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

A000292 Tetrahedral (or triangular pyramidal) numbers: a(n) = C(n+2,3) = n*(n+1)*(n+2)/6.

Original entry on oeis.org

0, 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969, 1140, 1330, 1540, 1771, 2024, 2300, 2600, 2925, 3276, 3654, 4060, 4495, 4960, 5456, 5984, 6545, 7140, 7770, 8436, 9139, 9880, 10660, 11480, 12341, 13244, 14190, 15180
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n) is the number of balls in a triangular pyramid in which each edge contains n balls.
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012).
Also (1/6)*(n^3 + 3*n^2 + 2*n) is the number of ways to color the vertices of a triangle using <= n colors, allowing rotations and reflections. Group is the dihedral group D_6 with cycle index (x1^3 + 2*x3 + 3*x1*x2)/6.
Also the convolution of the natural numbers with themselves. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Feb 01 2001
Connected with the Eulerian numbers (1, 4, 1) via 1*a(n-2) + 4*a(n-1) + 1*a(n) = n^3. - Gottfried Helms, Apr 15 2002
a(n) is sum of all the possible products p*q where (p,q) are ordered pairs and p + q = n + 1. E.g., a(5) = 5 + 8 + 9 + 8 + 5 = 35. - Amarnath Murthy, May 29 2003
Number of labeled graphs on n+3 nodes that are triangles. - Jon Perry, Jun 14 2003
Number of permutations of n+3 which have exactly 1 descent and avoid the pattern 1324. - Mike Zabrocki, Nov 05 2004
Schlaefli symbol for this polyhedron: {3,3}.
Transform of n^2 under the Riordan array (1/(1-x^2), x). - Paul Barry, Apr 16 2005
a(n) is a perfect square only for n = {1, 2, 48}. E.g., a(48) = 19600 = 140^2. - Alexander Adamchuk, Nov 24 2006
a(n+1) is the number of terms in the expansion of (a_1 + a_2 + a_3 + a_4)^n. - Sergio Falcon, Feb 12 2007 [Corrected by Graeme McRae, Aug 28 2007]
a(n+1) is the number of terms in the complete homogeneous symmetric polynomial of degree n in 3 variables. - Richard Barnes, Sep 06 2017
This is also the average "permutation entropy", sum((pi(n)-n)^2)/n!, over the set of all possible n! permutations pi. - Jeff Boscole (jazzerciser(AT)hotmail.com), Mar 20 2007
a(n) = (d/dx)(S(n, x), x)|A049310.%20-%20_Wolfdieter%20Lang">{x = 2}. First derivative of Chebyshev S-polynomials evaluated at x = 2. See A049310. - _Wolfdieter Lang, Apr 04 2007
If X is an n-set and Y a fixed (n-1)-subset of X then a(n-2) is equal to the number of 3-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Complement of A145397; A023533(a(n))=1; A014306(a(n))=0. - Reinhard Zumkeller, Oct 14 2008
Equals row sums of triangle A152205. - Gary W. Adamson, Nov 29 2008
a(n) is the number of gifts received from the lyricist's true love up to and including day n in the song "The Twelve Days of Christmas". a(12) = 364, almost the number of days in the year. - Bernard Hill (bernard(AT)braeburn.co.uk), Dec 05 2008
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF2 denominators of A156925. See A157703 for background information. - Johannes W. Meijer, Mar 07 2009
Starting with 1 = row sums of triangle A158823. - Gary W. Adamson, Mar 28 2009
Wiener index of the path with n edges. - Eric W. Weisstein, Apr 30 2009
This is a 'Matryoshka doll' sequence with alpha=0, the multiplicative counterpart is A000178: seq(add(add(i,i=alpha..k),k=alpha..n),n=alpha..50). - Peter Luschny, Jul 14 2009
a(n) is the number of nondecreasing triples of numbers from a set of size n, and it is the number of strictly increasing triples of numbers from a set of size n+2. - Samuel Savitz, Sep 12 2009 [Corrected and enhanced by Markus Sigg, Sep 24 2023]
a(n) is the number of ordered sequences of 4 nonnegative integers that sum to n. E.g., a(2) = 10 because 2 = 2 + 0 + 0 + 0 = 1 + 1 + 0 + 0 = 0 + 2 + 0 + 0 = 1 + 0 + 1 + 0 = 0 + 1 + 1 + 0 = 0 + 0 + 2 + 0 = 1 + 0 + 0 + 1 = 0 + 1 + 0 + 1 = 0 + 0 + 1 + 1 = 0 + 0 + 0 + 2. - Artur Jasinski, Nov 30 2009
a(n) corresponds to the total number of steps to memorize n verses by the technique described in A173964. - Ibrahima Faye (ifaye2001(AT)yahoo.fr), Feb 22 2010
The number of (n+2)-bit numbers which contain two runs of 1's in their binary expansion. - Vladimir Shevelev, Jul 30 2010
a(n) is also, starting at the second term, the number of triangles formed in n-gons by intersecting diagonals with three diagonal endpoints (see the first column of the table in Sommars link). - Alexandre Wajnberg, Aug 21 2010
Column sums of:
1 4 9 16 25...
1 4 9...
1...
..............
--------------
1 4 10 20 35...
From Johannes W. Meijer, May 20 2011: (Start)
The Ca3, Ca4, Gi3 and Gi4 triangle sums (see A180662 for their definitions) of the Connell-Pol triangle A159797 are linear sums of shifted versions of the duplicated tetrahedral numbers, e.g., Gi3(n) = 17*a(n) + 19*a(n-1) and Gi4(n) = 5*a(n) + a(n-1).
Furthermore the Kn3, Kn4, Ca3, Ca4, Gi3 and Gi4 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above. (End)
a(n-2)=N_0(n), n >= 1, with a(-1):=0, is the number of vertices of n planes in generic position in three-dimensional space. See a comment under A000125 for general arrangement. Comment to Arnold's problem 1990-11, see the Arnold reference, p. 506. - Wolfdieter Lang, May 27 2011
We consider optimal proper vertex colorings of a graph G. Assume that the labeling, i.e., coloring starts with 1. By optimality we mean that the maximum label used is the minimum of the maximum integer label used across all possible labelings of G. Let S=Sum of the differences |l(v) - l(u)|, the sum being over all edges uv of G and l(w) is the label associated with a vertex w of G. We say G admits unique labeling if all possible labelings of G is S-invariant and yields the same integer partition of S. With an offset this sequence gives the S-values for the complete graph on n vertices, n = 2, 3, ... . - K.V.Iyer, Jul 08 2011
Central term of commutator of transverse Virasoro operators in 4-D case for relativistic quantum open strings (ref. Zwiebach). - Tom Copeland, Sep 13 2011
Appears as a coefficient of a Sturm-Liouville operator in the Ovsienko reference on page 43. - Tom Copeland, Sep 13 2011
For n > 0: a(n) is the number of triples (u,v,w) with 1 <= u <= v <= w <= n, cf. A200737. - Reinhard Zumkeller, Nov 21 2011
Regarding the second comment above by Amarnath Murthy (May 29 2003), see A181118 which gives the sequence of ordered pairs. - L. Edson Jeffery, Dec 17 2011
The dimension of the space spanned by the 3-form v[ijk] that couples to M2-brane worldsheets wrapping 3-cycles inside tori (ref. Green, Miller, Vanhove eq. 3.9). - Stephen Crowley, Jan 05 2012
a(n+1) is the number of 2 X 2 matrices with all terms in {0, 1, ..., n} and (sum of terms) = n. Also, a(n+1) is the number of 2 X 2 matrices with all terms in {0, 1, ..., n} and (sum of terms) = 3*n. - Clark Kimberling, Mar 19 2012
Using n + 4 consecutive triangular numbers t(1), t(2), ..., t(n+4), where n is the n-th term of this sequence, create a polygon by connecting points (t(1), t(2)) to (t(2), t(3)), (t(2), t(3)) to (t(3), t(4)), ..., (t(1), t(2)) to (t(n+3), t(n+4)). The area of this polygon will be one-half of each term in this sequence. - J. M. Bergot, May 05 2012
Pisano period lengths: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17,108, 19, 40, ... . (The Pisano sequence modulo m is the auxiliary sequence p(n) = a(n) mod m, n >= 1, for some m. p(n) is periodic for all sequences with rational g.f., like this one, and others. The lengths of the period of p(n) are quoted here for m>=1.) - R. J. Mathar, Aug 10 2012
a(n) is the maximum possible number of rooted triples consistent with any phylogenetic tree (level-0 phylogenetic network) containing exactly n+2 leaves. - Jesper Jansson, Sep 10 2012
For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 4, 1, 2, 8, 2, 3, 3, 3, 4, 7, 4, 5, 2, 5, 6, 6, 6, 7, 1, 7, 8, 5, 8, 9, 9, 9}, which just rephrases the Pisano period length above. - Ant King, Oct 18 2012
a(n) is the number of functions f from {1, 2, 3} to {1, 2, ..., n + 4} such that f(1) + 1 < f(2) and f(2) + 1 < f(3). - Dennis P. Walsh, Nov 27 2012
a(n) is the Szeged index of the path graph with n+1 vertices; see the Diudea et al. reference, p. 155, Eq. (5.8). - Emeric Deutsch, Aug 01 2013
