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

A001405 a(n) = binomial(n, floor(n/2)).

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, 924, 1716, 3432, 6435, 12870, 24310, 48620, 92378, 184756, 352716, 705432, 1352078, 2704156, 5200300, 10400600, 20058300, 40116600, 77558760, 155117520, 300540195, 601080390, 1166803110
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Sperner's theorem says that this is the maximal number of subsets of an n-set such that no one contains another.
When computed from index -1, [seq(binomial(n,floor(n/2)), n = -1..30)]; -> [1,1,1,2,3,6,10,20,35,70,126,...] and convolved with aerated Catalan numbers [seq(((n+1) mod 2)*binomial(n,n/2)/((n/2)+1), n = 0..30)]; -> [1,0,1,0,2,0,5,0,14,0,42,0,132,0,...] shifts left by one: [1,1,2,3,6,10,20,35,70,126,252,...] and if again convolved with aerated Catalan numbers, gives A037952 apart from the initial term. - Antti Karttunen, Jun 05 2001 [This is correct because the g.f.'s satisfy (1+x*g001405(x))*g126120(x) = g001405(x) and g001405(x)*g126120(x) = g037952(x)/x. - R. J. Mathar, Sep 23 2021]
Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Gives for n >= 1 the maximum absolute column sum norm of the inverse of the Vandermonde matrix (a_ij) i=0..n-1, j=0..n-1 with a_00=1 and a_ij=i^j for (i,j) != (0,0). - Torsten Muetze, Feb 06 2004
Image of Catalan numbers A000108 under the Riordan array (1/(1-2x),-x/(1-2x)) or A065109. - Paul Barry, Jan 27 2005
Number of left factors of Dyck paths, consisting of n steps. Example: a(4)=6 because we have UDUD, UDUU, UUDD, UUDU, UUUD and UUUU, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Apr 23 2005
Number of dispersed Dyck paths of length n; they are defined as concatenations of Dyck paths and (1,0)-steps on the x-axis; equivalently, Motzkin paths with no (1,0)-steps at positive height. Example: a(4)=6 because we have HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD, where U=(1,1), H=(1,0), and D=(1,-1). - Emeric Deutsch, Jun 04 2011
a(n) is odd iff n=2^k-1. - Jon Perry, May 05 2005
An inverse Chebyshev transform of binomial(1,n)=(1,1,0,0,0,...) where g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), with c(x) the g.f. of A000108. - Paul Barry, May 13 2005
In a random walk on the number line, starting at 0 and with 0 absorbing after the first step, number of ways of ending up at a positive integer after n steps. - Joshua Zucker, Jul 31 2005
Maximum number of sums of the form Sum_{i=1..n} e(i)*a(i) that are congruent to 0 mod q, where e_i=0 or 1 and gcd(a_i,q)=1, provided that q > ceiling(n/2). - Ralf Stephan, Apr 27 2003
Also the number of standard tableaux of height <= 2. - Mike Zabrocki, Mar 24 2007
Hankel transform of this sequence forms A000012 = [1,1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
A001263 * [1, -2, 3, -4, 5, ...] = [1, -1, -2, 3, 6, -10, -20, 35, 70, -126, ...]. - Gary W. Adamson, Jan 02 2008
Equals right border of triangle A153585. - Gary W. Adamson, Dec 28 2008
Second binomial transform of A168491. - Philippe Deléham, Nov 27 2009
a(n) is also the number of distinct strings of length n, each of which is a prefix of a string of balanced parentheses; see example. - Lee A. Newberg, Apr 26 2010
Number of symmetric balanced strings of n pairs of parentheses; see example. - Joerg Arndt, Jul 25 2011
a(n) is the number of permutation patterns modulo 2. - Olivier Gérard, Feb 25 2011
For n >= 2, a(n-1) is the number of incongruent two-color bracelets of 2*n-1 beads, n of which are black (A007123), having a diameter of symmetry. - Vladimir Shevelev, May 03 2011
The number of permutations of n elements where p(k-2) < p(k) for all k. - Joerg Arndt, Jul 23 2011
Also size of the equivalence class of S_{n+1} containing the identity permutation under transformations of positionally adjacent elements of the form abc <--> cba where a < b < c, cf. A210668. - Tom Roby, May 15 2012
a(n) is the number of symmetric Dyck paths of length 2n. - Matt Watson, Sep 26 2012
a(n) is divisible by A000108(floor(n/2)) = abs(A129996(n-2)). - Paul Curtz, Oct 23 2012
a(n) is the number of permutations of length n avoiding both 213 and 231 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Number of symmetric standard Young tableaux of shape (n,n). - Ran Pan, Apr 10 2015
From Luciano Ancora, May 09 2015: (Start)
Also "stepped path" in the array formed by partial sums of the all 1's sequence (or a Pascal's triangle displayed as a square). Example:
