

A007318


Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(nk)!), 0 <= k <= n.
(Formerly M0082)


2037



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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,5


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 Shihso suanshu circa 1100. Another prominent early use was in Chu ShihChieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, AlKaraji discovered the binomial triangle "some time soon after 1007", and AlSamawal published it in the Albahir 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 kelement subsets of an nelement set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k1,n1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument  see Feller).
Binomial(n1,k1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k1,k1) 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+k1,n1) 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(i1,j) + A(i,j1) 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(n3,k1) 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(n3,k1) 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
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 kdigit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n1>=2 gives the number of kdigit (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+k1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k1, 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+31,3)=10 3digit numbers whose digits sum to 3 (see A052217) and also binomial(4+21,2)=10 4digit 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)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{nk}(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_{n1k}(x_1,x_2,...,x_{n1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n1).
Binomial(2n,2k) = sigma_{nk}(x_1,x_2,...,x_n), where x_i = tan^2((2i1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{nk}(x_1,x_2,...,x_n), where x_i = tan^2((2i1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(nk) and S(n,k) = binomial(n,k)*s(nk), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(nk), 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+n2,m1), 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
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,nk)}binomial(k,j)*binomial(nk,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 ConwaySloane, 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*(nk))!) 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(n1,m) is the right column, and the other entries are T(m,k) = T(m1,k1) + T(m1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n1).  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_(n1)(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
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,...,n1. 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+k1,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/(k1)  Sum_{n=k..m} 1/binomial(n,k))) = m!/((mk+1)!*(k2)!). Note: k/(k1) 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
The binomial coefficients binomial(n,k) give the number of individuals of the kth 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^(n1)  1 to the right of the pattern from 0 to 2^(n1)  1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1'scounting 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)
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.
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 kth 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,k1)}, including all of the elements of {A(n,k1)}. 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 kth diagonal of Pascal's Triangle. See examples.  Gregory L. Simay, Aug 06 2018
Binomial(n1,k) is also the number of permutations avoiding both 213 and 312 with k ascents.  Lara Pudwell, Dec 19 2018
Binomial(n1,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 kth exterior power of a vector space of dimension n.  Stefano Spezia, Dec 22 2018
C(n,k1) is the number of unoriented colorings of the facets (or vertices) of an ndimensional 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 wellknown characterization of the prime numbers: Consider the entries in the kth 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 (16231662).  Amiram Eldar, Jun 11 2021
Consider the nth 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 nth 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 selfinverse operation, and therefore the converse is true.  Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an ndimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1dedges) from a given vertex.  Eitan Y. Levine, May 01 2023
C(n+k1,k1) is the number of vertices at an L1 distance from a given vertex in an infinitedimensional 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+k1,k1) is the number of subsets of tokens with a total value of n.  Eitan Y. Levine, Jun 11 2023
Numbers in the kth column, i.e., numbers of the form C(n,k) for n >= k, are known as ksimplex numbers.  Pontus von Brömssen, Jun 26 2023
Let r(k) be the kth row and c(k) the kth 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
GCD( {binomial(n,k)}_{k=1..n1} ) = A014963(n), i.e. the GCD is p if n = p^b for b >= 1 and prime p, otherwise the GCD is 1. See the Balak Ram (1909) link (proving this claim, but not available online), the Carl McTague (2015) link (proving a generalization), and the "PDF formula proof" linked file.  Eitan Y. Levine, Aug 09 2023
The following alternating sum can produce any integer power: Sum_{i=0..nCeiling(n/k)} (1)^i * C((ni)k,n) * C(n,i) = k^n. Interestingly, replacing the second and fourth appearances of n in this formula by m for any 0 <= m < n, makes the sum zero: Sum_{i=0..nCeiling(m/k)} (1)^i * C((ni)k,m) * C(n,i) = 0.  Eitan Y. Levine, Aug 31 2023


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. 6874.
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 5648007388.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
P. 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. AddisonWesley, Reading, MA, 2nd. ed., 1994, p. 155.
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. 6061.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
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, 19861992.
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, AddisonWesley, Reading, MA, 1996, p. 143.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115118.
Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.