Also the number of permutations of length n that can be sorted by a single block transposition. - Vincent Vatter, Aug 21 2013
From J. M. Bergot, Sep 10 2013: (Start)
a(n) is the 3 X 3 matrix determinant
| C(n,1) C(n,2) C(n,3) |
| C(n+1,1) C(n+1,2) C(n+1,3) |
| C(n+2,1) C(n+2,2) C(n+2,3) |
(End)
In physics, a(n)/2 is the trace of the spin operator S_z^2 for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and the sum of their squares is 10/2 = a(3)/2. - Stanislav Sykora, Nov 06 2013
a(n+1) = (n+1)*(n+2)*(n+3)/6 is also the dimension of the Hilbert space of homogeneous polynomials of degree n. - L. Edson Jeffery, Dec 12 2013
For n >= 4, a(n-3) is the number of permutations of 1,2...,n with the distribution of up (1) - down (0) elements 0...0111 (n-4 zeros), or, equivalently, a(n-3) is up-down coefficient {n,7} (see comment in A060351). - Vladimir Shevelev, Feb 15 2014
a(n) is one-half the area of the region created by plotting the points (n^2,(n+1)^2). A line connects points (n^2,(n+1)^2) and ((n+1)^2, (n+2)^2) and a line is drawn from (0,1) to each increasing point. From (0,1) to (4,9) the area is 2; from (0,1) to (9,16) the area is 8; further areas are 20,40,70,...,2*a(n). - J. M. Bergot, May 29 2014
Beukers and Top prove that no tetrahedral number > 1 equals a square pyramidal number A000330. - Jonathan Sondow, Jun 21 2014
a(n+1) is for n >= 1 the number of nondecreasing n-letter words over the alphabet [4] = {1, 2, 3, 4} (or any other four distinct numbers). a(2+1) = 10 from the words 11, 22, 33, 44, 12, 13, 14, 23, 24, 34; which is also the maximal number of distinct elements in a symmetric 4 X 4 matrix. Inspired by the Jul 20 2014 comment by R. J. Cano on A000582. - Wolfdieter Lang, Jul 29 2014
Degree of the q-polynomial counting the orbits of plane partitions under the action of the symmetric group S3. Orbit-counting generating function is Product_{i <= j <= k <= n} ( (1 - q^(i + j + k - 1))/(1 - q^(i + j + k - 2)) ). See q-TSPP reference. - Olivier Gérard, Feb 25 2015
Row lengths of tables A248141 and A248147. - Reinhard Zumkeller, Oct 02 2014
If n is even then a(n) = Sum_{k=1..n/2} (2k)^2. If n is odd then a(n) = Sum_{k=0..(n-1)/2} (1+2k)^2. This can be illustrated as stacking boxes inside a square pyramid on plateaus of edge lengths 2k or 2k+1, respectively. The largest k are the 2k X 2k or (2k+1) X (2k+1) base. - R. K. Guy, Feb 26 2015
Draw n lines in general position in the plane. Any three define a triangle, so in all we see C(n,3) = a(n-2) triangles (6 lines produce 4 triangles, and so on). - Terry Stickels, Jul 21 2015
a(n-2) = fallfac(n,3)/3!, n >= 3, is also the number of independent components of an antisymmetric tensor of rank 3 and dimension n. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Number of compositions (ordered partitions) of n+3 into exactly 4 parts. - Juergen Will, Jan 02 2016
Number of weak compositions (ordered weak partitions) of n-1 into exactly 4 parts. - Juergen Will, Jan 02 2016
For n >= 2 gives the number of multiplications of two nonzero matrix elements in calculating the product of two upper n X n triangular matrices. - John M. Coffey, Jun 23 2016
Terms a(4n+1), n >= 0, are odd, all others are even. The 2-adic valuation of the subsequence of every other term, a(2n+1), n >= 0, yields the ruler sequence A007814. Sequence A275019 gives the 2-adic valuation of a(n). - M. F. Hasler, Dec 05 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 12 2017
C(n+2,3) is the number of ways to select 1 triple among n+2 objects, thus a(n) is the coefficient of x1^(n-1)*x3 in exponential Bell polynomial B_{n+2}(x1,x2,...), hence its link with A050534 and A001296 (see formula). - Cyril Damamme, Feb 26 2018
a(n) is also the number of 3-cycles in the (n+4)-path complement graph. - Eric W. Weisstein, Apr 11 2018
a(n) is the general number of all geodetic graphs of diameter n homeomorphic to a complete graph K4. - Carlos Enrique Frasser, May 24 2018
a(n) + 4*a(n-1) + a(n-2) = n^3 = A000578(n), for n >= 0 (extending the a(n) formula given in the name). This is the Worpitzky identity for cubes. (Number of components of the decomposition of a rank 3 tensor in dimension n >= 1 into symmetric, mixed and antisymmetric parts). For a(n-2) see my Dec 10 2015 comment. - Wolfdieter Lang, Jul 16 2019
a(n) also gives the total number of regular triangles of length k (in some length unit), with k from {1, 2, ..., n}, in the matchstick arrangement with enclosing triangle of length n, but only triangles with the orientation of the enclosing triangle are counted. Row sums of unsigned A122432(n-1, k-1), for n >= 1. See the Andrew Howroyd comment in A085691. - Wolfdieter Lang, Apr 06 2020
a(n) is the number of bigrassmannian permutations on n+1 elements, i.e., permutations which have a unique left descent, and a unique right descent. - Rafael Mrden, Aug 21 2020
a(n-2) is the number of chiral pairs of colorings of the edges or vertices of a triangle using n or fewer colors. - Robert A. Russell, Oct 20 2020
a(n-2) is the number of subsets of {1,2,...,n} whose diameters are their size. For example, for n=4, a(2)=4 and the sets are {1,3}, {2,4}, {1,2,4}, {1,3,4}. - Enrique Navarrete, Dec 26 2020
For n>1, a(n-2) is the number of subsets of {1,2,...,n} in which the second largest element is the size of the subset. For example, for n=4, a(2)=4 and the sets are {2,3}, {2,4}, {1,3,4}, {2,3,4}. - Enrique Navarrete, Jan 02 2021
a(n) is the number of binary strings of length n+2 with exactly three 0's. - Enrique Navarrete, Jan 15 2021
From Tom Copeland, Jun 07 2021: (Start)
Aside from the zero, this sequence is the fourth diagonal of the Pascal matrix A007318 and the only nonvanishing diagonal (fourth) of the matrix representation IM = (A132440)^3/3! of the differential operator D^3/3!, when acting on the row vector of coefficients of an o.g.f., or power series.
M = e^{IM} is the lower triangular matrix of coefficients of the Appell polynomial sequence p_n(x) = e^{D^3/3!} x^n = e^{b. D} x^n = (b. + x)^n = Sum_{k=0..n} binomial(n,k) b_n x^{n-k}, where the (b.)^n = b_n have the e.g.f. e^{b.t} = e^{t^3/3!}, which is that for A025035 aerated with double zeros, the first column of M.
See A099174 and A000332 for analogous relationships for the third and fifth diagonals of the Pascal matrix. (End)
a(n) is the number of circles with a radius of integer length >= 1 and center at a grid point in an n X n grid. - Albert Swafford, Jun 11 2021
Maximum Wiener index over all connected graphs with n+1 vertices. - Allan Bickle, Jul 09 2022
The third Euler row (1,4,1) has an additional connection with the tetrahedral numbers besides the n^3 identity stated above: a^2(n) + 4*a^2(n+1) + a^2(n+2) = a(n^2+4n+4), which can be shown with algebra. E.g., a^2(2) + 4*a^2(3) + a^2(4) = 16 + 400 + 400 = a(16). Although an analogous thing happens with the (1,1) row of Euler's triangle and triangular numbers C(n+1,2) = A000217(n) = T(n), namely both T(n-1) + T(n) = n^2 and T^2(n-1) + T^2(n) = T(n^2) are true, only one (the usual identity) still holds for the Euler row (1,11,11,1) and the C(n,4) numbers in A000332. That is, the dot product of (1,11,11,1) with the squares of 4 consecutive terms of A000332 is not generally a term of A000332. - Richard Peterson, Aug 21 2022
For n > 1, a(n-2) is the number of solutions of the Diophantine equation x1 + x2 + x3 + x4 + x5 = n, subject to the constraints 0 <= x1, 1 <= x2, 2 <= x3, 0 <= x4 <= 1, 0 <= x5 and x5 is even. - Daniel Checa, Nov 03 2022
a(n+1) is also the number of vertices of the generalized Pitman-Stanley polytope with parameters 2, n, and vector (1,1, ... ,1), which is integrally equivalent to a flow polytope over the grid graph having 2 rows and n columns. - William T. Dugan, Sep 18 2023
a(n) is the number of binary words of length (n+1) containing exactly one substring 01. a(2) = 4: 001, 010, 011, 101. - Nordine Fahssi, Dec 09 2024
a(n) is the number of directed bishop moves on an n X n chessboard, identified under rotations (0, 90, 180 and 270 degree) and all reflections. - Hilko Koning, Aug 27 2025

