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.

Showing 1-7 of 7 results.

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

Original entry on oeis.org

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

Views

Author

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

Keywords

Comments

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

Examples

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

References

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

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

A027907 Triangle of trinomial coefficients T(n,k) (n >= 0, 0 <= k <= 2*n), read by rows: n-th row is obtained by expanding (1 + x + x^2)^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1, 1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1, 1, 7, 28, 77, 161, 266, 357, 393, 357, 266, 161, 77, 28, 7, 1, 1, 8, 36, 112, 266
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

When the rows are centered about their midpoints, each term is the sum of the three terms directly above it (assuming the undefined terms in the previous row are zeros). - N. J. A. Sloane, Dec 23 2021
T(n,k) = number of integer strings s(0),...,s(n) such that s(0)=0, s(n)=k, s(i) = s(i-1) + c, where c is 0, 1 or 2. Columns of T include A002426, A005717 and A014531.
Also number of ordered trees having n+1 leaves, all at level three and n+k+3 edges. Example: T(3,5)=3 because we have three ordered trees with 4 leaves, all at level three and 11 edges: the root r has three children; from one of these children two paths of length two are hanging (i.e., 3 possibilities) while from each of the other two children one path of length two is hanging. Diagonal sums are the tribonacci numbers; more precisely: Sum_{i=0..floor(2*n/3)} T(n-i,i) = A000073(n+2). - Emeric Deutsch, Jan 03 2004
T(n,k) = A111808(n,k) for 0 <= k <= n and T(n, 2*n-k) = A111808(n,k) for 0 <= k < n. - Reinhard Zumkeller, Aug 17 2005
The trinomial coefficients, T(n,i), are the absolute value of the coefficients of the chromatic polynomial of P_2 X P_n factored with x*(x-1)^i terms. Example: The chromatic polynomial of P_2 X P_2 is: x*(x-1) - 2*x*(x-1)^2 + x*(x-1)^3 and so T(1,0)=1, T(1,1)=2 and T(1,1) = 1. - Thomas J. Pfaff (tpfaff(AT)ithaca.edu), Oct 02 2006
T(n,k) is the number of distinct ways in which k unlabeled objects can be distributed in n labeled urns allowing at most 2 objects to fall into each urn. - N-E. Fahssi, Mar 16 2008
T(n,k) is the number of compositions of k into n parts p, each part 0 <= p <= 2. Adding 1 to each part, as a corollary, T(n,k) is the number of compositions of n+k into n parts p where 1 <= p <= 3. E.g., T(2,3)=2 since 5 = 3+2 = 2+3. - Steffen Eger, Jun 10 2011
Number of lattice paths from (0,0) to (n,k) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Number of lattice paths from (0,0) to (2*n-k,k) using steps (2,0), (1,1), (0,2). - Werner Schulte, Jan 25 2017
T(n,k) is number of distinct ways to sum the integers -1, 0 , and 1 n times to obtain n-k, where T(n,0) = T(n,2*n+1) = 1. - William Boyles, Apr 23 2017
T(n-1,k-1) is the number of 2-compositions of n with 0's having k parts; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
T(n,k) is the number of ways to obtain a sum of n+k when throwing a 3-sided die n times. Follows from the "T(n,k) is the number of compositions of n+k into n parts p where 1 <= p <= 3" comment above. - Feryal Alayont, Dec 30 2024

Examples

			The triangle T(n, k) begins:
  n\k 0   1   2   3   4   5   6   7   8   9 10 11 12
  0:  1
  1:  1   1   1
  2:  1   2   3   2   1
  3:  1   3   6   7   6   3   1
  4:  1   4  10  16  19  16  10   4   1
  5:  1   5  15  30  45  51  45  30  15   5  1
  6:  1   6  21  50  90 126 141 126  90  50 21  6  1
Concatenated rows:
G.f. = 1 + (x^2+x+1)*x + (x^2+x+1)^2*x^4 + (x^2+x+1)^3*x^9 + ...
     = 1 + (x + x^2 + x^3) + (x^4 + 2*x^5 + 3*x^6 + 2*x^7 + x^8) +
  (x^9 + 3*x^10 + 6*x^11 + 7*x^12 + 6*x^13 + 3*x^14 + x^15) + ... .
As a centered triangle, this begins:
           1
        1  1  1
     1  2  3  2  1
  1  3  6  7  6  3  1