[1], [1], 1, 1, 1, 1, 1, ... A000012
1, [2], [3], 4, 5, 6, 7, ...
1, 3, [6], [10], 15, 21, 28, ...
1, 4, 10, [20], [35], 56, 84, ...
1, 5, 15, 35, [70], [126], 210, ...
Sequences in second formula are the mixed diagonals shown in this array. (End)
a(n) = A265848(n,n). - Reinhard Zumkeller, Dec 24 2015
The constant Sum_{n >= 0} a(n)/n! is 1 + A130820. - Peter Bala, Jul 02 2016
Number of meanders (walks starting at the origin and ending at any altitude >= 0 that may touch but never go below the x-axis) with n steps from {-1,1}. - David Nguyen, Dec 20 2016
a(n) is also the number of paths of n steps (either up or down by 1) that end at the maximal value achieved along the path. - Winston Luo, Jun 01 2017
Number of binary n-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions. - Juan A. Olmos, Dec 21 2017
Equivalently, a(n) is the number of subsets of {1,...,n} containing as many even numbers as odd numbers. - Gus Wiseman, Mar 17 2018
a(n) is the number of Dyck paths with semilength = n+1, returns to the x-axis = floor((n+3)/2) and up movements in odd positions = floor((n+3)/2). Example: a(4)=6, U=up movement in odd position, u=up movement in even position, d=down movement, -=return to x-axis: Uududd-Ud-Ud-, Ud-Uudd-Uudd-, Uudd-Uudd-Ud-, Ud-Ud-Uududd-, Uudd-Ud-Uudd-, Ud-Uududd-Ud-. - Roger Ford, Dec 29 2017
Let C_n(R, H) denote the transition matrix from the ribbon basis to the homogeneous basis of the graded component of the algebra of noncommutative symmetric functions of order n. Letting I(2^(n-1)) denote the identity matrix of order 2^(n-1), it has been conjectured that the dimension of the kernel of C_n(R, H) - I(2^(n-1)) is always equal to a(n-1). - John M. Campbell, Mar 30 2018
The number of U-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are U-equivalent iff the positions of pattern U are identical in these paths. - Sergey Kirgizov, Apr 08 2018
All binary self-dual codes of length 2n, for n > 0, must contain at least a(n) codewords of weight n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 2n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself n times. A permutation equivalent code can be constructed by augmenting two identity matrices of length n together. - Nathan J. Russell, Nov 25 2018
Closed under addition. - Torlach Rush, Apr 18 2019
The sequence starting (1, 2, 3, 6, ...) is the invert transform of A097331: (1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, ...). - Gary W. Adamson, Feb 22 2020
From Gary W. Adamson, Feb 24 2020: (Start)
The sequence is the culminating limit of an infinite set of sequences with convergents of 2*cos(Pi/N), N = (3, 5, 7, 9, ...).
The first few such sequences are:
N = 3: (1, 1, 1, 1, 1, 1, 1, 1, ...)
N = 5: (1, 1, 2, 3, 5, 8, 13, 21, ...) = A000045
N = 7: (1, 1, 2, 3, 6, 10, 19, 33, ...) = A028495, a(n)/a(n-1) tends to 1.801937...
N = 9 (1, 1, 2, 3, 6, 10, 20, 35, ...) = A061551, a(n)/a(n_1) tends to 1.879385...
...
In the limit one gets the current sequence with ratio 2. (End)
a(n) is also the number of monotone lattice paths from (0,0) to (floor(n/2),ceiling(n/2)). These are the number of Grand Dyck paths when n is even. - Nachum Dershowitz, Aug 12 2020
The maximum number of preimages that a permutation of length n+1 can have under the consecutive-132-avoiding stack-sorting map. - Colin Defant, Aug 28 2020
Counts faro permutations of length n. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one. - Sergey Kirgizov, Jan 12 2021
Per "Sperner's Theorem", the largest possible familes of finite sets none of which contain any other sets in the family. - Renzo Benedetti, May 26 2021
a(n-1) are the incomplete, primitive Dyck paths of n steps without a first return: paths of U and D steps starting at the origin, never touching the horizontal axis later on, and ending above the horizontal axis. n=1: {U}, n=2: {UU}, n=3: {UUU, UUD}, n=4: {UUUU, UUUD, UUDU}, n=5: {UUUUU, UUUUD, UUUDD, UUDUU, UUUDU, UUDUD}. For comparison: A037952 counts incomplete Dyck paths with n steps with any number of intermediate returns to the horizontal axis, ending above the horizontal axis. - R. J. Mathar, Sep 24 2021
a(n) is the number of noncrossing partitions of [n] whose nontrivial blocks are of type {a,b}, with a <= n/2, b > n/2. - Francesca Aicardi, May 29 2022
Maximal coefficient of (1+x)^n. - Vaclav Kotesovec, Dec 30 2022
Sums of lower-left-to-upper-right diagonals of the Catalan Triangle A001263. - Howard A. Landman, Sep 16 2024