Examples

			a(2) = 3*4*5/6 = 10, the number of balls in a pyramid of 3 layers of balls, 6 in a triangle at the bottom, 3 in the middle layer and 1 on top.
Consider the square array
  1  2  3  4  5  6 ...
  2  4  6  8 10 12 ...
  3  6  9 12 16 20 ...
  4  8 12 16 20 24 ...
  5 10 15 20 25 30 ...
  ...
then a(n) = sum of n-th antidiagonal. - _Amarnath Murthy_, Apr 06 2003
G.f. = x + 4*x^2 + 10*x^3 + 20*x^4 + 35*x^5 + 56*x^6 + 84*x^7 + 120*x^8 + 165*x^9 + ...
Example for a(3+1) = 20 nondecreasing 3-letter words over {1,2,3,4}: 111, 222, 333; 444, 112, 113, 114, 223, 224, 122, 224, 133, 233, 144, 244, 344; 123, 124, 134, 234.  4 + 4*3 + 4 = 20. - _Wolfdieter Lang_, Jul 29 2014
Example for a(4-2) = 4 independent components of a rank 3 antisymmetric tensor A of dimension 4: A(1,2,3), A(1,2,4), A(1,3,4) and A(2,3,4). - _Wolfdieter Lang_, Dec 10 2015

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_0.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, pp. 44, 70.
  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 4.
  • M. V. Diudea, I. Gutman, and J. Lorentz, Molecular Topology, Nova Science, 2001, Huntington, N.Y. pp. 152-156.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, pp. 292-293.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • V. Ovsienko and S. Tabachnikov, Projective Differential Geometry Old and New, Cambridge Tracts in Mathematics (no. 165), Cambridge Univ. Press, 2005.
  • Kenneth A Ross, First Digits of Squares and Cubes, Math. Mag. 85 (2012) 36-42. doi:10.4169/math.mag.85.1.36.
  • 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).
  • A. Szenes, The combinatorics of the Verlinde formulas (N.J. Hitchin et al., ed.), in Vector bundles in algebraic geometry, Cambridge, 1995.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 11-13.
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 126-127.
  • B. Zwiebach, A First Course in String Theory, Cambridge, 2004; see p. 226.

Links

Crossrefs

Bisections give A000447 and A002492.
Sums of 2 consecutive terms give A000330.
a(3n-3) = A006566(n). A000447(n) = a(2n-2). A002492(n) = a(2n+1).
Column 0 of triangle A094415.
Cf. A000217 (first differences), A001044, (see above example), A061552, A040977, A133111, A133112, A152205, A158823, A156925, A157703, A173964, A058187, A190717, A190718, A100440, A181118, A222716.
Partial sums are A000332. - Jonathan Vos Post, Mar 27 2011
Cf. A216499 (the analogous sequence for level-1 phylogenetic networks).
Cf. A068980 (partitions), A231303 (spin physics).
Cf. similar sequences listed in A237616.
Cf. A104712 (second column, if offset is 2).
Cf. A145397 (non-tetrahedral numbers). - Daniel Forgues, Apr 11 2015
Cf. A127324.
Cf. A007814, A275019 (2-adic valuation).
Cf. A000578 (cubes), A005900 (octahedral numbers), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Cf. A002817 (4-cycle count of \bar P_{n+4}), A060446 (5-cycle count of \bar P_{n+3}), A302695 (6-cycle count of \bar P_{n+5})
Row 2 of A325000 (simplex facets and vertices) and A327084 (simplex edges and ridges).
Cf. A085691 (matchsticks), A122432 (unsigned row sums).
Cf. (triangle colorings) A006527 (oriented), A000290 (achiral), A327085 (chiral simplex edges and ridges).
Row 3 of A321791 (cycles of n colors using k or fewer colors).
Cf. A007318, A025035, A099174.
The Wiener indices of powers of paths for k = 1..6 are given in A000292, A002623, A014125, A122046, A122047, and A175724, respectively.

Programs

  • GAP
    a:=n->Binomial(n+2,3);; A000292:=List([0..50],n->a(n)); # Muniru A Asiru, Feb 28 2018
  • Haskell
    a000292 n = n * (n + 1) * (n + 2) `div` 6
a000292_list = scanl1 (+) a000217_list
-- Reinhard Zumkeller, Jun 16 2013, Feb 09 2012, Nov 21 2011
  • Magma
    [n*(n+1)*(n+2)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 03 2014
  • Maple
    a:=n->n*(n+1)*(n+2)/6; seq(a(n), n=0..50);
A000292 := n->binomial(n+2,3); seq(A000292(n), n=0..50);
isA000292 := proc(n)
    option remember;
    local a,i ;
    for i from iroot(6*n,3)-1 do
        a := A000292(i) ;
        if a > n then
            return false;
        elif a = n then
            return true;
        end if;
    end do:
end proc: # R. J. Mathar, Aug 14 2024
  • Mathematica
    Table[Binomial[n + 2, 3], {n, 0, 20}] (* Zerinvary Lajos, Jan 31 2010 *)
Accumulate[Accumulate[Range[0, 50]]] (* Harvey P. Dale, Dec 10 2011 *)
Table[n (n + 1)(n + 2)/6, {n,0,100}] (* Wesley Ivan Hurt, Sep 25 2013 *)
Nest[Accumulate, Range[0, 50], 2] (* Harvey P. Dale, May 24 2017 *)
Binomial[Range[20] + 1, 3] (* Eric W. Weisstein, Sep 08 2017 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 4, 10}, 20] (* Eric W. Weisstein, Sep 08 2017 *)
CoefficientList[Series[x/(-1 + x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 08 2017 *)
Table[Range[n].Range[n,1,-1],{n,0,50}] (* Harvey P. Dale, Mar 02 2024 *)
  • Maxima
    A000292(n):=n*(n+1)*(n+2)/6$ makelist(A000292(n),n,0,60); /* Martin Ettl, Oct 24 2012 */
  • PARI
    a(n) = (n) * (n+1) * (n+2) / 6  \\ corrected by Harry J. Smith, Dec 22 2008
  • PARI
    a=vector(10000);a[2]=1;for(i=3,#a,a[i]=a[i-2]+i*i); \\ Stanislav Sykora, Nov 07 2013
  • PARI
    is(n)=my(k=sqrtnint(6*n,3)); k*(k+1)*(k+2)==6*n \\ Charles R Greathouse IV, Dec 13 2016
  • Python
    # Compare A000217.
def A000292():
    x, y, z = 1, 1, 1
    yield 0
    while True:
        yield x
        x, y, z = x + y + z + 1, y + z + 1, z + 1
a = A000292(); print([next(a) for i in range(45)]) # Peter Luschny, Aug 03 2019