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • D. C. Fielder and C. O. Alford, Pascal's triangle: top gun or just one of the gang?, in G E Bergum et al., eds., Applications of Fibonacci Numbers Vol. 4 1991 pp. 77-90 (Kluwer).
  • L. Kleinrock, Uniform permutation of sequences, JPL Space Programs Summary, Vol. 37-64-III, Apr 30, 1970, pp. 32-43.

Links

Crossrefs

Columns of T include A002426, A005717, A014531, A005581, A005712, etc. See also A035000, A008287.
First differences are in A025177. Pairwise sums are in A025564.
Cf. A007318 (Pascal's triangle); A000073, A001006, A123531, A055217, A027908, A027913, A027914, A027023.

Programs

  • Haskell
    a027907 n k = a027907_tabf !! n !! k
a027907_row n = a027907_tabf !! n
a027907_tabf = [1] : iterate f [1, 1, 1] where
   f row = zipWith3 (((+) .) . (+))
                    (row ++ [0, 0]) ([0] ++ row ++ [0]) ([0, 0] ++ row)
a027907_list = concat a027907_tabf
-- Reinhard Zumkeller, Jul 06 2014, Jan 22 2013, Apr 02 2011
  • Maple
    A027907 := proc(n,k) expand((1+x+x^2)^n) ; coeftayl(%,x=0,k) ; end proc:
seq(seq(A027907(n,k),k=0..2*n),n=0..5) ; # R. J. Mathar, Jun 13 2011
T := (n,k) -> simplify(GegenbauerC(`if`(kPeter Luschny, May 08 2016
  • Mathematica
    Table[CoefficientList[Series[(Sum[x^i, {i, 0, 2}])^n, {x, 0, 2 n}], x], {n, 0, 10}] // Grid (* Geoffrey Critzer, Mar 31 2010 *)
Table[Sum[Binomial[n, i]Binomial[n - i, k - 2i], {i, 0, n}], {n, 0, 10}, {k, 0, 2n}] (* Adi Dani, May 07 2011 *)
T[ n_, k_] := If[ n < 0, 0, Coefficient[ (1 + x + x^2)^n, x, k]]; (* Michael Somos, Nov 08 2016 *)
Flatten[DeleteCases[#,0]&/@CellularAutomaton[{Total[#] &, {}, 1}, {{1}, 0}, 8] ] (* Giorgos Kalogeropoulos, Nov 09 2021 *)
  • Maxima
    trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
create_list(trinomial(n,k),n,0,8,k,0,2*n); /* Emanuele Munarini, Mar 15 2011 */
  • Maxima
    create_list(ultraspherical(k,-n,-1/2),n,0,6,k,0,2*n); /* Emanuele Munarini, Oct 18 2016 */
  • PARI
    {T(n, k) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, k))}; /* Michael Somos, Jun 27 2003 */