Examples

			For n = 4, the a(4) = 6 distinct strings of length 4, each of which is a prefix of a string of balanced parentheses, are ((((, (((), (()(, ()((, ()(), and (()). - _Lee A. Newberg_, Apr 26 2010
There are a(5)=10 symmetric balanced strings of 5 pairs of parentheses:
[ 1] ((((()))))
[ 2] (((()())))
[ 3] ((()()()))
[ 4] ((())(()))
[ 5] (()()()())
[ 6] (()(())())
[ 7] (())()(())
[ 8] ()()()()()
[ 9] ()((()))()
[10] ()(()())() - _Joerg Arndt_, Jul 25 2011
G.f. = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 20*x^6 + 35*x^7 + 70*x^8 + ...
The a(4)=6 binary 4-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions are 0000, 1100, 1001, 0110, 0011, 1111. - _Juan A. Olmos_, Dec 21 2017

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.
  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 1999; see p. 135.
  • K. Engel, Sperner Theory, Camb. Univ. Press, 1997; Theorem 1.1.1.
  • P. Frankl, Extremal sets systems, Chap. 24 of R. L. Graham et al., eds, Handbook of Combinatorics, North-Holland.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.16(b), p. 452.

Links

Crossrefs

Row sums of Catalan triangle A053121 and of symmetric Dyck paths A088855.
Enumerates the structures encoded by A061854 and A061855.
First differences are in A037952.
Apparently a(n) = lim_{k->infinity} A094718(k, n).
Partial sums are in A036256. Column k=2 of A182172. Column k=1 of A335570.
Bisections: A000984 (even part), A001700 (odd part).
Cf. A000712, A001006, A001700, A005773, A005817, A007578, A007579, A022916, A022917 (permutation patterns mod k), A049401, A051920, A063886, A130820, A132815, A153585, A239241, A265848.
Cf. A097331.
Cf. A000045, A028495, A061551
Cf. A107373, A340567, A340568, A340569 (popularity of certain patterns in faro permutations).

Programs

  • GAP
    List([0..40],n->Binomial(n,Int(n/2))); # Muniru A Asiru, Apr 08 2018
  • Haskell
    a001405 n = a007318_row n !! (n `div` 2) -- Reinhard Zumkeller, Nov 09 2011
  • Magma
    [Binomial(n, Floor(n/2)): n in [0..40]]; // Vincenzo Librandi, Nov 16 2014
  • Maple
    A001405 := n->binomial(n, floor(n/2)): seq(A001405(n), n=0..33);
  • Mathematica
    Table[Binomial[n, Floor[n/2]], {n, 0, 40}] (* Stefan Steinerberger, Apr 08 2006 *)
Table[DifferenceRoot[Function[{a,n},{-4 n a[n]-2 a[1+n]+(2+n) a[2+n] == 0,a[1] == 1,a[2] == 1}]][n], {n, 30}] (* Luciano Ancora, Jul 08 2015 *)
Array[Binomial[#,Floor[#/2]]&,40,0] (* Harvey P. Dale, Mar 05 2018 *)
  • Maxima
    A001405(n):=binomial(n,floor(n/2))$
makelist(A001405(n),n,0,30); /* Martin Ettl, Nov 01 2012 */
  • PARI
    a(n) = binomial(n, n\2);
  • PARI
    first(n) = x='x+O('x^n); Vec((-1+2*x+sqrt(1-4*x^2))/(2*x-4*x^2)) \\ Iain Fox, Dec 20 2017 (edited by Iain Fox, May 07 2018)
  • Python
    from math import comb
def A001405(n): return comb(n,n//2) # Chai Wah Wu, Jun 07 2022