Formula

a(n) = C(n+2,3) = n*(n+1)*(n+2)/6 (see the name).
G.f.: x / (1 - x)^4.
a(n) = -a(-4 - n) for all in Z.
a(n) = Sum_{k=0..n} A000217(k) = Sum_{k=1..n} Sum_{j=0..k} j, partial sums of the triangular numbers.
a(2n)= A002492(n). a(2n+1)=A000447(n+1).
a(n) = Sum_{1 <= i <= j <= n} |i - j|. - Amarnath Murthy, Aug 05 2002
a(n) = (n+3)*a(n-1)/n. - Ralf Stephan, Apr 26 2003
Sums of three consecutive terms give A006003. - Ralf Stephan, Apr 26 2003
Determinant of the n X n symmetric Pascal matrix M_(i, j) = C(i+j+2, i). - Benoit Cloitre, Aug 19 2003
The sum of a series constructed by the products of the index and the length of the series (n) minus the index (i): a(n) = sum[i(n-i)]. - Martin Steven McCormick (mathseq(AT)wazer.net), Apr 06 2005
a(n) = Sum_{k=0..floor((n-1)/2)} (n-2k)^2 [offset 0]; a(n+1) = Sum_{k=0..n} k^2*(1-(-1)^(n+k-1))/2 [offset 0]. - Paul Barry, Apr 16 2005
a(n) = -A108299(n+5, 6) = A108299(n+6, 7). - Reinhard Zumkeller, Jun 01 2005
a(n) = -A110555(n+4, 3). - Reinhard Zumkeller, Jul 27 2005
Values of the Verlinde formula for SL_2, with g = 2: a(n) = Sum_{j=1..n-1} n/(2*sin^2(j*Pi/n)). - Simone Severini, Sep 25 2006
a(n-1) = (1/(1!*2!))*Sum_{1 <= x_1, x_2 <= n} |det V(x_1, x_2)| = (1/2)*Sum_{1 <= i,j <= n} |i-j|, where V(x_1, x_2) is the Vandermonde matrix of order 2. Column 2 of A133112. - Peter Bala, Sep 13 2007
Starting with 1 = binomial transform of [1, 3, 3, 1, ...]; e.g., a(4) = 20 = (1, 3, 3, 1) dot (1, 3, 3, 1) = (1 + 9 + 9 + 1). - Gary W. Adamson, Nov 04 2007
a(n) = A006503(n) - A002378(n). - Reinhard Zumkeller, Sep 24 2008
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n >= 4. - Jaume Oliver Lafont, Nov 18 2008
Sum_{n>=1} 1/a(n) = 3/2, case x = 1 in Gradstein-Ryshik 1.513.7. - R. J. Mathar, Jan 27 2009
E.g.f.:((x^3)/6 + x^2 + x)*exp(x). - Geoffrey Critzer, Feb 21 2009
Limit_{n -> oo} A171973(n)/a(n) = sqrt(2)/2. - Reinhard Zumkeller, Jan 20 2010
With offset 1, a(n) = (1/6)*floor(n^5/(n^2 + 1)). - Gary Detlefs, Feb 14 2010
a(n) = Sum_{k = 1..n} k*(n-k+1). - Vladimir Shevelev, Jul 30 2010
a(n) = (3*n^2 + 6*n + 2)/(6*(h(n+2) - h(n-1))), n > 0, where h(n) is the n-th harmonic number. - Gary Detlefs, Jul 01 2011
a(n) = coefficient of x^2 in the Maclaurin expansion of 1 + 1/(x+1) + 1/(x+1)^2 + 1/(x+1)^3 + ... + 1/(x+1)^n. - Francesco Daddi, Aug 02 2011
a(n) = coefficient of x^4 in the Maclaurin expansion of sin(x)*exp((n+1)*x). - Francesco Daddi, Aug 04 2011
a(n) = 2*A002415(n+1)/(n+1). - Tom Copeland, Sep 13 2011
a(n) = A004006(n) - n - 1. - Reinhard Zumkeller, Mar 31 2012
a(n) = (A007531(n) + A027480(n) + A007290(n))/11. - J. M. Bergot, May 28 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 1. - Ant King, Oct 18 2012
G.f.: x*U(0) where U(k) = 1 + 2*x*(k+2)/( 2*k+1 - x*(2*k+1)*(2*k+5)/(x*(2*k+5)+(2*k+2)/U(k+1) )); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Dec 01 2012
a(n^2 - 1) = (1/2)*(a(n^2 - n - 2) + a(n^2 + n - 2)) and
a(n^2 + n - 2) - a(n^2 - 1) = a(n-1)*(3*n^2 - 2) = 10*A024166(n-1), by Berselli's formula in A222716. - Jonathan Sondow, Mar 04 2013
G.f.: x + 4*x^2/(Q(0)-4*x) where Q(k) = 1 + k*(x+1) + 4*x - x*(k+1)*(k+5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n+1) = det(C(i+3,j+2), 1 <= i,j <= n), where C(n,k) are binomial coefficients. - Mircea Merca, Apr 06 2013
a(n) = a(n-2) + n^2, for n > 1. - Ivan N. Ianakiev, Apr 16 2013
a(2n) = 4*(a(n-1) + a(n)), for n > 0. - Ivan N. Ianakiev, Apr 26 2013
G.f.: x*G(0)/2, where G(k) = 1 + 1/(1 - x/(x + (k+1)/(k+4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 02 2013
a(n) = n + 2*a(n-1) - a(n-2), with a(0) = a(-1) = 0. - Richard R. Forberg, Jul 11 2013
a(n)*(m+1)^3 + a(m)*(n+1) = a(n*m + n + m), for any nonnegative integers m and n. This is a 3D analog of Euler's theorem about triangular numbers, namely t(n)*(2m+1)^2 + t(m) = t(2nm + n + m), where t(n) is the n-th triangular number. - Ivan N. Ianakiev, Aug 20 2013
Sum_{n>=0} a(n)/(n+1)! = 2*e/3 = 1.8121878856393... . Sum_{n>=1} a(n)/n! = 13*e/6 = 5.88961062832... . - Richard R. Forberg, Dec 25 2013
a(n+1) = A023855(n+1) + A023856(n). - Wesley Ivan Hurt, Sep 24 2013
a(n) = A024916(n) + A076664(n), n >= 1. - Omar E. Pol, Feb 11 2014
a(n) = A212560(n) - A059722(n). - J. M. Bergot, Mar 08 2014
Sum_{n>=1} (-1)^(n + 1)/a(n) = 12*log(2) - 15/2 = 0.8177661667... See A242024, A242023. - Richard R. Forberg, Aug 11 2014
3/(Sum_{n>=m} 1/a(n)) = A002378(m), for m > 0. - Richard R. Forberg, Aug 12 2014
a(n) = Sum_{i=1..n} Sum_{j=i..n} min(i,j). - Enrique Pérez Herrero, Dec 03 2014
Arithmetic mean of Square pyramidal number and Triangular number: a(n) = (A000330(n) + A000217(n))/2. - Luciano Ancora, Mar 14 2015
a(k*n) = a(k)*a(n) + 4*a(k-1)*a(n-1) + a(k-2)*a(n-2). - Robert Israel, Apr 20 2015
Dirichlet g.f.: (zeta(s-3) + 3*zeta(s-2) + 2*zeta(s-1))/6. - Ilya Gutkovskiy, Jul 01 2016
a(n) = A080851(1,n-1) - R. J. Mathar, Jul 28 2016
a(n) = (A000578(n+1) - (n+1) ) / 6. - Zhandos Mambetaliyev, Nov 24 2016
G.f.: x/(1 - x)^4 = (x * r(x) * r(x^2) * r(x^4) * r(x^8) * ...), where r(x) = (1 + x)^4 = (1 + 4x + 6x^2 + 4x^3 + x^4); and x/(1 - x)^4 = (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...) where r(x) = (1 + x + x^2)^4. - Gary W. Adamson, Jan 23 2017
a(n) = A000332(n+3) - A000332(n+2). - Bruce J. Nicholson, Apr 08 2017
a(n) = A001296(n) - A050534(n+1). - Cyril Damamme, Feb 26 2018
a(n) = Sum_{k=1..n} (-1)^(n-k)*A122432(n-1, k-1), for n >= 1, and a(0) = 0. - Wolfdieter Lang, Apr 06 2020
From Robert A. Russell, Oct 20 2020: (Start)
a(n) = A006527(n) - a(n-2) = (A006527(n) + A000290(n)) / 2 = a(n-2) + A000290(n).
a(n-2) = A006527(n) - a(n) = (A006527(n) - A000290(n)) / 2 = a(n) - A000290(n).
a(n) = 1*C(n,1) + 2*C(n,2) + 1*C(n,3), where the coefficient of C(n,k) is the number of unoriented triangle colorings using exactly k colors.
a(n-2) = 1*C(n,3), where the coefficient of C(n,k) is the number of chiral pairs of triangle colorings using exactly k colors.
a(n-2) = A327085(2,n). (End)
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(sqrt(2)*Pi)/(3*sqrt(2)*Pi).
Product_{n>=2} (1 - 1/a(n)) = sqrt(2)*sinh(sqrt(2)*Pi)/(33*Pi). (End)
a(n) = A002623(n-1) + A002623(n-2), for n>1. - Ivan N. Ianakiev, Nov 14 2021

Extensions

Corrected and edited by Daniel Forgues, May 14 2010

A000332 Binomial coefficient binomial(n,4) = n*(n-1)*(n-2)*(n-3)/24.

