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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A134287 Fifth column of triangle A103371 (without leading zeros).

Original entry on oeis.org

1, 30, 315, 1960, 8820, 31752, 97020, 261360, 637065, 1431430, 3006003, 5962320, 11262160, 20391840, 35581680, 60093504, 98590905, 157608990, 246142435, 376372920, 564559380, 832117000, 1206913500, 1724814000, 2431508625
Offset: 0

Views

Author

Wolfdieter Lang, Nov 13 2007

Keywords

Comments

Kekulé numbers for certain benzenoids.
a(n) = K(L(n))*K(O(2,4,n)) with the Cyvin and Gutman Kekulé number notation. See p. 62 for the L(n) structure with K(L(n))=n+1 and p. 105 (i) for the O(k,m,n) structure and its Kekulé number. This corresponds to an essentially disconnected 7-tier benzenoid structure similar to the 6-tier structure shown on p. 230, nr. 23 (see A108647).
a(n-5), n >= 5, is the number of ways to put n identical objects into m=5 of altogether n distinguishable boxes (n-5 boxes stay empty).

Examples

			a(2)=315 because n=7 identical balls can be put into m=5 of n=7 distinguishable boxes in binomial(7,5)*(5!/(4!*1!)+ 5!/(3!*2!)) = 21*(5+10) = 315 ways. The m=5 part partitions of 7, namely (1^4,3) and (1^3,2^2) specify the filling of each of the 21 possible five box choices. - _Wolfdieter Lang_, Nov 13 2007
		

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.

Crossrefs

Cf. A108647 (fourth column of triangle A103371).

Programs

  • Haskell
    a134287 = flip a103371 4 . (+ 4)  -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    [5*Binomial(n+5, 5)^2/(n+5): n in [0..30]]; // G. C. Greubel, Oct 28 2022
    
  • Maple
    seq(binomial(n+4,4)^2*(n+5)/5, n=0..24); # Peter Luschny, Jan 13 2014
  • Mathematica
    CoefficientList[Series[(1 + 20 x + 60 x^2 + 40 x^3 + 5 x^4)/(1 - x)^10, {x, 0, 24}], x]
  • MuPAD
    5*binomial(n+5,5)^2/(n+5) $ n = 0..35; // Zerinvary Lajos, May 09 2008
    
  • PARI
    a(n) = 5*binomial(n+5, 5)^2/(n+5); \\ Michel Marcus, Jan 07 2014
    
  • SageMath
    [5*binomial(n+5,5)^2/(n+5) for n in range(31)] # G. C. Greubel, Oct 28 2022

Formula

a(n) = A103371(n+4,4), n >= 0.
a(n) = ((n+1)*(n+2)*(n+3)*(n+4))^2*(n+5)/2880, n >= 0. 2880 = 4!*5! = A010790(4).
G.f.: (1+20*x+60*x^2+40*x^3+5*x^4)/(1-x)^10. Numerator polynomial from fifth row of triangle A132813.
a(n) = 5*C(n+5,5)^2/(n+5), n >= 0. - Zerinvary Lajos, May 09 2008
a(n) = (C(n+6,6)*C(n+5,4)+5*C(n+5,6)*C(n+5,4))/(n+5). - Gary Detlefs, Jan 06 2014
From Amiram Eldar, May 31 2022: (Start)
Sum_{n>=0} 1/a(n) = 350*Pi^2/3 - 13805/12.
Sum_{n>=0} (-1)^n/a(n) = 5*Pi^2 + 640*log(2)/3 - 785/4. (End)
E.g.f.: (2880 + 83520*x + 368640*x^2 + 529920*x^3 + 330120*x^4 + 102024*x^5 + 16616*x^6 + 1432*x^7 + 61*x^8 + x^9)*exp(x)/2880. - G. C. Greubel, Oct 28 2022

A001700 a(n) = binomial(2*n+1, n+1): number of ways to put n+1 indistinguishable balls into n+1 distinguishable boxes = number of (n+1)-st degree monomials in n+1 variables = number of monotone maps from 1..n+1 to 1..n+1.

