+ 2*j,k+j)/b(n+2*j,j)); \\ G. C. Greubel, Nov 14 2019, Mar 03 2021
  • Sage
    b=binomial;
def T(n, k): return b(n,k)*product(b(n+2*j,k+j)/b(n+2*j,j) for j in (1..7))
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Nov 14 2019, Mar 03 2021

Formula

T(n,k) = binomial(n,k)*Product_{j=1..7} j!*(n+j)!/((k+j)!*(n-k+j)!).

Extensions

Edited by G. C. Greubel, Nov 14 2019

A174109 Triangle read by rows: T(n, k) = c(n, q)/(c(k, q)*c(n-k, q)), where c(n, q) = Product_{j=1..n} (j+q)!/(j-1)! and q = 8.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 1, 55, 55, 1, 1, 220, 1210, 220, 1, 1, 715, 15730, 15730, 715, 1, 1, 2002, 143143, 572572, 143143, 2002, 1, 1, 5005, 1002001, 13026013, 13026013, 1002001, 5005, 1, 1, 11440, 5725720, 208416208, 677352676, 208416208, 5725720, 11440, 1
Offset: 0

Views

Author

Roger L. Bagula, Mar 08 2010

Keywords

Comments

Triangle of generalized binomial coefficients (n,k)A342889.%20-%20_N.%20J.%20A.%20Sloane">9; cf. A342889. - _N. J. A. Sloane, Apr 03 2021
Row sums are: {1, 2, 12, 112, 1652, 32892, 862864, 28066040, 1105659414, 51177188350, 2734044648194, ...}.
These sequences (q >= 2) are a generalization of A056939.

Examples

			Triangle begins as:
  1.
  1,     1.
  1,    10,       1.
  1,    55,      55,         1.
  1,   220,    1210,       220,         1.
  1,   715,   15730,     15730,       715,         1.
  1,  2002,  143143,    572572,    143143,      2002,       1.
  1,  5005, 1002001,  13026013,  13026013,   1002001,    5005,     1.
  1, 11440, 5725720, 208416208, 677352676, 208416208, 5725720, 11440, 1.

Links

Crossrefs

Cf. A056939 (q=2), A056940 (q=3), A056941 (q=4), A142465 (q=5), A142467 (q=6), A142468 (q=7), this sequence (q=8).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Mathematica
    c[n_, q_]:= Product[i+j, {j, 0, q}, {i, 1, n}];
T[n_, m_, q_] = c[n, q]/(c[m, q]*c[n - m, q]);
Table[T[n, k, 8], {n, 0, 10}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, Apr 13 2019 *)
T[n_, k_, q_]:= (BarnesG[n+q+2]*BarnesG[k+1]*BarnesG[n-k+1]*BarnesG[q+2] )/(BarnesG[n-k+q+2]*BarnesG[k+q+2]*BarnesG[n+1]);
Table[T[n, k, 8], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Apr 13 2019 *)
  • PARI
    {c(m,q) = prod(j=1,m, (j+q)!/(j-1)!)};
for(n=0,10, for(k=0,n, print1(c(n, 8)/(c(k, 8)*c(n-k, 8)), ", "))) \\ G. C. Greubel, Apr 13 2019

Formula

T(n, k, q) = c(n, q)/(c(k, q)*c(n-k, q)) where c(n, q) = Product_{j=1..n} (j+q)!/(j-1)! and q=8.
T(n, k, q) = (G(q+2)*G(k+1)*G(n+q+2)*G(n-k+1))/(G(n+1)*G(k+q+2)*G(n-k+q+ 2)), where G(x) is the Barnes G-function (see A000178). - G. C. Greubel, Apr 13 2019

Extensions

Edited by G. C. Greubel, Apr 13 2019

A342889 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_10 (n >= 0, 0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 11, 1, 1, 66, 66, 1, 1, 286, 1716, 286, 1, 1, 1001, 26026, 26026, 1001, 1, 1, 3003, 273273, 1184183, 273273, 3003, 1, 1, 8008, 2186184, 33157124, 33157124, 2186184, 8008, 1, 1, 19448, 14158144, 644195552, 2254684432, 644195552, 14158144, 19448, 1
Offset: 0

Views

Author

N. J. A. Sloane, Apr 01 2021

Keywords

Examples

			Triangle begins:
  [1],
  [1, 1],
  [1, 11, 1],
  [1, 66, 66, 1],
  [1, 286, 1716, 286, 1],
  [1, 1001, 26026, 26026, 1001, 1],
  [1, 3003, 273273, 1184183, 273273, 3003, 1],
  [1, 8008, 2186184, 33157124, 33157124, 2186184, 8008, 1],
...

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Links

Crossrefs

Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Maple
    # Generalized binomial coefficient:
GBC := proc(n,k,m) local a,j;
a := mul((binomial(n+m-j,m)/binomial(j+m-1,m)),j=1..k);
end;
# Returns first M rows of triangle:
GBCT := proc(m,M) local a,b,n,k; global GBC;
a:=[];
for n from 0 to M do
  b:=[seq(GBC(n,k,m),k=0..n)];
  a:=[op(a),b];
od: a; end;
GBCT(10,12);
  • Mathematica
    f[n_, k_, m_] := Product[Binomial[n + m - j, m]/Binomial[j + m - 1, m], {j, k}]; Table[f[n, k, 10], {n, 0, 8}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 25 2023 *)
  • PARI
    f(n, k, m) = prod(j=1, k, binomial(n-j+m, m)/binomial(j-1+m, m));
T(n, k) = f(n, k, 10); \\ Seiichi Manyama, Apr 02 2021

Formula

The generalized binomial coefficient (n,k)m = Product{j=1..k} binomial(n+m-j,m)/binomial(j+m-1,m).

A342890 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_11 (n >= 0, 0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 12, 1, 1, 78, 78, 1, 1, 364, 2366, 364, 1, 1, 1365, 41405, 41405, 1365, 1, 1, 4368, 496860, 2318680, 496860, 4368, 1, 1, 12376, 4504864, 78835120, 78835120, 4504864, 12376, 1, 1, 31824, 32821152, 1837984512, 6892441920, 1837984512, 32821152, 31824, 1
Offset: 0

Views

Author

N. J. A. Sloane, Apr 01 2021

Keywords

Comments

For references, links, programs, etc., see earlier sequences in this series, especially A342889.

Examples

			Triangle begins:
  [1],
  [1, 1],
  [1, 12, 1],
  [1, 78, 78, 1],
  [1, 364, 2366, 364, 1],
  [1, 1365, 41405, 41405, 1365, 1],
  [1, 4368, 496860, 2318680, 496860, 4368, 1],
  [1, 12376, 4504864, 78835120, 78835120, 4504864, 12376, 1],
  [1, 31824, 32821152, 1837984512, 6892441920, 1837984512, 32821152, 31824, 1],
...

Links

Crossrefs

Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • PARI
    f(n, k, m) = prod(j=1, k, binomial(n-j+m, m)/binomial(j-1+m, m));
T(n, k) = f(n, k, 11); \\ Seiichi Manyama, Apr 02 2021

Formula

The generalized binomial coefficient (n,k)m = Product{j=1..k} binomial(n+m-j,m)/binomial(j+m-1,m).
Showing 1-10 of 15 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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