Formula

a(n) = max_{k=0..n} binomial(n, k).
a(2*n) = A000984(n), a(2*n+1) = A001700(n).
By symmetry, a(n) = binomial(n, ceiling(n/2)). - Labos Elemer, Mar 20 2003
P-recursive with recurrence: a(0) = 1, a(1) = 1, and for n >= 2, (n+1)*a(n) = 2*a(n-1) + 4*(n-1)*a(n-2). - Peter Bala, Feb 28 2011
G.f.: (1+x*c(x^2))/sqrt(1-4*x^2) = 1/(1 - x - x^2*c(x^2)); where c(x) = g.f. for Catalan numbers A000108.
G.f.: (-1 + 2*x + sqrt(1-4*x^2))/(2*x - 4*x^2). - Lee A. Newberg, Apr 26 2010
G.f.: 1/(1 - x - x^2/(1 - x^2/(1 - x^2/(1 - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 12 2009
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = Sum_{k = 0..2*m} (-1)^k*a(k)*a(2*m-k). - Len Smiley, Dec 09 2001
G.f.: (sqrt((1+2*x)/(1-2*x)) - 1)/(2*x). - Vladeta Jovovic, Apr 28 2003
The o.g.f. A(x) satisfies A(x) + x*A^2(x) = 1/(1-2*x). - Peter Bala, Feb 28 2011
E.g.f.: BesselI(0, 2*x) + BesselI(1, 2*x). - Vladeta Jovovic, Apr 28 2003
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = 2*a(2*m) - c(m), where c(m)=A000108(m) are the Catalan numbers. - Christopher Hanusa (chanusa(AT)washington.edu), Nov 25 2003
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*A000108(k). - Paul Barry, Jan 27 2005
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(1, n-2*k). - Paul Barry, May 13 2005
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..floor((n+1)/2)} (binomial(n+1, k)*(cos((n-2*k+1)*Pi/2) + sin((n-2*k+1)*Pi/2))).
a(n) = Sum_{k=0..n+1}, (binomial(n+1, (n-k+1)/2)*(1-(-1)^(n-k))*(cos(k*Pi/2) + sin(k*Pi))/2). (End)
a(n) = Sum_{k=floor(n/2)..n} (binomial(n,n-k) - binomial(n,n-k-1)). - Paul Barry, Sep 06 2007
Inverse binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double inverse binomial transform of A001700. Row sums of triangle A132815. - Gary W. Adamson, Aug 31 2007
a(n) = Sum_{k=0..n} A120730(n,k). - Philippe Deléham, Oct 16 2008
a(n) = Sum_{k = 0..floor(n/2)} (binomial(n,k) - binomial(n,k-1)). - Nishant Doshi (doshinikki2004(AT)gmail.com), Apr 06 2009
Sum_{n>=0} a(n)/10^(n+1) = 0.1123724... = (sqrt(3)-sqrt(2))/(2*sqrt(2)); Sum_{n>=0} a(n)/100^(n+1) = 0.0101020306102035... = (sqrt(51)-sqrt(49))/(2*sqrt(49)). - Mark Dols, Jul 15 2010
Conjectured: a(n) = 2^n*2F1(1/2,-n;2;2), useful for number of paths in 1-d for which the coordinate is never negative. - Benjamin Phillabaum, Feb 20 2011
a(2*m+1) = (2*m+1)*a(2*m)/(m+1), e.g., a(7) = (7/4)*a(6) = (7/4)*20 = 35. - Jon Perry, Jan 20 2011
From Peter Bala, Feb 28 2011: (Start)
Let F(x) be the logarithmic derivative of the o.g.f. A(x). Then 1+x*F(x) is the o.g.f. for A027306.
Let G(x) be the logarithmic derivative of 1+x*A(x). Then x*G(x) is the o.g.f. for A058622. (End)
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal; and V = the vector [1,0,0,0,...]. a(n) = M^n*V, leftmost term. - Gary W. Adamson, Jun 13 2011
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal. a(n) = M^n_{1,1}. - Corrected by Gary W. Adamson, Jan 30 2012
a(n) = A007318(n, floor(n/2)). - Reinhard Zumkeller, Nov 09 2011
a(n+1) = Sum_{k=0..n} a(n-k)*A097331(k) = a(n) + Sum_{k=0..(n-1)/2} A000108(k)*a(n-2*k-1). - Philippe Deléham, Nov 27 2011
a(n) = A214282(n) - A214283(n), for n > 0. - Reinhard Zumkeller, Jul 14 2012
a(n) = Sum_{k=0..n} A168511(n,k)*(-1)^(n-k). - Philippe Deléham, Mar 19 2013
a(n+2*p-2) = Sum_{k=0..floor(n/2)} A009766(n-k+p-1, k+p-1) + binomial(n+2*p-2, p-2), for p >= 1. - Johannes W. Meijer, Aug 02 2013
O.g.f.: (1-x*c(x^2))/(1-2*x), with the o.g.f. c(x) of Catalan numbers A000108. See the rewritten formula given by Lee A. Newberg above. This is the o.g.f. for the row sums the Riordan triangle A053121. - Wolfdieter Lang, Sep 22 2013
a(n) ~ 2^n / sqrt(Pi * n/2). - Charles R Greathouse IV, Oct 23 2015
a(n) = 2^n*hypergeom([1/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
a(2*k) = Sum_{i=0..k} binomial(k, i)*binomial(k, i), a(2*k+1) = Sum_{i=0..k} binomial(k+1, i)*binomial(k, i). - Juan A. Olmos, Dec 21 2017
a(0) = 1, a(n) = 2 * a(n-1) for even n, a(n) = (2*n/(n+1)) * a(n-1) for odd n. - James East, Sep 25 2019
a(n) = A037952(n) + A000108(n/2) where A(.)=0 for non-integer argument. - R. J. Mathar, Sep 23 2021
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=0} 1/a(n) = 2*Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 - 2*Pi/(9*sqrt(3)). (End)
For k>2, Sum_{n>=0} a(n)/k^n = (sqrt((k+2)/(k-2)) - 1)*k/2. - Vaclav Kotesovec, May 13 2022
From Peter Bala, Mar 24 2023: (Start)
a(n) = Sum_{k = 0..n+1} (-1)^(k+binomial(n+2,2)) * k/(n+1) * binomial(n+1,k)^2.
(n + 1)*(2*n - 1)*a(n) = (-1)^(n+1)*2*a(n-1) + 4*(n - 1)*(2*n + 1)*a(n-2) with a(0) = a(1) = 1. (End)
a(n) = Integral_{x=-2..2} x^n*W(x)*dx, n>=0, where W(x) = sqrt((2+x)/(2-x))/(2*Pi) is a positive function on x=(-2,2) and is singular at x = 2. Therefore a(n) is a positive definite sequence. - Karol A. Penson, May 12 2025