Original entry on oeis.org

0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of intersection points of diagonals of convex n-gon where no more than two diagonals intersect at any point in the interior.
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation. - Ignacio Larrosa Cañestro, Apr 09 2002. [See Les Reid link for proof. - N. J. A. Sloane, Apr 02 2016] [See Peter Kagey link for alternate proof. - Sameer Gauria, Jul 29 2025]
Start from cubane and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink on chemistry. - Robert G. Wilson v, Aug 02 2002
For n>0, a(n) = (-1/8)*(coefficient of x in Zagier's polynomial P_(2n,n)). (Zagier's polynomials are used by PARI/GP for acceleration of alternating or positive series.)
Figurate numbers based on the 4-dimensional regular convex polytope called the regular 4-simplex, pentachoron, 5-cell, pentatope or 4-hypertetrahedron with Schlaefli symbol {3,3,3}. a(n)=((n*(n-1)*(n-2)*(n-3))/4!). - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 07 2009
Maximal number of crossings that can be created by connecting n vertices with straight lines. - Cameron Redsell-Montgomerie (credsell(AT)uoguelph.ca), Jan 30 2007
If X is an n-set and Y a fixed (n-1)-subset of X then a(n) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Product of four consecutive numbers divided by 24. - Artur Jasinski, Dec 02 2007
The only prime in this sequence is 5. - Artur Jasinski, Dec 02 2007
For strings consisting entirely of 0's and 1's, the number of distinct arrangements of four 1's such that 1's are not adjacent. The shortest possible string is 7 characters, of which there is only one solution: 1010101, corresponding to a(5). An eight-character string has 5 solutions, nine has 15, ten has 35 and so on, congruent to A000332. - Gil Broussard, Mar 19 2008
For a(n)>0, a(n) is pentagonal if and only if 3 does not divide n. All terms belong to the generalized pentagonal sequence (A001318). Cf. A000326, A145919, A145920. - Matthew Vandermast, Oct 28 2008
Nonzero terms = row sums of triangle A158824. - Gary W. Adamson, Mar 28 2009
Except for the 4 initial 0's, is equivalent to the partial sums of the tetrahedral numbers A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009
If the first 3 zeros are disregarded, that is, if one looks at binomial(n+3, 4) with n>=0, then it becomes a 'Matryoshka doll' sequence with alpha=0: seq(add(add(add(i,i=alpha..k),k=alpha..n),n=alpha..m),m=alpha..50). - Peter Luschny, Jul 14 2009
For n>=1, a(n) is the number of n-digit numbers the binary expansion of which contains two runs of 0's. - Vladimir Shevelev, Jul 30 2010
For n>0, a(n) is the number of crossing set partitions of {1,2,..,n} into n-2 blocks. - Peter Luschny, Apr 29 2011
The Kn3, Ca3 and Gi3 triangle sums of A139600 are related to the sequence given above, e.g., Gi3(n) = 2*A000332(n+3) - A000332(n+2) + 7*A000332(n+1). For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 29 2011
For n > 3, a(n) is the hyper-Wiener index of the path graph on n-2 vertices. - Emeric Deutsch, Feb 15 2012
Except for the four initial zeros, number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts. - V.J. Pohjola, Aug 31 2012
a(n+3) is the number of different ways to color the faces (or the vertices) of a regular tetrahedron with n colors if we count mirror images as the same.
a(n) = fallfac(n,4)/4! is also the number of independent components of an antisymmetric tensor of rank 4 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Does not satisfy Benford's law [Ross, 2012] - N. J. A. Sloane, Feb 12 2017
Number of chiral pairs of colorings of the vertices (or faces) of a regular tetrahedron with n available colors. Chiral colorings come in pairs, each the reflection of the other. - Robert A. Russell, Jan 22 2020
From Mircea Dan Rus, Aug 26 2020: (Start)
a(n+3) is the number of lattice rectangles (squares included) in a staircase of order n; this is obtained by stacking n rows of consecutive unit lattice squares, aligned either to the left or to the right, which consist of 1, 2, 3, ..., n squares and which are stacked either in the increasing or in the decreasing order of their lengths. Below, there is a staircase or order 4 which contains a(7) = 35 rectangles. [See the Teofil Bogdan and Mircea Dan Rus link, problem 3, under A004320]
_
||
|||_
|||_|_
|||_|_|
(End)
a(n+4) is the number of strings of length n on an ordered alphabet of 5 letters where the characters in the word are in nondecreasing order. E.g., number of length-2 words is 15: aa,ab,ac,ad,ae,bb,bc,bd,be,cc,cd,ce,dd,de,ee. - Jim Nastos, Jan 18 2021
From Tom Copeland, Jun 07 2021: (Start)
Aside from the zeros, this is the fifth diagonal of the Pascal matrix A007318, the only nonvanishing diagonal (fifth) of the matrix representation IM = (A132440)^4/4! of the differential operator D^4/4!, when acting on the row vector of coefficients of an o.g.f., or power series.
M = e^{IM} is the matrix of coefficients of the Appell sequence p_n(x) = e^{D^4/4!} x^n = e^{b. D} x^n = (b. + x)^n = Sum_{k=0..n} binomial(n,k) b_n x^{n-k}, where the (b.)^n = b_n have the e.g.f. e^{b.t} = e^{t^4/4!}, which is that for A025036 aerated with triple zeros, the first column of M.
See A099174 and A000292 for analogous relationships for the third and fourth diagonals of the Pascal matrix. (End)
For integer m and positive integer r >= 3, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (3 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024

Examples

			a(5) = 5 from the five independent components of an antisymmetric tensor A of rank 4 and dimension 5, namely A(1,2,3,4), A(1,2,3,5), A(1,2,4,5), A(1,3,4,5) and A(2,3,4,5). See the Dec 10 2015 comment. - _Wolfdieter Lang_, Dec 10 2015

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 74, Problem 8.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 70.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 7.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 294.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Charles W. Trigg, Mathematical Quickies, New York: Dover Publications, Inc., 1985, p. 53, #191.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 127.

Links

Crossrefs

binomial(n, k): A161680 (k = 2), A000389 (k = 5), A000579 (k = 6), A000580 (k = 7), A000581 (k = 8), A000582 (k = 9).
Cf. A053134, A053126, A075733, A006322, A006325.
Cf. also A000583, A014820, A092181, A092182, A092183.
Cf. A000217, A000292, A007318 (column k = 4).
Cf. A158824.
Cf. A006008 (Number of ways to color the faces (or vertices) of a regular tetrahedron with n colors when mirror images are counted as two).
Cf. A104712 (third column, k=4).
See A269747 for a 3-D analog.
Cf. A006008 (oriented), A006003 (achiral) tetrahedron colorings.
Row 3 of A325000, col. 4 of A007318.
Cf. A000292, A025036, A099174.

Programs

  • GAP
    A000332 := List([1..10^2], n -> Binomial(n, 4)); # Muniru A Asiru, Oct 16 2017
  • Magma
    [Binomial(n,4): n in [0..50]]; // Vincenzo Librandi, Nov 23 2014
  • Maple
    A000332 := n->binomial(n,4); [seq(binomial(n,4), n=0..100)];
  • Mathematica
    Table[ Binomial[n, 4], {n, 0, 45} ] (* corrected by Harvey P. Dale, Aug 22 2011 *)
Table[(n-4)(n-3)(n-2)(n-1)/24, {n, 100}] (* Artur Jasinski, Dec 02 2007 *)
LinearRecurrence[{5,-10,10,-5,1}, {0,0,0,0,1}, 45] (* Harvey P. Dale, Aug 22 2011 *)
CoefficientList[Series[x^4 / (1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 23 2014 *)
  • PARI
    a(n)=binomial(n,4);
  • Python
    # Starts at a(3), i.e. computes n*(n+1)*(n+2)*(n+3)/24
# which is more in line with A000217 and A000292.
def A000332():
    x, y, z, u = 1, 1, 1, 1
    yield 0
    while True:
        yield x
        x, y, z, u = x + y + z + u + 1, y + z + u + 1, z + u + 1, u + 1
a = A000332(); print([next(a) for i in range(41)]) # Peter Luschny, Aug 03 2019
  • Python
    print([n*(n-1)*(n-2)*(n-3)//24 for n in range(50)])
# Gennady Eremin, Feb 06 2022