Formula

G.f.: 1/(1-z*(1+w+w^2)).
T(n,k) = Sum_{r=0..floor(k/3)} (-1)^r*binomial(n, r)*binomial(k-3*r+n-1, n-1).
Recurrence: T(0,0) = 1; T(n,k) = T(n-1,k-2) + T(n-1,k-1) + T(n-1,k-0), with T(n,k) = 0 if k < 0 or k > 2*n:
T(i,0) = T(i, 2*i) = 1 for i >= 0, T(i, 1) = T(i, 2*i-1) = i for i >= 1 and for i >= 2 and 2 <= j <= i-2, T(i, j) = T(i-1, j-2) + T(i-1, j-1) + T(i-1, j).
The row sums are powers of 3 (A000244). - Gerald McGarvey, Aug 14 2004
T(n,k) = Sum_{i=0..floor(k/2)} binomial(n, 2*i+n-k) * binomial(2*i+n-k, i). - Ralf Stephan, Jan 26 2005
T(n,k) = Sum_{j=0..n} binomial(n, j) * binomial(j, k-j). - Paul Barry, May 21 2005
T(n,k) = Sum_{j=0..n} binomial(k-j, j) * binomial(n, k-j). - Paul Barry, Nov 04 2005
From Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006: (Start)
T(n,k) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(2*n-2*j, k-j); (G. E. Andrews (1990)) obtained by expanding ((1+x)^2 - x)^n.
T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(n-j, k-2*j); obtained by expanding ((1+x) + x^2)^n.
T(n,k) = (-1)^k*Sum_{j=0..n} (-3)^j * binomial(n,j) * binomial(2*n-2*j, k-j); obtained by expanding ((1-x)^2 + 3*x)^n.
T(n,k) = (1/2)^k * Sum_{j=0..n} 3^j * binomial(n,j) * binomial(2*n-2*j, k-2*j); obtained by expanding ((1+x/2)^2 + (3/4)*x^2)^n.
T(n,k) = (2^k/4^n) * Sum_{j=0..n} 3^j * binomial(n,j) * binomial(2*n-2*j, k); obtained by expanding ((1/2+x)^2 + 3/4)^n using T(n,k) = T(2*n-k). (End)
From Paul D. Hanna, Apr 18 2012: (Start)
Let A(x) be the g.f. of the flattened sequence, then:
G.f.: A(x) = Sum_{n>=0} x^(n^2) * (1+x+x^2)^n.
G.f.: A(x) = Sum_{n>=0} x^n*(1+x+x^2)^n * Product_{k=1..n} (1 - (1+x+x^2) * x^(4*k-3)) / (1 - (1+x+x^2)*x^(4*k-1)).
G.f.: A(x) = 1/(1 - x*(1+x+x^2)/(1 + x*(1-x^2)*(1+x+x^2)/(1 - x^5*(1+x+x^2)/(1 + x^3*(1-x^4)*(1+x+x^2)/(1 - x^9*(1+x+x^2)/(1 + x^5*(1-x^6)*(1+x+x^2)/(1 - x^13* (1+x+x^2)/(1 + x^7*(1-x^8)*(1+x+x^2)/(1 - ...))))))))), a continued fraction.
(End)
Triangle: G.f. = Sum_{n>=0} (1+x+x^2)^n * x^(n^2) * y^n. - Daniel Forgues, Mar 16 2015
From Peter Luschny, May 08 2016: (Start)
T(n+1,n)/(n+1) = A001006(n) (Motzkin) for n>=0.
T(n,k) = H(n, k) if k < n else H(n, 2*n-k) where H(n,k) = binomial(n,k)*hypergeom([(1-k)/2, -k/2], [n-k+1], 4).
T(n,k) = GegenbauerC(m, -n, -1/2) where m=k if k < n else 2*n-k. (End)
T(n,k) = (-1)^k * C(2*n,k) * hypergeom([-k, -(2*n-k)], [-n+1/2], 3/4), for all k with 0 <= k <= 2n. - Robert S. Maier, Jun 13 2023
T(n,n) = Sum_{k=0..2*n} (-1)^k*(T(n,k))^2 and T(2*n,2*n) = Sum_{k=0..2*n} (T(n,k))^2 for n >= 0. - Werner Schulte, Nov 08 2016
T(n,n) = A002426(n), central trinomial coefficients. - M. F. Hasler, Nov 02 2019
Sum_{k=0..n-1} T(n, 2*k) = (3^n-1)/2. - Tony Foster III, Oct 06 2020