Original entry on oeis.org

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Views

Author

Keywords

Comments

To show for example that C(2n+1, n+1) is the number of monotone maps from 1..n + 1 to 1..n + 1, notice that we can describe such a map by a nondecreasing sequence of length n + 1 with entries from 1 to n + 1. The number k of increases in this sequence is anywhere from 0 to n. We can specify these increases by throwing k balls into n+1 boxes, so the total is Sum_{k = 0..n} C((n+1) + k - 1, k) = C(2*n+1, n+1).
Also number of ordered partitions (or compositions) of n + 1 into n + 1 parts. E.g., a(2) = 10: 003, 030, 300, 012, 021, 102, 120, 210, 201, 111. - Mambetov Bektur (bektur1987(AT)mail.ru), Apr 17 2003
Also number of walks of length n on square lattice, starting at origin, staying in first and second quadrants. - David W. Wilson, May 05 2001. (E.g., for n = 2 there are 10 walks, all starting at 0, 0: 0, 1 -> 0, 0; 0, 1 -> 1, 1; 0, 1 -> 0, 2; 1, 0 -> 0, 0; 1, 0 -> 1, 1; 1, 0 -> 2, 0; 1, 0 -> 1, -1; -1, 0 -> 0, 0; -1, 0 -> -1, 1; -1, 0-> -2, 0.)
Also total number of leaves in all ordered trees with n + 1 edges.
Also number of digitally balanced numbers [A031443] from 2^(2*n+1) to 2^(2*n+2). - Naohiro Nomoto, Apr 07 2001
Also number of ordered trees with 2*n + 2 edges having root of even degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of paths of length 2*d(G) connecting two neighboring nodes in optimal chordal graph of degree 4, G(2*d(G)^2 + 2*d(G) + 1, 2d(G) + 1), where d(G) = diameter of graph G. - S. Bujnowski (slawb(AT)atr.bydgoszcz.pl), Feb 11 2002
Define an array by m(1, j) = 1, m(i, 1) = i, m(i, j) = m(i, j-1) + m(i-1, j); then a(n) = m(n, n), diagonal of A165257 - Benoit Cloitre, May 07 2002
Also the numerator of the constant term in the expansion of cos^(2*n)(x) or sin^(2*n)(x) when the denominator is 2^(2*n-1). - Robert G. Wilson v
Consider the expansion of cos^n(x) as a linear combination of cosines of multiple angles. If n is odd, then the expansion is a combination of a*cos((2*k-1)*x)/2^(n-1) for all 2*k - 1 <= n. If n is even, then the expansion is a combination of a*cos(2k*x)/2^(n-1) terms plus a constant. "The constant term, [a(n)/2^(2n-1)], is due to the fact that [cos^2n(x)] is never negative, i.e., electrical engineers would say the average or 'dc value' of [cos^(2*n)(x)] is [a(n)/2^(2*n-1)]. The dc value of [cos^(2*n-1)(x)] on the other hand, is zero because it is symmetrical about the horizontal axis, i.e., it is negative and positive equally." Nahin[62] - Robert G. Wilson v, Aug 01 2002
Also number of times a fixed Dyck word of length 2*k occurs in all Dyck words of length 2*n + 2*k. Example: if the fixed Dyck word is xyxy (k = 2), then it occurs a(1) = 3 times in the 5 Dyck words of length 6 (n = 1): (xy[xy)xy], xyxxyy, xxyyxy, x(xyxy)y, xxxyyy (placed between parentheses). - Emeric Deutsch, Jan 02 2003
a(n+1) is the determinant of the n X n matrix m(i, j) = binomial(2*n-i, j). - Benoit Cloitre, Aug 26 2003