Formula

a(n) = n*(n-1)*(n-2)*(n-3)/24.
G.f.: x^4/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = n*a(n-1)/(n-4). - Benoit Cloitre, Apr 26 2003, R. J. Mathar, Jul 07 2009
a(n) = Sum_{k=1..n-3} Sum_{i=1..k} i*(i+1)/2. - Benoit Cloitre, Jun 15 2003
Convolution of natural numbers {1, 2, 3, 4, ...} and A000217, the triangular numbers {1, 3, 6, 10, ...}. - Jon Perry, Jun 25 2003
a(n) = A110555(n+1,4). - Reinhard Zumkeller, Jul 27 2005
a(n+1) = ((n^5-(n-1)^5) - (n^3-(n-1)^3))/24 - (n^5-(n-1)^5-1)/30; a(n) = A006322(n-2)-A006325(n-1). - Xavier Acloque, Oct 20 2003; R. J. Mathar, Jul 07 2009
a(4*n+2) = Pyr(n+4, 4*n+2) where the polygonal pyramidal numbers are defined for integers A>2 and B>=0 by Pyr(A, B) = B-th A-gonal pyramid number = ((A-2)*B^3 + 3*B^2 - (A-5)*B)/6; For all positive integers i and the pentagonal number function P(x) = x*(3*x-1)/2: a(3*i-2) = P(P(i)) and a(3*i-1) = P(P(i) + i); 1 + 24*a(n) = (n^2 + 3*n + 1)^2. - Jonathan Vos Post, Nov 15 2004
First differences of A000389(n). - Alexander Adamchuk, Dec 19 2004
For n > 3, the sum of the first n-2 tetrahedral numbers (A000292). - Martin Steven McCormick (mathseq(AT)wazer.net), Apr 06 2005 [Corrected by Doug Bell, Jun 25 2017]
Starting (1, 5, 15, 35, ...), = binomial transform of [1, 4, 6, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
Sum_{n>=4} 1/a(n) = 4/3, from the Taylor expansion of (1-x)^3*log(1-x) in the limit x->1. - R. J. Mathar, Jan 27 2009
A034263(n) = (n+1)*a(n+4) - Sum_{i=0..n+3} a(i). Also A132458(n) = a(n)^2 - a(n-1)^2 for n>0. - Bruno Berselli, Dec 29 2010
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=1. - Harvey P. Dale, Aug 22 2011
a(n) = (binomial(n-1,2)^2 - binomial(n-1,2))/6. - Gary Detlefs, Nov 20 2011
a(n) = Sum_{k=1..n-2} Sum_{i=1..k} i*(n-k-2). - Wesley Ivan Hurt, Sep 25 2013
a(n) = (A000217(A000217(n-2) - 1))/3 = ((((n-2)^2 + (n-2))/2)^2 - (((n-2)^2 + (n-2))/2))/(2*3). - Raphie Frank, Jan 16 2014
Sum_{n>=0} a(n)/n! = e/24. Sum_{n>=3} a(n)/(n-3)! = 73*e/24. See A067764 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
Sum_{n>=4} (-1)^(n+1)/a(n) = 32*log(2) - 64/3 = A242023 = 0.847376444589... . - Richard R. Forberg, Aug 11 2014
4/(Sum_{n>=m} 1/a(n)) = A027480(m-3), for m>=4. - Richard R. Forberg, Aug 12 2014
E.g.f.: x^4*exp(x)/24. - Robert Israel, Nov 23 2014
a(n+3) = C(n,1) + 3*C(n,2) + 3*C(n,3) + C(n,4). Each term indicates the number of ways to use n colors to color a tetrahedron with exactly 1, 2, 3, or 4 colors.
a(n) = A080852(1,n-4). - R. J. Mathar, Jul 28 2016
From Gary W. Adamson, Feb 06 2017: (Start)
G.f.: Starting (1, 5, 14, ...), x/(1-x)^5 can be written
as (x * r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = (1+x)^5;
as (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...) where r(x) = (1+x+x^2)^5;
as (x * r(x) * r(x^4) * r(x^16) * r(x^64) * ...) where r(x) = (1+x+x^2+x^3)^5;
... (as a conjectured infinite set). (End)
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = A006008(n) - a(n+3) = (A006008(n) - A006003(n)) / 2 = a(n+3) - A006003(n).
a(n+3) = A006008(n) - a(n) = (A006008(n) + A006003(n)) / 2 = a(n) + A006003(n).
a(n) = A007318(n,4).
a(n+3) = A325000(3,n). (End)
Product_{n>=5} (1 - 1/a(n)) = cosh(sqrt(15)*Pi/2)/(100*Pi). - Amiram Eldar, Jan 21 2021

Extensions

Some formulas that referred to another offset corrected by R. J. Mathar, Jul 07 2009

A327084 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 10, 11, 1, 5, 20, 66, 34, 1, 6, 35, 276, 792, 156, 1, 7, 56, 900, 10688, 25506, 1044, 1, 8, 84, 2451, 90005, 1601952, 2302938, 12346, 1, 9, 120, 5831, 533358, 43571400, 892341888, 591901884, 274668
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
A(n,k) is also the number of unoriented colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of unoriented colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  1  2   3     4     5      6       7       8        9       10 ...
  1  4  10    20    35     56      84     120      165      220 ...
  1 11  66   276   900   2451    5831   12496    24651    45475 ...
  1 34 792 10688 90005 533358 2437848 9156288 29522961 84293770 ...
  ...
For A(2,3) = 10, the nine achiral colorings are AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, and CCC. The chiral pair is ABC-ACB.

Links

Crossrefs

Cf. A327083 (oriented), A327085 (chiral), A327086 (achiral), A327088 (exactly k colors), A325000 (vertices, facets), A337884 (faces, peaks), A337408 (orthotope edges, orthoplex ridges), A337412 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000027, A000292, A063842(n-1), A063843.
Cf. A063841 (k-multigraphs on n nodes).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
compress[x : {{, } ...}] := (s = Sort[x]; For[i = Length[s], i > 1, i -= 1, If[s[[i, 1]] == s[[i-1,1]], s[[i-1,2]] += s[[i,2]]; s = Delete[s,i], Null]]; s)
CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] & /@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
row[n_Integer] := row[n] = Factor[Total[pc[#] j^Total[CycleX[#]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
array[n_, k_] := row[n] /. j -> k
Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
(* Using Fripertinger's exponent per Andrew Howroyd code in A063841: *)
pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
array[n_,k_] := Total[pc[#]k^ex[#] &/@ IntegerPartitions[n+1]]/(n+1)!
Table[array[n,d-n+1], {d,10}, {n,d}] // Flatten
(* Another program (translated from Andrew Howroyd's PARI code): *)
permcount[v_] := Module[{m=1, s=0, k=0, t}, For[i=1, i <= Length[v], i++, t = v[[i]]; k = If[i>1 && t == v[[i-1]], k+1, 1]; m *= t*k; s += t]; s!/m];
edges[v_] := Sum[GCD[v[[i]], v[[j]]], {i, 2, Length[v]}, {j, 1, i-1}] + Total[Quotient[v, 2]];
T[n_, k_] := Module[{s = 0}, Do[s += permcount[p]*k^edges[p], {p, IntegerPartitions[n+1]}]; s/(n+1)!];
Table[T[n-k+1, k], {n, 1, 9}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jan 08 2021 *)
  • PARI
    permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m}
edges(v) = {sum(i=2, #v, sum(j=1, i-1, gcd(v[i], v[j]))) + sum(i=1, #v, v[i]\2)}
T(n, k) = {my(s=0); forpart(p=n+1, s+=permcount(p)*k^edges(p)); s/(n+1)!} \\ Andrew Howroyd, Sep 06 2019
  • Python
    from itertools import combinations
from math import prod, gcd, factorial
from fractions import Fraction
from sympy.utilities.iterables import partitions
def A327084_T(n,k): return int(sum(Fraction(k**(sum(p[r]*p[s]*gcd(r,s) for r,s in combinations(p.keys(),2))+sum((q>>1)*r+(q*r*(r-1)>>1) for q, r in p.items())),prod(q**r*factorial(r) for q, r in p.items())) for p in partitions(n+1))) # Chai Wah Wu, Jul 09 2024

Formula

The algorithm used in the Mathematica program below assigns each permutation of the vertices to a partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327088(n,j) * binomial(k,j).
A(n,k) = A327083(n,k) - A327085(n,k) = (A327083(n,k) + A327086(n,k)) / 2 = A327085(n,k) + A327086(n,k).
A(n,k) = A063841(n+1,k-1).

A327085 Array read by descending antidiagonals: A(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 4, 21, 6, 0, 0, 10, 140, 405, 28, 0, 0, 20, 575, 7904, 17154, 252, 0, 0, 35, 1785, 76880, 1415648, 1920375, 4726, 0, 0, 56, 4606, 486522, 41453650, 855834880, 547375212, 150324, 0
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n-2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. The chiral colorings of its edges come in pairs, each the reflection of the other.
A(n,k) is also the number of chiral pairs of colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of chiral pairs of colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  0 0   0    0     0      0       0       0        0        0         0 ...
  0 0   1    4    10     20      35      56       84      120       165 ...
  0 1  21  140   575   1785    4606   10416    21330    40425     71995 ...
  0 6 405 7904 76880 486522 2300305 8806336 28725192 82626270 214744629 ...
  ...
For A(2,3) = 1, the chiral pair is ABC-ACB.

Links

Crossrefs

Cf. A327083 (oriented), A327084 (unoriented), A327086 (achiral), A327089 (exactly k colors), A325000(n,k-n) (vertices, facets), A337885 (faces, peaks), A337409 (orthotope edges, orthoplex ridges), A337413 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000004, A000292(n-2), A337899, A331352.