A008287 Triangle of quadrinomial coefficients, row n is the sequence of coefficients of (1 + x + x^2 + x^3)^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 3, 6, 10, 12, 12, 10, 6, 3, 1, 1, 4, 10, 20, 31, 40, 44, 40, 31, 20, 10, 4, 1, 1, 5, 15, 35, 65, 101, 135, 155, 155, 135, 101, 65, 35, 15, 5, 1, 1, 6, 21, 56, 120, 216, 336, 456, 546, 580, 546, 456, 336, 216, 120, 56, 21, 6, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Coefficient of x^k in (1 + x + x^2 + x^3)^n is the number of distinct ways in which k unlabeled objects can be distributed in n labeled urns allowing at most 3 objects to fall in each urn. - N-E. Fahssi, Mar 16 2008
Rows of A008287 mod 2 converted to decimal equals A177882. - Vladimir Shevelev, Jan 02 2011
T(n,k) is the number of compositions of k into n parts p, each part 0<=p<=3. Adding 1 to each part, as a corollary, T(n,k) is the number of compositions of n+k into n parts p where 1<=p<=4. E.g., T(2,3)=4 since 3=0+3=3+0=1+2=2+1. In general, the entry (n,k) of the (l+1)-nomial triangle gives the number of compositions of k into n parts p, each part 0<=p<=l. - Steffen Eger, Jun 18 2011
Number of lattice paths from (0,0) to (n,k) using steps (1,0), (1,1), (1,2), (1,3). - Joerg Arndt, Jul 05 2011
T(n-1,k-1) is the number of 3-compositions of n with zeros having k parts; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 16 2020
T(n,k) is the number of ways to obtain a sum of n+k when throwing a 4-sided die n times. Follows from the "T(n,k) is the number of compositions of n+k into n parts p where 1 <= p <= 4" comment above. - Feryal Alayont, Dec 30 2024

Examples

			Triangle begins
  1;
  1,1,1,1;
  1,2,3,4,3,2,1;
  1,3,6,10,12,12,10,6,3,1; ...

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • D. C. Fielder and C. O. Alford, Pascal's triangle: top gun or just one of the gang?, in G E Bergum et al., eds., Applications of Fibonacci Numbers Vol. 4 1991 pp. 77-90 (Kluwer).

Links

Crossrefs

Cf. A007318, A027907, A177882.

Programs

  • Haskell
    a008287 n = a008287_list !! n
a008287_list = concat $ iterate ([1,1,1,1] *) [1]
instance Num a => Num [a] where
   fromInteger k = [fromInteger k]
   (p:ps) + (q:qs) = p + q : ps + qs
   ps + qs         = ps ++ qs
   (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
    *                = []
-- Reinhard Zumkeller, Apr 02 2011
  • Maple
    #Define the r-nomial coefficients for r = 1, 2, 3, ...
rnomial := (r,n,k) -> add((-1)^i*binomial(n,i)*binomial(n+k-1-r*i,n-1), i = 0..floor(k/r)):
#Display the 4-nomials as a table
r := 4:  rows := 10:
for n from 0 to rows do
seq(rnomial(r,n,k), k = 0..(r-1)*n)
end do;
# Peter Bala, Sep 07 2013
# second Maple program:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))((1+x+x^2+x^3)^n):
seq(T(n), n=0..10);  # Alois P. Heinz, Aug 17 2018
  • Mathematica
    Flatten[Table[CoefficientList[(1 + x + x^2 + x^3)^n, x], {n, 0, 10}]] (* T. D. Noe, Apr 04 2011 *)
T[n_, k_] := Sum[Binomial[n, i] Binomial[n, k-2i], {i, 0, k/2}]; Table[T[n, k], {n, 0, 6}, {k, 0, 3n}] // Flatten (* Jean-François Alcover, Feb 02 2018 *)
  • Maxima
    quadrinomial(n,k):=coeff(expand((1+x+x^2+x^3)^n),x,k);
create_list(quadrinomial(n,k),n,0,8,k,0,3*n); /* Emanuele Munarini, Mar 15 2011 */