a(n-1) = (2*n)!/(2*n!*n!), formula in [Davenport] used by Gauss for the special case prime p = 4*n + 1: x = a(n-1) mod p and y = x*(2n)! mod p are solutions of p = x^2 + y^2. - Frank Ellermann. Example: For prime 29 = 4*7 + 1 use a(7-1) = 1716 = (2*7)!/(2*7!*7!), 5 = 1716 mod 29 and 2 = 5*(2*7)! mod 29, then 29 = 5*5 + 2*2.
The number of compositions of 2*n, say c_1 + c_2 + ... + c_k = 2n, satisfy that Sum_{i = 1..j} c_i < 2*j for all j = 1..k, or equivalently, the number of subsets, say S, of [2*n-1] = {1, 2, ..., 2*n-1} with at least n elements such that if 2k is in S, then there must be at least k elements in S smaller than 2k. E.g., a(2) = 3 because we can write 4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 1 + 2 + 1. - Ricky X. F. Chen (ricky_chen(AT)mail.nankai.edu.cn), Jul 30 2006
The number of walks of length 2*n + 1 on an infinite linear lattice that begin at the origin and end at node (1). Also the number of paths on a square lattice from the origin to (n+1, n) that use steps (1,0) and (0,1). Also number of binary numbers of length 2*n + 1 with n + 1 ones and n zeros. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
If Y is a 3-subset of an 2*n-set X then, for n >= 3, a(n-1) is the number of n-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
Also the number of rankings (preferential arrangements) of n unlabeled elements onto n levels when empty levels are allowed. - Thomas Wieder, May 24 2008
Also the Catalan transform of A000225 shifted one index, i.e., dropping A000225(0). - R. J. Mathar, Nov 11 2008
With offset 1. The number of solutions in nonnegative integers to X1 + X2 + ... + Xn = n. The number of terms in the expansion of (X1 + X2 + ... + Xn)^n. The coefficient of x^n in the expansion of (1 + x + x^2 + ...)^n. The number of distinct image sets of all functions taking [n] into [n]. - Geoffrey Critzer, Feb 22 2009
The Hankel transform of the aerated sequence 1, 0, 3, 0, 10, 0, ... is 1, 3, 3, 5, 5, 7, 7, ... (A109613(n+1)). - Paul Barry, Apr 21 2009
Also the number of distinct network topologies for a network of n items with 1 to n - 1 unidirectional connections to other objects in the network. - Anthony Bachler, May 05 2010
Equals INVERT transform of the Catalan numbers starting with offset 1. E.g.: a(3) = 35 = (1, 2, 5) dot (10, 3, 1) + 14 = 21 + 14 = 35. - Gary W. Adamson, May 15 2009
The integral of 1/(1+x^2)^(n+1) is given by a(n)/2^(2*n - 1) * (x/(1 + x^2)^n*P(x) + arctan(x)), where P(x) is a monic polynomial of degree 2*n - 2 with rational coefficients. - Christiaan van de Woestijne, Jan 25 2011
a(n) is the number of Schroder paths of semilength n in which the (2,0)-steps at level 0 come in 2 colors and there are no (2,0)-steps at a higher level. Example: a(2) = 10 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 2^2 = 4 paths of shape HH, 2 paths of shape HUD, 2 paths of shape UDH, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
a(n) is the number of Motzkin paths of length n in which the (1,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=35 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 3^3 = 27 paths of shape HHH, 3 paths of shape HUD, 3 paths of shape UDH, and 2 paths of shape UHD. - Emeric Deutsch, May 02 2011