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
compress[x : {{, } ...}] := (s = Sort[x]; For[i = Length[s], i > 1, i -= 1, If[s[[i, 1]] == s[[i-1,1]], s[[i-1,2]] += s[[i,2]]; s = Delete[s,i], Null]]; s)
CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] & /@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
row[n_Integer] := row[n] = Factor[Total[If[EvenQ[Total[1-Mod[#,2]]], 1, -1] pc[#] j^Total[CycleX[#]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
array[n_, k_] := row[n] /. j -> k
Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
(* Using Fripertinger's exponent per Andrew Howroyd's code in A063841: *)
pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
array[n_,k_] := Total[If[EvenQ[Total[1-Mod[#,2]]],1,-1] pc[#]k^ex[#] &/@ IntegerPartitions[n+1]]/(n+1)!
Table[array[n,d-n+1], {d,10}, {n,d}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the vertices to a partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327089(n,j) * binomial(k,j).
A(n,k) = A327083(n,k) - A327084(n,k) = (A327083(n,k) - A327086(n,k)) / 2 = A327084(n,k) - A327086(n,k).

A325001 Array read by descending antidiagonals: A(n,k) is the number of achiral colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 5, 1, 5, 16, 15, 6, 1, 6, 25, 34, 21, 7, 1, 7, 36, 65, 56, 28, 8, 1, 8, 49, 111, 125, 84, 36, 9, 1, 9, 64, 175, 246, 210, 120, 45, 10, 1, 10, 81, 260, 441, 461, 330, 165, 55, 11, 1, 11, 100, 369, 736, 917, 792, 495, 220, 66, 12, 1
Offset: 1

Views

Author

Robert A. Russell, Mar 23 2019

Keywords

Comments

For n=1, the figure is a line segment with two vertices. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with four triangular faces. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Each of its n+1 facets is a regular (n-1)-dimensional simplex. An achiral coloring is the same as its reflection.

Examples

			The array begins with A(1,1):
  1  2  3   4   5    6    7     8     9    10    11     12     13 ...
  1  4  9  16  25   36   49    64    81   100   121    144    169 ...
  1  5 15  34  65  111  175   260   369   505   671    870   1105 ...
  1  6 21  56 125  246  441   736  1161  1750  2541   3576   4901 ...
  1  7 28  84 210  461  917  1688  2919  4795  7546  11452  16848 ...
  1  8 36 120 330  792 1715  3424  6399 11320 19118  31032  48672 ...
  1  9 45 165 495 1287 3003  6434 12861 24265 43593  75087 124683 ...
  1 10 55 220 715 2002 5005 11440 24309 48610 92323 167740 293215 ...
  ...
For A(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.

Links

Crossrefs

Cf. A324999 (oriented), A325000 (unoriented), A325000(n,k-n) (chiral), A325003 (exactly k colors), A327086 (edges, ridges), A337886 (faces, peaks), A325007 (orthotope facets, orthoplex vertices), A325015 (orthoplex facets, orthotope vertices).
Rows 1-4 are A000027, A000290, A006003, A132366(n-1).
Column 2 is A162880.

Programs

  • Mathematica
    Table[Binomial[d+1,n+1] - Binomial[d+1-n,n+1], {d,1,15}, {n,1,d}] // Flatten

Formula

A(n,k) = binomial(n+k,n+1) - binomial(k,n+1).
A(n,k) = Sum_{j=1..n} A325003(n,j) * binomial(k,j).
A(n,k) = 2*A325000(n,k) - A324999(n,k) = A324999(n,k) - 2*A325000(n,k-n) = A325000(n,k) - A325000(n,k-n).
G.f. for row n: (x - x^(n+1)) / (1-x)^(n+2).
Linear recurrence for row n: A(n,k) = Sum_{j=1..n+1} -binomial(j-n-2,j) * A(n,k-j).
G.f. for column k: (1 - (1-x^2)^k) / (x*(1-x)^k).

A325005 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the facets of a regular n-dimensional orthotope using up to k colors.

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 10, 21, 10, 1, 15, 55, 56, 15, 1, 21, 120, 220, 126, 21, 1, 28, 231, 680, 715, 252, 28, 1, 36, 406, 1771, 3060, 2002, 462, 36, 1, 45, 666, 4060, 10626, 11628, 5005, 792, 45, 1, 55, 1035, 8436, 31465, 53130, 38760, 11440, 1287, 55, 1
Offset: 1

Views

Author

Robert A. Russell, Mar 23 2019

Keywords

Comments

Also called hypercube, n-dimensional cube, and measure polytope. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is a cube with six square faces. For n=4, the figure is a tesseract with eight cubic facets. The Schläfli symbol, {4,3,...,3}, of the regular n-dimensional orthotope (n>1) consists of a four followed by n-2 threes. Each of its 2n facets is an (n-1)-dimensional orthotope. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
Also the number of unoriented colorings of the vertices of a regular n-dimensional orthoplex using up to k colors.

Examples

			Array begins with A(1,1):
1  3    6    10     15      21       28        36        45         55 ...
1  6   21    55    120     231      406       666      1035       1540 ...
1 10   56   220    680    1771     4060      8436     16215      29260 ...
1 15  126   715   3060   10626    31465     82251    194580     424270 ...
1 21  252  2002  11628   53130   201376    658008   1906884    5006386 ...
1 28  462  5005  38760  230230  1107568   4496388  15890700   50063860 ...
1 36  792 11440 116280  888030  5379616  26978328 115775100  436270780 ...
1 45 1287 24310 319770 3108105 23535820 145008513 752538150 3381098545 ...
For A(1,2) = 3, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses one color for each vertex.

Links

Crossrefs

Cf. A325004 (oriented), A325006 (chiral), A325007 (achiral), A325009 (exactly k colors).
Other n-dimensional polytopes: A325000 (simplex), A325013 (orthoplex).
Rows 1-3 are A000217, A002817, A198833.

Programs

  • Mathematica
    Table[Binomial[Binomial[d-n+2,2]+n-1,n],{d,1,11},{n,1,d}] // Flatten

Formula

A(n,k) = binomial(n + binomial(k+1,2) - 1, n).
A(n,k) = Sum_{j=1..2n} A325009(n,j) * binomial(k,j).
A(n,k) = A325004(n,k) - A325006(n,k) = (A325004(n,k) + A325007(n,k)) / 2 = A325006(n,k) + A325007(n,k).
G.f. for row n: Sum_{j=1..2n} A325009(n,j) * x^j / (1-x)^(j+1).
Linear recurrence for row n: T(n,k) = Sum_{j=0..2n} binomial(-2-j,2n-j) * T(n,k-1-j).
G.f. for column k: 1/(1-x)^binomial(k+1,2) - 1.

A324999 Array read by descending antidiagonals: A(n,k) is the number of oriented colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 4, 1, 9, 4, 1, 16, 11, 5, 1, 25, 24, 15, 6, 1, 36, 45, 36, 21, 7, 1, 49, 76, 75, 56, 28, 8, 1, 64, 119, 141, 127, 84, 36, 9, 1, 81, 176, 245, 258, 210, 120, 45, 10, 1, 100, 249, 400, 483, 463, 330, 165, 55, 11, 1, 121, 340, 621, 848, 931, 792, 495, 220, 66, 12, 1
Offset: 1

Views

Author

Robert A. Russell, Mar 23 2019

Keywords

Comments

For n=1, the figure is a line segment with two vertices. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with four triangular faces. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Each of its n+1 facets is a regular (n-1)-dimensional simplex. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.

Examples

			The array begins with A(1,1):
  1  4  9  16  25   36   49    64    81   100   121    144    169    196 ...
  1  4 11  24  45   76  119   176   249   340   451    584    741    924 ...
  1  5 15  36  75  141  245   400   621   925  1331   1860   2535   3381 ...
  1  6 21  56 127  258  483   848  1413  2254  3465   5160   7475  10570 ...
  1  7 28  84 210  463  931  1744  3087  5215  8470  13300  20280  30135 ...
  1  8 36 120 330  792 1717  3440  6471 11560 19778  32616  52104  80952 ...
  1  9 45 165 495 1287 3003  6436 12879 24355 43923  76077 127257 206493 ...
  1 10 55 220 715 2002 5005 11440 24311 48630 92433 168180 294645 499422 ...
  ...
For A(1,2) = 4, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses two colors. For A(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.

Links

Crossrefs

Cf. A325000 (unoriented), A325000(n,k-n) (chiral), A325001 (achiral), A325002 (exactly k colors), A327083 (edges, ridges), A337883 (faces, peaks), A325004 (orthotope facets, orthoplex vertices), A325012 (orthoplex facets, orthotope vertices).
Rows 1-4 are A000290, A006527, A006008, A337895.