A037952 a(n) = binomial(n, floor((n-1)/2)).

Original entry on oeis.org

0, 1, 1, 3, 4, 10, 15, 35, 56, 126, 210, 462, 792, 1716, 3003, 6435, 11440, 24310, 43758, 92378, 167960, 352716, 646646, 1352078, 2496144, 5200300, 9657700, 20058300, 37442160, 77558760, 145422675, 300540195, 565722720, 1166803110, 2203961430, 4537567650
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

First differences of central binomial coefficients: a(n) = A001405(n+1) - A001405(n).
The maximum size of an intersecting (or proper) antichain on an n-set. - Vladeta Jovovic, Dec 27 2000
Number of ordered trees with n+1 edges, having root of degree at least 2 and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
a(n)=number of Dyck (n+1)-paths that are symmetric but not prime. A prime Dyck path is one that returns to the x-axis only at its terminal point. For example a(3)=3 counts UDUUDDUD, UUDDUUDD, UDUDUDUD. - David Callan, Dec 09 2004
Number of involutions of [n+2] containing the pattern 132 exactly once. For example, a(3)=3 because we have 1'3'2'45, 42'5'13' and 52'4'3'1 (the entries corresponding to the pattern 132 are "primed"). - Emeric Deutsch, Nov 17 2005
Also number of ways to put n eggs in floor(n/2) baskets where order of the baskets matters and all baskets have at least 1 egg. - Ben Paul Thurston, Sep 30 2006
For n >= 1 the number of standard Young tableaux with shapes corresponding to partitions into at most 2 distinct parts. - Joerg Arndt, Oct 25 2012
It seems that 3, 4, 10, ... are Colbourn's Covering Array Numbers CAN(2,k,2). - Ryan Dougherty, May 27 2015
Essentially the same as A007007. - Georg Fischer, Oct 02 2018
a(n) is the number of subsets of {1,2,...,n} that contain exactly 1 more odd than even elements. For example, for n = 6, a(6) = 15 and the 15 sets are {1}, {3}, {5}, {1,2,3}, {1,2,5}, {1,3,4}, {1,3,6}, {1,4,5}, {1,5,6}, {2,3,5}, {3,4,5}, {3,5,6}, {1,2,3,4,5}, {1,2,3,5,6}, {1,3,4,5,6}. - Enrique Navarrete, Dec 21 2019
a(n) is the number of lattice paths of n steps taken from the step set {U=(1,1), D=(1,-1)} that start at the origin, never go below the x-axis, and end strictly above the x-axis; more succinctly, proper left factors of Dyck paths. For example, a(3)=3 counts UUU, UUD, UDU, and a(4)=4 counts UUUU, UUUD, UUDU, UDUU. - David Callan and Emeric Deutsch, Jan 25 2021
For n >= 3, a(n) is also the number of pinnacle sets in the (n-2)-Plummer-Toft graph. - Eric W. Weisstein, Sep 11 2024