Formula

n-th row is formed by expanding (1+x+x^2+x^3)^n.
From Vladimir Shevelev, Dec 15 2010: (Start)
T(n,0) = 1; T(n,3*n) = 1; T(n,k) = T(n,3*n-k);
T(n,k) = 0, iff k<0 or k>3*n; Sum{k=0..3*n} T(n,k) = 4^n; Sum{k=0..3*n}((-1)^k)*T(n,k)=0 for n > 0; [corrected by Werner Schulte, Sep 09 2015]
T(n,k) = Sum{i=0..floor(k/2)} C(n,i)*C(n,k-2*i);
T(n+1,k) = T(n,k-3)+T(n,k-2)+T(n,k-1)+T(n,k). (End)
T(n,k) = Sum_{i = 0..floor(k/4)} (-1)^i*C(n,i)*C(n+k-1-4*i,n-1) for n >= 0 and 0 <= k <= 3*n. - Peter Bala, Sep 07 2013
G.f.: 1/(1 - ( x + y*x + y^2*x +y^3*x )). - Geoffrey Critzer, Feb 05 2014
T(n,k) = Sum_{j=0..k} (-2)^j*binomial(n,j)*binomial(3*n-2*j,k-j) for n >= 0 and 0 <= k <= 3*n (conjectured). - Werner Schulte, Sep 09 2015

A112467 Riordan array ((1-2x)/(1-x), x/(1-x)).

Original entry on oeis.org

1, -1, 1, -1, 0, 1, -1, -1, 1, 1, -1, -2, 0, 2, 1, -1, -3, -2, 2, 3, 1, -1, -4, -5, 0, 5, 4, 1, -1, -5, -9, -5, 5, 9, 5, 1, -1, -6, -14, -14, 0, 14, 14, 6, 1, -1, -7, -20, -28, -14, 14, 28, 20, 7, 1, -1, -8, -27, -48, -42, 0, 42, 48, 27, 8, 1, -1, -9, -35, -75, -90, -42, 42, 90, 75, 35, 9, 1, -1, -10, -44, -110, -165, -132, 0, 132, 165, 110
Offset: 0

Views

Author

Paul Barry, Sep 06 2005

Keywords

Comments

Row sums are A000007. Diagonal sums are -F(n-2). Inverse is A112468. T(2n,n)=0.
(-1,1)-Pascal triangle. - Philippe Deléham, Aug 07 2006
Apart from initial term, same as A008482. - Philippe Deléham, Nov 07 2006
Each column equals the cumulative sum of the previous column. - Mats Granvik, Mar 15 2010
Reading along antidiagonals generates in essence rows of A192174. - Paul Curtz, Oct 02 2011
Triangle T(n,k), read by rows, given by (-1,2,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 01 2011

Examples

			Triangle starts:
    1;
   -1,  1;
   -1,  0,   1;
   -1, -1,   1,   1;
   -1, -2,   0,   2,   1;
   -1, -3,  -2,   2,   3,   1;
   -1, -4,  -5,   0,   5,   4,  1;
   -1, -5,  -9,  -5,   5,   9,  5,  1;
   -1, -6, -14, -14,   0,  14, 14,  6,  1;
   -1, -7, -20, -28, -14,  14, 28, 20,  7,  1;
   -1, -8, -27, -48, -42,   0, 42, 48, 27,  8, 1;
   -1, -9, -35, -75, -90, -42, 42, 90, 75, 35, 9, 1;
  ...
From _Paul Barry_, Apr 08 2011: (Start)
Production matrix begins:
   1,  1,
  -2, -1,  1,
   2,  0, -1,  1,
  -2,  0,  0, -1,  1,
   2,  0,  0,  0, -1,  1,
  -2,  0,  0,  0,  0, -1,  1,
   2,  0,  0,  0,  0,  0, -1,  1
  ... (End)

Links

Crossrefs

Same first 3 rows as in A054525.
Cf. A008482, A037012, A080232, A112466, A112467, A174293, A174294, A174295, A174296, A174297.