Also number of digitally balanced numbers having length 2*(n + 1) in binary representation: a(n) = #{m: A070939(A031443(m)) = 2*(n + 1)}. - Reinhard Zumkeller, Jun 08 2011
a(n) equals 2^(2*n + 3) times the coefficient of Pi in 2F1([1/2, n+2]; [3/2]; -1). - John M. Campbell, Jul 17 2011
For positive n, a(n) equals 4^(n+2) times the coefficient of Pi^2 in Integral_{x = 0..Pi/2} x sin^(2*n + 2)x. - John M. Campbell, Jul 19 2011 [Apparently, the contributor means Integral_{x = 0..Pi/2} x * (sin(x))^(2*n + 2).]
a(n-1) = C(2*n, n)/2 is the number of ways to assign 2*n people into 2 (unlabeled) groups of size n. - Dennis P. Walsh, Nov 09 2011
Equals row sums of triangle A205945. - Gary W. Adamson, Feb 01 2012
a(n-1) gives the number of n-regular sequences defined by Erdős and Gallai in 1960 in connection with the degree sequences of simple graphs. - Matuszka Tamás, Mar 06 2013
a(n) is the sum of falling diagonals of squares in the comment in A085812 (equivalent to the Cloitre formula of Aug 2002). - John Molokach, Sep 26 2013
For n > 0: largest terms of Zigzag matrices as defined in A088961. - Reinhard Zumkeller, Oct 25 2013
Also the number of different possible win/loss round sequences (from the perspective of the eventual winner) in a "best of 2*n + 1" two-player game. For example, a(2) = 10 means there are 10 different win/loss sequences in a "best of 5" game (like a tennis match in which the first player to win 3 sets, out of a maximum of 5, wins the match); the 10 sequences are WWW, WWLW, WWLLW, WLWW, WLWLW, WLLWW, LWWW, LWWLW, LWLWW, LLWWW. See also A072600. - Philippe Beaudoin, May 14 2014; corrected by Jon E. Schoenfield, Nov 23 2014
When adding 1 to the beginning of the sequence: Convolving a(n)/2^n with itself equals 2^(n+1). For example, when n = 4: convolving {1, 1/1, 3/2, 10/4, 35/8, 126/16} with itself is 32 = 2^5. - Bob Selcoe, Jul 16 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to a family of arrays associated to the Catalan A000108 (t = 1), and Riordan, or Motzkin sums A005043 (t = 0), with the o.g.f. [1 - sqrt(1 - 4x/(1 + (1 - t)x))]/2 and inverse x*(1 - x)/[1 + (t - 1)*x*(1 - x)]. See A091867 for more info on this family. Here is t = -3 (mod signs in the results).
Let C(x) = [1 - sqrt(1-4x)]/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1 + t*x) with inverse P(x, -t).
O.g.f: G(x) = [-1 + sqrt(1 + 4*x/(1 - 4*x))]/2 = -C[P(-x, 4)].
Inverse o.g.f: Ginv(x) = x*(1 + x)/(1 + 4*x*(1 + x)) = -P(Cinv(-x), -4) (shifted signed A001792). A088218(x) = 1 + G(x).
Equals A001813/2 omitting the leading 1 there. (End)
Placing n distinguishable balls into n indistinguishable boxes gives A000110(n) (the number of set partitions). - N. J. A. Sloane, Jun 19 2015
The sequence is the INVERTi transform of A049027: (1, 4, 17, 74, 326, ...). - Gary W. Adamson, Jun 23 2015
a(n) is the number of compositions of 2*n + 2 such that the sum of the elements at odd positions is equal to the sum of the elements at even positions. a(2) = 10 because there are 10 such compositions of 6: (3, 3), (1, 3, 2), (2, 3, 1), (1, 1, 2, 2), (1, 2, 2, 1), (2, 2, 1, 1), (2, 1, 1, 2), (1, 2, 1, 1, 1), (1, 1, 1, 2, 1), (1, 1, 1, 1, 1, 1). - Ran Pan, Oct 08 2015