Links

Crossrefs

Cf. A007007, A032263, A014495 (partial sums), A001405 (partial sums + 1).
Cf. A035951, A035953, A035954, A035955, A035956, A035957.
Cf. A051303, A051304, A051305, A051306, A051307.
Cf. A047171, A036256, A051920.
Cf. A265848.

Programs

  • Haskell
    a037952 n = a037952_list !! n
a037952_list = zipWith (-) (tail a001405_list) a001405_list
-- Reinhard Zumkeller, Mar 04 2012
  • Magma
    [Binomial(n, Floor((n-1)/2)): n in [0..40]]; // G. C. Greubel, Jun 21 2022
  • Maple
    a:= n-> binomial(n, floor((n-1)/2)):
seq(a(n), n=0..35);  # Alois P. Heinz, Sep 19 2017
  • Mathematica
    Table[ Binomial[n, Floor[n/2]], {n, 0, 35}]//Differences (* Jean-François Alcover, Jun 10 2013 *)
f[n_] := Binomial[n, Floor[(n-1)/2]]; Array[f, 35, 0] (* Robert G. Wilson v, Nov 13 2014 *)
  • PARI
    a(n) = binomial(n, (n-1)\2); \\ Altug Alkan, Oct 03 2018
  • SageMath
    [binomial(n, (n-1)//2) for n in (0..40)] # G. C. Greubel, Jun 21 2022

Formula

E.g.f.: BesselI(1, 2*x) + BesselI(2, 2*x). - Vladeta Jovovic, Apr 28 2003
O.g.f.: (1-sqrt(1-4x^2))/(x - 2x^2 + x*sqrt(1-4x^2)).
Convolution of A001405 and A126120 shifted right: g001405(x)*g126120(x) = g037952(x)/x. - Philippe Deléham, Mar 17 2007
D-finite with recurrence: (n+2)*a(n) + (-n-2)*a(n-1) + 2*(-2*n+1)*a(n-2) + 4*(n-2)*a(n-3) = 0. - R. J. Mathar, Jan 25 2013. Proved by Robert Israel, Nov 13 2014
For n > 0: a(n) = A265848(n,0). - Reinhard Zumkeller, Dec 24 2015
a(n) = binomial(n, (n-2)/2) = A001791(n/2), n even; a(n) = binomial(n, (n+1)/2) = A001700((n-1)/2), n odd. - Enrique Navarrete, Dec 21 2019
From R. J. Mathar, Sep 23 2021: (Start)
A001405(n) = a(n) + A000108(n/2), where A(.)=0 for non-integer arguments.
a(n) = Sum_{m=1..n} A053121(n,m) [comment Callan-Deutsch].
a(2n+1) = A000984(n+1)/2. (End)
a(n) = Sum_{k=2..n} A143359(n,k). [Callan's 2004 comment]. - R. J. Mathar, Sep 24 2021
From Amiram Eldar, Sep 27 2024: (Start)
Sum_{n>=1} 1/a(n) = 1 + Pi/sqrt(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = (3 - Pi/sqrt(3))/9. (End)