LINKS

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
A. Farina, S. Giompapa, A. Graziano, A. Liburdi, M. Ravanelli and F. Zirilli, TartagliaPascal's triangle: a historical perspective with applications, Signal, Image and Video Processing, Vol. 7, No. 1 (January 2013), pp. 173188.
Subhash Kak, The golden mean and the physics of aesthetics, in: B. Yadav and M. Mohan (eds.), Ancient Indian Leaps into Mathematics, Birkhäuser, Boston, MA, 2009, pp. 111119; arXiv preprint, arXiv:physics/0411195 [physics.histph], 2004.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003. [Cached copy, with permission (pdf only)]
Jack Ramsay, On Arithmetical Triangles, The Pulse of Long Island, June 1965 [Mentions application to design of antenna arrays. Annotated scan.]
Louis W. Shapiro, Seyoum Getu, WenJin Woan and Leon C. Woodson, The Riordan group, Discrete Applied Math., Vol. 34 (1991), pp. 229239.
Christopher Stover and Eric W. Weisstein, Composition. From MathWorld  A Wolfram Web Resource.


FORMULA

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n1, k) + C(n1, k1).
The triangle is symmetric: C(n,k) = C(n,nk).
a(n+1, m) = a(n, m) + a(n, m1), a(n, 1) := 0, a(n, m) := 0, n<m; a(0, 0)=1.
C(n, k) = n!/(k!(nk)!) if 0<=k<=n, otherwise 0.
G.f.: 1/(1yx*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1xy) = 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/(1x)^(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 mth power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n1, k1) + m*T(n1, k), where n is the rowindex and k is the column; also T(n, k) = m^(nk)*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 ith partition of (n+1); let d(i) = the number of different parts of the ith partition of (n+1); let m(i, j) = multiplicity of the jth part of the ith 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
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^(nk)*T(nk,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^(nk)*T(nk,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(t+p1, t) = Sum_{i=0..t} C(i+p2, i) = Sum_{i=1..p} C(i+t2, t1). 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
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*k1))/(1(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1 x*(1+x)/(1+ x*(1x)*(1+x)/(1 x^3*(1+x)/(1+ x^2*(1x^2)*(1+x)/(1 x^5*(1+x)/(1+ x^3*(1x^3)*(1+x)/(1 x^7*(1+x)/(1+ x^4*(1x^4)*(1+x)/(1 ...))))))))).
These formulas are due to (1) a qseries identity and (2) a partial elliptic theta function expression. (End)
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
Row n (n >= 0) of the triangle is given by the nth 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_{n1}(x) + P_{n2}(x), n > 1.  L. Edson Jeffery, Aug 12 2013
For a closedform formula for arbitrary left and right borders of Pascallike triangles see A228196.  Boris Putievskiy, Aug 18 2013
(1+x)^n = Sum_{k=0..n} (1)^(nk)*binomial(n,k)*Sum_{i=0..k} k^(ni)*binomial(k,i)*x^(ni)/(ni)!.  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^2x^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+k1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k).  Richard R. Forberg, Jan 01 2014
Multiply each nth 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^MI)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
B) P(:xD:) = exp(dP:xD:) = exp[(e^MI):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 ndim 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)
Recurrence equation: T(n,k) = T(n1,k)*(n + k)/(n  k)  T(n1,k1) 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 welldefined, 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,bc+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 StammWilbrandt, 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/(1x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k.  Gary W. Adamson, Oct 17 2016
BoasBuck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the BoasBuck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(nk))*Sum_{j=k..n1} 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((p1)/a, b) == (1)^b * fact_a(a*ba+1)/fact_a(a*b) (mod p), where fact_n denotes the nth multifactorial, a divides p1, and the denominator of the fraction on the right side of the equation represents the modular inverse.  Isaac Saffold, Jan 07 2019
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)


EXAMPLE

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 kiterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive kiterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive kiterations 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
BoasBuck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the BoasBuck comment above.  Wolfdieter Lang, Nov 12 2018


MAPLE



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 *)


PROG

(Axiom)  (start)
)set expose add constructor OutputForm
pascal(0, n) == 1
pascal(n, n) == 1
pascal(i, j  0 < i and i < j) == pascal(i1, j1) + pascal(i, j1)
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)
(Python) # See Hobson link. Further programs:
from math import prod, factorial
def C(n, k): return prod(range(n, nk, 1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
(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
(Maxima) create_list(binomial(n, k), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */
(Sage) def C(n, k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014
(Magma) /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
(GAP) Flat(List([0..12], n>List([0..n], k>Binomial(n, k)))); # Stefano Spezia, Dec 22 2018


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


KEYWORD



AUTHOR



EXTENSIONS

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


STATUS

approved