a(n-1) is also the Schur function of the partition (n) of n evaluated at x_1 = x_2 = ... = x_n = 1, i.e., the number of semistandard Young tableaux of shape (n) (weakly increasing rows with n boxes with numbers from {1, 2, ..., n}). - Wolfdieter Lang, Oct 11 2015
Also the number of ordered (rooted planar) forests with a total of n+1 edges and no trivial trees. - Nachum Dershowitz, Mar 30 2016
a(n) is the number of sets (i1,...in) of length n so that n >= i1 >= i2 >= ...>= in >= 1. For instance, n=3 as there are only 10 such sets (3,3,3) (3,3,2) (3,3,1) (3,2,2) (3,2,1) (3,1,1) (2,2,2) (2,2,1) (2,1,1) (1,1,1,) 3,2,1 is each used 10 times respectively. - Anton Zakharov, Jul 04 2016
The repeated middle term in the odd rows of Pascal's triangle, or half the central binomial coefficient in the even rows of Pascal's triangle, n >= 2. - Enrique Navarrete, Feb 12 2018
a(n) is the number of walks of length 2n+1 from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n+1 from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Total number of nodes summed over all Dyck paths of semilength n. - Alois P. Heinz, Mar 08 2020
a(n-1) is the determinant of the n X n matrix m(i, j) = binomial(n+i-1, j). - Fabio Visonà, May 21 2022
Let X_i be iid standard Gaussian random variable N(0,1), and S_n be the partial sum S_n = X_1+...+X_n. Then P(S_1>0,S_2>0,...,S_n>0) = a(n+1)/2^(2n-1) = a(n+1) / A004171(n+1). For example, P(S_1>0) = 1/2, P(S_1>0,S_2>0) = 3/8, P(S_1>0,S_2>0,S_3>0) = 5/16, etc. This probability is also equal to the volume of the region x_1 > 0, x_2 > -x_1, x_3 > -(x_1+x_2), ..., x_n > -(x_1+x_2+...+x_(n-1)) in the hypercube [-1/2, 1/2]^n. This also holds for the Cauchy distribution and other stable distributions with mean 0, skew 0 and scale 1. - Xiaohan Zhang, Nov 01 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 132, 213, and 321. - Lara Pudwell, Apr 10 2023
Number of vectors in (Z_>=0)^(n+1) such that the sum of the components is n+1. binomial(2*n-1, n) provides this property for n. - Michael Richard, Jun 12 2023
Also number of discrete negations on the finite chain L_n={0,1,...,n-1,n}, i.e., monotone decreasing unary operators such that N(0)=n and N(n)=0. - Marc Munar, Oct 10 2023
a(n) is the number of Dyck paths of semilength n+1 having one of its peaks marked. - Juan B. Gil, Jan 03 2024
a(n) is the dimension of the (n+1)-st symmetric power of an (n+1)-dimensional vector space. - Mehmet A. Ates, Feb 15 2024
a(n) is the independence number of the twisted odd graph O^(sigma)(n+2). - _Miquel A. Fiol, Aug 26 2024
a(n) is the number of non-descending sequences with length n and the last number is less or equal to n. a(n) is also the number of integer partitions (of any positive integer) with length n and largest part is less or equal to n. - Zlatko Damijanic, Dec 06 2024
a(n) is the number of triangulations of a once-punctured (n+1)-gon [from Fontaine & Plamondon's Theorem 3.6]. - Esther Banaian, May 06 2025