Programs

  • Magma
    [n eq 0 select 1 else (2*k-n)*Binomial(n,k)/n: k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 04 2019
  • Maple
    seq(seq( `if`(n=0, 1, (2*k-n)*binomial(n,k)/n), k=0..n), n=0..10); # G. C. Greubel, Dec 04 2019
  • Mathematica
    T[n_, k_]= If[n==0, 1, ((2*k-n)/n)*Binomial[n, k]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Roger L. Bagula, Feb 16 2009; modified by G. C. Greubel, Dec 04 2019 *)
  • PARI
    T(n, k) = if(n==0, 1, (2*k-n)*binomial(n,k)/n ); \\ G. C. Greubel, Dec 04 2019
  • Sage
    def T(n, k):
    if (n==0): return 1
    else: return (2*k-n)*binomial(n,k)/n
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Dec 04 2019

Formula

Number triangle T(n, k) = binomial(n, n-k) - 2*binomial(n-1, n-k-1).
Sum_{k=0..n} T(n, k)*x^k = (x-1)*(x+1)^(n-1). - Philippe Deléham, Oct 03 2005
T(n,k) = ((2*k-n)/n)*binomial(n, k), with T(0,0)=1. - Roger L. Bagula, Feb 16 2009; modified by G. C. Greubel, Dec 04 2019
T(n,k) = T(n-1,k-1) + T(n-1,k) with T(0,0)=1, T(1,0)=-1, T(n,k)=0 for k>n or for n<0. - Philippe Deléham, Nov 01 2011
G.f.: (1-2x)/(1-(1+y)*x). - Philippe Deléham, Dec 15 2011
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A133494(n), A081294(n), A005053(n), A067411(n), A199661(n), A083233(n) for x = 1, 2, 3, 4, 5, 6, 7, respectively. - Philippe Deléham, Dec 15 2011
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(-1 - x + x^2/2! + x^3/3!) = -1 - 2*x - 2*x^2/2! + 5*x^4/4! + 14*x^5/5! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
Sum_{k=0..n} T(n,k) = 0^n = A000007(n). - G. C. Greubel, Dec 04 2019

A349813 Triangle read by rows: row 1 is [3]; for n >= 1, row n gives coefficients of expansion of (-3 - x + x^2 + 3*x^3)*(1 + x + x^2 + x^3)^(n-1) in order of increasing powers of x.

Original entry on oeis.org

3, -3, -1, 1, 3, -3, -4, -3, 0, 3, 4, 3, -3, -7, -10, -10, -4, 4, 10, 10, 7, 3, -3, -10, -20, -30, -31, -20, 0, 20, 31, 30, 20, 10, 3, -3, -13, -33, -63, -91, -101, -81, -31, 31, 81, 101, 91, 63, 33, 13, 3, -3, -16, -49, -112, -200, -288, -336, -304, -182, 0, 182, 304, 336, 288, 200, 112, 49, 16, 3
Offset: 0

Views

Author

N. J. A. Sloane, Dec 23 2021

Keywords

Comments

The row polynomials can be further factorized, since -3 - x + x^2 + 3*x^3 = -(1-x)*(3 + 4*x + 3*x^2) and 1 + x + x^2 + x^3 = (1+x)*(1+x^2).
The rule for constructing this triangle (ignoring row 0) is the same as that for A008287: each number is the sum of the four numbers immediately above it in the previous row. Here row 1 is [-3, -1, 1, 3] instead of [1, 1, 1, 1].

Examples

			Triangle begins:
   3;
  -3,  -1,   1,   3;
  -3,  -4,  -3,   0,   3,    4,   3;
  -3,  -7, -10, -10,  -4,    4,  10,  10,  7,  3;
  -3, -10, -20, -30, -31,  -20,   0,  20, 31, 30,  20, 10,  3;
  -3, -13, -33, -63, -91, -101, -81, -31, 31, 81, 101, 91, 63, 33, 13, 3;
  ...

Links

  • Jack Ramsay, On Arithmetical Triangles, The Pulse of Long Island, June 1965 [Mentions application to design of antenna arrays. Annotated scan.]

Crossrefs

Cf. A007318, A008287, A349812, A349815, A349819.
The right half of the triangle gives A349814.

Programs

  • Maple
    t1:=-3-x+x^2+3*x^3;
m:=1+x+x^2+x^3;
lprint([3]);
for n from 1 to 12 do
w1:=expand(t1*m^(n-1));
w4:=series(w1,x,3*n+1);
w5:=seriestolist(w4);
lprint(w5);
od:

A349815 Triangle read by rows: row 1 is [1]; for n >= 1, row n gives coefficients of expansion of (-1 - x + x^2 + x^3)*(1 + x + x^2 + x^3)^(n-1) in order of increasing powers of x.