A085163 Permutation of natural numbers induced by the Catalan bijection gma085163 acting on symbolless S-expressions encoded by A014486/A063171.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 5, 8, 9, 11, 14, 16, 19, 17, 20, 12, 10, 18, 21, 15, 13, 22, 23, 25, 28, 30, 33, 37, 39, 42, 44, 56, 51, 53, 47, 60, 45, 48, 54, 57, 61, 31, 34, 26, 24, 46, 49, 38, 32, 59, 58, 62, 40, 29, 55, 35, 43, 27, 50, 63, 52, 41, 36, 64, 65, 67, 70, 72, 75
Offset: 0

Views

Author

Antti Karttunen, Jun 23 2003

Keywords

Links

Crossrefs

Inverse: A085164. a(n) = A085161(A057508(n)). Occurs in A073200. Cf. also A085171, A085172. Scheme-function app-to-xrt given in A085203.
Number of fixed points 1, 1, 2, 2, 3, 4, 7, 11, 21, 36, 71, ... in range [A014137(n-1)..A014138(n-1)] of this permutation appears to be A051920 after the initial two ones.

A085164 Inverse permutation to A085163.

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 5, 6, 8, 9, 17, 10, 16, 21, 11, 20, 12, 14, 18, 13, 15, 19, 22, 23, 45, 24, 44, 58, 25, 54, 26, 42, 49, 27, 43, 56, 63, 28, 48, 29, 53, 62, 30, 57, 31, 37, 46, 35, 38, 47, 59, 33, 61, 34, 39, 55, 32, 40, 51, 50, 36, 41, 52, 60, 64, 65, 129, 66, 128, 170
Offset: 0

Views

Author

Antti Karttunen, Jun 23 2003

Keywords

Links

Crossrefs

Inverse: A085163. a(n) = A057508(A085161(n)). Occurs in A073200. Cf. also A085171, A085172.
Number of fixed points: A051920. See comment at A085163.

A006481 Euler characteristics of polytopes.

Original entry on oeis.org

1, 2, 3, 4, 5, 11, 21, 36, 57, 127, 253, 463, 793, 1717, 3433, 6436, 11441, 24311, 48621, 92379, 167961, 352717, 705433, 1352079, 2496145, 5200301, 10400601, 20058301, 37442161
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Very like A051920. Cf. A320996.

Programs

Formula

Numbers suggest that for n not divisible by 4, a(n) = C(n, [n/2]) + 1 and C(n, [(n-1)/2]) + 1 otherwise (see A051920 and A037952+1). - Ralf Stephan, Jun 07 2005

A193242 a(n) = (C(n, floor(n/2)) + 2)^n for n >= 0.

Original entry on oeis.org

1, 3, 16, 125, 4096, 248832, 113379904, 94931877133, 722204136308736, 9223372036854775808, 1117730665547154976408576, 214633637635011206805784100864, 397495155639882245867698528490622976, 1135797931555041090259334993227408493600768
Offset: 0

Views

Author

Alexander R. Povolotsky, Feb 11 2013

Keywords

Comments

If the terms in each row of a Pascal's triangle (see the tabular presentation of A007318) 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 are concatenated (if necessary) and considered as palindromes, represented in different bases, then A051920(n) for n>=0 could be considered as the smallest base radix for which those palindromes are composed of single digits/letters. Those palindromes will look like: 1, 11, 121, 1331, 14641, 15AA51, 16FKF61, 17LZZL71, ... . Conversion of such palindromes from their above mentioned bases to decimal yields this sequence of the consecutive ascending powers. Such powers are enumerations of the rows in a Pascal's triangle, counting from 0, namely: 1, 3, 16, 125, 4096, 248832, 113379904, 94931877133, ... (that is 1^0, 3^1, 4^2, 5^3, 8^4, 12^5, 22^6, 37^7, ...). In general those powers could be described as (A051920(n)+1)^n for n >= 0. Another property of the palindromes discussed above is that their digits/letters sum to 2^n.
Also (as noted by Robert Munafo) the terms of this sequence are a(n) = (A001405(n)+2)^n for n>=0.

Links

Crossrefs

Cf. A007318, A051920, A001405.

Programs

  • Mathematica
    Table[(Binomial[n, Floor[n/2]] + 2)^n, {n,0,20}] (* G. C. Greubel, Feb 20 2017 *)
  • PARI
    for(n=0,20, print1((binomial(n, floor(n/2)) + 2)^n, ", ")) \\ G. C. Greubel, Feb 20 2017

Formula

a(n) = (A051920(n) + 1)^n.
a(n) = (A001405(n) + 2)^n.