Programs

  • Mathematica
    Table[Binomial[d+1,n+1] + Binomial[d+1-n,n+1], {d,1,15}, {n,1,d}] // Flatten

Formula

A(n,k) = binomial(n+k,n+1) + binomial(k,n+1).
A(n,k) = Sum_{j=1..n+1} A325002(n,j) * binomial(k,j).
A(n,k) = A325000(n,k) + A325000(n,k-n) = 2*A325000(n,k) - A325001(n,k) = 2*A325000(n,k-n) + A325001(n,k).
G.f. for row n: (x + x^(n+1)) / (1-x)^(n+2).
Linear recurrence for row n: A(n,k) = Sum_{j=1..n+2} -binomial(j-n-3,j) * A(n,k-j).
G.f. for column k: (1 - 2*(1-x)^k + (1-x^2)^k) / (x*(1-x)^k) - 2*k.

A325013 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the facets of a regular n-dimensional orthoplex using up to k colors.

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 10, 21, 22, 1, 15, 55, 267, 402, 1, 21, 120, 1996, 132102, 1228158, 1, 28, 231, 10375, 11756666, 484086357207, 400507806843728, 1, 36, 406, 41406, 405385550, 4805323147589984, 74515759884862073604656433, 527471432057653004017274030725792, 1
Offset: 1

Views

Author

Robert A. Russell, May 27 2019

Keywords

Comments

Also called cross polytope and hyperoctahedron. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is an octahedron with eight triangular faces. For n=4, the figure is a 16-cell with sixteen tetrahedral facets. The Schläfli symbol, {3,...,3,4}, of the regular n-dimensional orthoplex (n>1) consists of n-2 threes followed by a four. Each of its 2^n facets is an (n-1)-dimensional simplex. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
Also the number of unoriented colorings of the vertices of a regular n-dimensional orthotope (cube) using up to k colors.

Examples

			Array begins with A(1,1):
1   3      6       10        15         21          28           36 ...
1   6     21       55       120        231         406          666 ...
1  22    267     1996     10375      41406      135877       384112 ...
1 402 132102 11756666 405385550 7416923886 86986719477 735192450952 ...
For A(2,2)=6, two squares have all edges the same color, two have three edges the same color, one has opposite edges the same color, and one has opposite edges different colors.

Links

Crossrefs

Cf. A325012 (oriented), A325014 (chiral), A325015 (achiral), A325017 (exactly k colors).
Other n-dimensional polytopes: A325000 (simplex), A325005 (orthotope).
Rows 1-4 are A000217, A002817, A128766, A128767; column 2 is A000616.
Cf. A000048, A001037.

Programs

  • Mathematica
    a48[n_] := a48[n] = DivisorSum[NestWhile[#/2&, n, EvenQ], MoebiusMu[#]2^(n/#)&]/(2n); (* A000048 *)
a37[n_] := a37[n] = DivisorSum[n, MoebiusMu[n/#]2^#&]/n; (* A001037 *)
CI0[{n_Integer}] := CI0[{n}] = CI[Transpose[If[EvenQ[n], p2 = IntegerExponent[n, 2]; sub = Divisors[n/2^p2]; {2^(p2+1) sub, a48 /@ (2^p2 sub) }, sub = Divisors[n]; {sub, a37 /@ sub}]]] 2^(n-1); (* even perm. *)
CI1[{n_Integer}] := CI1[{n}] = CI[sub = Divisors[n]; Transpose[If[EvenQ[n], {sub, a37 /@ sub}, {2 sub, a48 /@ sub}]]] 2^(n-1); (* odd perm. *)
compress[x : {{, } ...}] := (s = Sort[x]; For[i = Length[s], i > 1, i -= 1, If[s[[i, 1]]==s[[i-1, 1]], s[[i-1, 2]] += s[[i, 2]]; s = Delete[s, i], Null]]; s)
cix[{a_, b_}, {c_, d_}] := {LCM[a, c], (a b c d)/LCM[a, c]};
Unprotect[Times]; Times[CI[a_List], CI[b_List]] :=  (* combine *) CI[compress[Flatten[Outer[cix, a, b, 1], 1]]]; Protect[Times];
CI0[p_List] := CI0[p] = Expand[CI0[Drop[p, -1]] CI0[{Last[p]}] + CI1[Drop[p, -1]] CI1[{Last[p]}]]
CI1[p_List] := CI1[p] = Expand[CI0[Drop[p, -1]] CI1[{Last[p]}] + CI1[Drop[p, -1]] CI0[{Last[p]}]]
pc[p_List] := Module[{ci,mb},mb = DeleteDuplicates[p]; ci = Count[p, #] & /@ mb; n!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
row[n_Integer] := row[n] = Factor[(Total[((CI0[#] + CI1[#]) pc[#]) & /@ IntegerPartitions[n]])/(n! 2^n)] /. CI[l_List] :> j^(Total[l][[2]])
array[n_, k_] := row[n] /. j -> k
Table[array[n, d-n+1], {d, 1, 10}, {n, 1, d}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n. It then determines the number of permutations for each partition and the cycle index for each partition.
A(n,k) = A325012(n,k) - A325014(n,k) = (A325012(n,k) + A325015(n,k)) / 2 = A325014(n,k) + A325015(n,k).
A(n,k) = Sum_{j=1..2^n} A325017(n,j) * binomial(k,j).

A337884 Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the triangular faces of a regular n-dimensional simplex using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 15, 34, 1, 5, 35, 792, 2136, 1, 6, 70, 10688, 4977909, 7013320, 1, 7, 126, 90005, 1533771392, 9930666709494, 1788782616656, 1, 8, 210, 533358, 132597435125, 234249157811872000, 12979877431438089379035, 53304527811667897248, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. An n-simplex has n+1 vertices. For n=2, the figure is a triangle with one triangular face. For n=3, the figure is a tetrahedron with 4 triangular faces. For higher n, the number of triangular faces is C(n+1,3).
Also the number of unoriented colorings of the peaks of a regular n-dimensional simplex. A peak of an n-simplex is an (n-3)-dimensional simplex.

Examples

			Table begins with T(2,1):
 1    2       3          4            5             6               7 ...
 1    5      15         35           70           126             210 ...
 1   34     792      10688        90005        533358         2437848 ...
 1 2136 4977909 1533771392 132597435125 5079767935320 110837593383153 ...
For T(3,4)=35, the 34 achiral arrangements are AAAA, AAAB, AAAC, AAAD, AABB, AABC, AABD, AACC, AACD, AADD, ABBB, ABBC, ABBD, ABCC, ABDD, ACCC, ACCD, ACDD, ADDD, BBBB, BBBC, BBBD, BBCC, BBCD, BBDD, BCCC, BCCD, BCDD, BDDD, CCCC, CCCD, CCDD, CDDD, and DDDD. The chiral pair is ABCD-ABDC.

Links

Crossrefs

Cf. A337883 (oriented), A337885 (chiral), A337886 (achiral), A051168 (binary Lyndon words).
Other elements: A325000 (vertices), A327084 (edges).
Other polytopes: A337888 (orthotope), A337892 (orthoplex).
Rows 2-4 are A000027, A000332(n+3), A063843.

Programs

  • Mathematica
    m=2; (* dimension of color element, here a triangular face *)
lw[n_,k_]:=lw[n, k]=DivisorSum[GCD[n,k],MoebiusMu[#]Binomial[n/#,k/#]&]/n (*A051168*)
cxx[{a_, b_},{c_, d_}]:={LCM[a, c], GCD[a, c] b d}
compress[x:{{, } ...}] := (s=Sort[x];For[i=Length[s],i>1,i-=1,If[s[[i,1]]==s[[i-1,1]], s[[i-1,2]]+=s[[i,2]]; s=Delete[s,i], Null]]; s)
combine[a : {{, } ...}, b : {{, } ...}] := Outer[cxx, a, b, 1]
CX[p_List, 0] := {{1, 1}} (* cycle index for partition p, m vertices *)
CX[{n_Integer}, m_] := If[2m>n, CX[{n}, n-m], CX[{n},m] = Table[{n/k, lw[n/k, m/k]}, {k, Reverse[Divisors[GCD[n, m]]]}]]
CX[p_List, m_Integer] := CX[p, m] = Module[{v = Total[p], q, r}, If[2 m > v, CX[p, v - m], q = Drop[p, -1]; r = Last[p]; compress[Flatten[Join[{{CX[q, m]}}, Table[combine[CX[q, m - j], CX[{r}, j]], {j, Min[m, r]}]], 2]]]]
pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
row[n_Integer] := row[n] = Factor[Total[pc[#] j^Total[CX[#, m+1]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
array[n_, k_] := row[n] /. j -> k
Table[array[n,d+m-n], {d,8}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the vertices to a partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition using a formula for binary Lyndon words. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337883(n,k) - A337885(n,k) = (A337883(n,k) + A337886(n,k)) / 2 = A337885(n,k) + A337886(n,k).
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