Original entry on oeis.org

1, -1, -1, 1, 1, -1, -2, -1, 0, 1, 2, 1, -1, -3, -4, -4, -2, 2, 4, 4, 3, 1, -1, -4, -8, -12, -13, -8, 0, 8, 13, 12, 8, 4, 1, -1, -5, -13, -25, -37, -41, -33, -13, 13, 33, 41, 37, 25, 13, 5, 1, -1, -6, -19, -44, -80, -116, -136, -124, -74, 0, 74, 124, 136, 116, 80, 44, 19, 6, 1, -1, -7, -26, -70, -149, -259, -376, -456, -450, -334, -124, 124, 334, 450, 456, 376, 259, 149, 70, 26, 7, 1
Offset: 0

Views

Author

N. J. A. Sloane, Dec 23 2021

Keywords

Comments

The row polynomials can be further factorized, since -3 - x + x^2 + 3*x^3 = -(1-x)*(1+x)^2 and 1 + x + x^2 + x^3 = (1+x)*(1+x^2).
The rule for constructing this triangle (ignoring row 0) is the same as that for A008287: each number is the sum of the four numbers immediately above it in the previous row. Here row 1 is [-1, -1, 1, 3] instead of [1, 1, 1, 1].
This is a companion to A008287 and A349813.

Examples

			Triangle begins:
   1;
  -1, -1,   1,   1;
  -1, -2,  -1,   0,   1,   2,   1;
  -1, -3,  -4,  -4,  -2,   2,   4,   4,  3,  1;
  -1, -4,  -8, -12, -13,  -8,   0,   8, 13, 12,  8,  4,  1;
  -1, -5, -13, -25, -37, -41, -33, -13, 13, 33, 41, 37, 25, 13, 5, 1;
  ...

Links

  • Jack Ramsay, On Arithmetical Triangles, The Pulse of Long Island, June 1965 [Mentions application to design of antenna arrays. Annotated scan.]

Crossrefs

Cf. A007318, A008287, A349812, A349813.
The right half of the triangle gives A349816. For the central nonzero entries see A349818.

Programs

  • Maple
    t1:=-1-x+x^2+x^3;
m:=1+x+x^2+x^3;
lprint([3]);
for n from 1 to 12 do
w1:=expand(t1*m^(n-1));
w4:=series(w1,x,3*n+1);
w5:=seriestolist(w4);
lprint(w5);
od:

A349818 Central column (ignoring the zeros) of A349815, or leading entries in rows of A349816.

Original entry on oeis.org

1, 1, 1, 2, 8, 13, 74, 124, 784, 1364, 9069, 16194, 111144, 202070, 1418196, 2612376, 18642208, 34682348, 250706548, 470066728, 3433030048, 6477333149, 47703926354, 90472092748, 670963514192, 1278016171132, 9534032792470, 18226719157820, 136658371126320
Offset: 0

Views

Author

N. J. A. Sloane, Dec 24 2021

Keywords

Comments

Note that the analogous sequence for A349812 is the Motzkin numbers A001006.

Links

Crossrefs

Cf. A001006, A349812, A349815, A349816.

Programs

  • PARI
    a(n) = if(n==0, 1, polcoef((-1 - x + x^2 + x^3)*(1 + x + x^2 + x^3)^(n-1), 3*n\2+1)) \\ Andrew Howroyd, Feb 28 2023

Formula

a(n) = [x^floor(3*n/2+1)](-1 - x + x^2 + 3*x^3)*(1 + x + x^2 + x^3)^(n-1) for n > 0. - Andrew Howroyd, Feb 28 2023

Extensions

Offset corrected and terms a(21) and beyond from Andrew Howroyd, Feb 28 2023
Showing 1-7 of 7 results.

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

Content is available under The OEIS End-User License Agreement.