Extensions

Corrected a(8) onward - G. C. Greubel, Feb 20 2017

A238727 Number T(n,k) of standard Young tableaux with n cells where k is the largest value in the last row; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 3, 0, 0, 1, 2, 7, 0, 0, 1, 3, 8, 14, 0, 0, 1, 4, 11, 19, 41, 0, 0, 1, 7, 19, 34, 64, 107, 0, 0, 1, 11, 32, 62, 119, 202, 337, 0, 0, 1, 21, 64, 131, 248, 418, 671, 1066, 0, 0, 1, 36, 124, 277, 545, 943, 1518, 2361, 3691
Offset: 0

Views

Author

Joerg Arndt and Alois P. Heinz, Mar 03 2014

Keywords

Comments

T(0,0) = 1 by convention.
Also the number of ballot sequences of length n having the last occurrence of the maximal value at position k.
T(n,3) = A051920(n-3) for n>3.
T(2*n,n) gives A246818.
Main diagonal gives A238728.
Row sums give A000085.

Examples

			The 10 tableaux with n=4 cells sorted by largest value in the last row:
  :[1 3 4]:[1 4] [1 2 4]:[1] [1 2] [1 3] [1 2 3] [1 2] [1 3] [1 2 3 4]:
  :[2]    :[2]   [3]    :[2] [3]   [2]   [4]     [3 4] [2 4]          :
  :       :[3]          :[3] [4]   [4]                                :
  :       :             :[4]                                          :
  : --2-- : -----3----- : ---------------------4--------------------- :
The 10 ballot sequences of length 4 sorted by the position of the last occurrence of the maximal value:
  [1, 2, 1, 1]  ->  2 } -- 1
  [1, 2, 3, 1]  ->  3 \ __ 2
  [1, 1, 2, 1]  ->  3 /
  [1, 2, 3, 4]  ->  4 \
  [1, 1, 2, 3]  ->  4  \
  [1, 2, 1, 3]  ->  4   \
  [1, 1, 1, 2]  ->  4    } 7
  [1, 1, 2, 2]  ->  4   /
  [1, 2, 1, 2]  ->  4  /
  [1, 1, 1, 1]  ->  4 /
thus row 4 = [0, 0, 1, 2, 7].
Triangle T(n,k) begins:
  00:   1;
  01:   0, 1;
  02:   0, 0, 2;
  03:   0, 0, 1,  3;
  04:   0, 0, 1,  2,   7;
  05:   0, 0, 1,  3,   8,  14;
  06:   0, 0, 1,  4,  11,  19,  41;
  07:   0, 0, 1,  7,  19,  34,  64, 107;
  08:   0, 0, 1, 11,  32,  62, 119, 202,  337;
  09:   0, 0, 1, 21,  64, 131, 248, 418,  671, 1066;
  10:   0, 0, 1, 36, 124, 277, 545, 943, 1518, 2361, 3691;

Links

Programs

  • Maple
    h:= proc(l) option remember; local n, s; n:= nops(l); s:= add(i, i=l);
     `if`(n=0, 1, add(`if`(il[i+1], h(subsop(i=l[i]-1, l)),
     `if`(i=n, (p->add(coeff(p,x,j)*x^`if`(j1, l[i]-1, [][]), l))), 0)), i=1..n))
    end:
g:= (n, i, l)-> `if`(n=0 or i=1, h([l[], 1$n]),
       add(g(n-i*j, i-1, [l[], i$j]), j=0..n/i)):
T:= n-> (p->seq(coeff(p, x, i), i=0..n))(g(n$2, [])):
seq(T(n), n=0..12);
  • Mathematica
    h[l_] := h[l] = With[{n = Length[l], s = Total[l]},
     If[n == 0, 1, Sum[If[i < n && l[[i]] > l[[i + 1]],
     h[ReplacePart[l, i -> l[[i]] - 1]], If[i == n, Function[p,
     Sum[Coefficient[p, x, j] x^If[j < s, s, j], {j, 0,
     Exponent[p, x]}]][h[ReplacePart[l, i -> If[l[[i]] > 1,
     l[[i]] - 1, Nothing]]]], 0]], {i, n}]]];
g[n_, i_, l_] := If[n == 0 || i == 1, h[Join[l, Table[1, {n}]]],
     Sum[g[n - i*j, i - 1, Join[l, Table[i, {j}]]], {j, 0, n/i}]];
T[n_] := CoefficientList[g[n, n, {}], x];
Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Aug 27 2021, after Maple code *)
Showing 1-8 of 8 results.

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