Examples

			There are a(2)=10 ways to put 3 indistinguishable balls into 3 distinguishable boxes, namely, (OOO)()(), ()(OOO)(), ()()(OOO), (OO)(O)(), (OO)()(O), (O)(OO)(), ()(OO)(O), (O)()(OO), ()(O)(OO), and (O)(O)(O). - _Dennis P. Walsh_, Apr 11 2012
a(2) = 10: Semistandard Young tableaux for partition (3) of 3 (the indeterminates x_i, i = 1, 2, 3 are omitted and only their indices are given): 111, 112, 113, 122, 123, 133, 222, 223, 233, 333. - _Wolfdieter Lang_, Oct 11 2015
		

References

  • H. Davenport, The Higher Arithmetic. Cambridge Univ. Press, 7th ed., 1999, ch. V.3 (p. 122).
  • A. Frosini, R. Pinzani, and S. Rinaldi, About half the middle binomial coefficient, Pure Math. Appl., 11 (2000), 497-508.
  • Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 449.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • Paul J. Nahin, "An Imaginary Tale, The Story of [Sqrt(-1)]," Princeton University Press, Princeton, NJ 1998, p. 62.
  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
  • 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).

Crossrefs

Equals A000984(n+1)/2.
a(n) = (2*n+1)*Catalan(n) [A000108] = A035324(n+1, 1) (first column of triangle).
Row sums of triangles A028364, A050166, A039598.
Bisections: a(2*k) = A002458(k), a(2*k+1) = A001448(k+1)/2, k >= 0.
Other versions of the same sequence: A088218, A110556, A138364.
Diagonals 1 and 2 of triangle A100257.
Second row of array A102539.
Column of array A073165.
Row sums of A103371. - Susanne Wienand, Oct 22 2011
Cf. A002054: C(2*n+1, n-1). - Bruno Berselli, Jan 20 2014

Programs

  • GAP
    List([0..30],n->Binomial(2*n+1,n+1)); # Muniru A Asiru, Feb 26 2019
  • Haskell
    a001700 n = a007318 (2*n+1) (n+1)  -- Reinhard Zumkeller, Oct 25 2013
    
  • Magma
    [Binomial(2*n, n)/2: n in [1..40]]; // Vincenzo Librandi, Nov 10 2014
    
  • Maple
    A001700 := n -> binomial(2*n+1,n+1); seq(A001700(n), n=0..20);
    A001700List := proc(m) local A, P, n; A := [1]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
    A := [op(A), P[-1]] od; A end: A001700List(27); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[ Binomial[2n + 1, n + 1], {n, 0, 23}]
    CoefficientList[ Series[2/((Sqrt[1 - 4 x] + 1)*Sqrt[1 - 4 x]), {x, 0, 22}], x] (* Robert G. Wilson v, Aug 08 2011 *)
  • Maxima
    B(n,a,x):=coeff(taylor(exp(x*t)*(t/(exp(t)-1))^a,t,0,20),t,n)*n!;
    makelist((-1)^(n)*B(n,n+1,-n-1)/n!,n,0,10); /* Vladimir Kruchinin, Apr 06 2016 */
    
  • PARI
    a(n)=binomial(2*n+1,n+1)
    
  • PARI
    z='z+O('z^50); Vec((1/sqrt(1-4*z)-1)/(2*z)) \\ Altug Alkan, Oct 11 2015
    
  • Python
    from _future_ import division
    A001700_list, b = [], 1
    for n in range(10**3):
        A001700_list.append(b)
        b = b*(4*n+6)//(n+2) # Chai Wah Wu, Jan 26 2016
    
  • Sage
    [rising_factorial(n+1,n+1)/factorial(n+1) for n in (0..22)] # Peter Luschny, Nov 07 2011
    

Formula

a(n-1) = binomial(2*n, n)/2 = A000984(n)/2 = (2*n)!/(2*n!*n!).
D-finite with recurrence: a(0) = 1, a(n) = 2*(2*n+1)*a(n-1)/(n+1) for n > 0.
G.f.: (1/sqrt(1 - 4*x) - 1)/(2*x).
L.g.f.: log((1 - sqrt(1 - 4*x))/(2*x)) = Sum_{n >= 0} a(n)*x^(n+1)/(n+1). - Vladimir Kruchinin, Aug 10 2010
G.f.: 2F1([1, 3/2]; [2]; 4*x). - Paul Barry, Jan 23 2009
G.f.: 1/(1 - 2*x - x/(1 - x/(1 - x/(1 - x/(1 - ... (continued fraction). - Paul Barry, May 06 2009
G.f.: c(x)^2/(1 - x*c(x)^2), c(x) the g.f. of A000108. - Paul Barry, Sep 07 2009
O.g.f.: c(x)/sqrt(1 - 4*x) = (2 - c(x))/(1 - 4*x), with c(x) the o.g.f. of A000108. Added second formula. - Wolfdieter Lang, Sep 02 2012
Convolution of A000108 (Catalan) and A000984 (central binomial): Sum_{k=0..n} C(k)*binomial(2*(n-k), n-k), C(k) Catalan. - Wolfdieter Lang, Dec 11 1999
a(n) = Sum_{k=0..n} C(n+k, k). - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n+1, k+1). - Benoit Cloitre, Oct 19 2002
a(n) = Sum_{k = 0..n+1} binomial(2*n+2, k)*cos((n - k + 1)*Pi). - Paul Barry, Nov 02 2004
a(n) = 4^n*binomial(n+1/2, n)/(n+1). - Paul Barry, May 10 2005
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n + 1)/(2*n + 1)! = BesselI(1, 2*x). - Michael Somos, Jun 22 2005
E.g.f. in Maple notation: exp(2*x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). Integral representation as n-th moment of a positive function on [0, 4]: a(n) = Integral_{x = 0..4} x^n * (x/(4 - x))^(1/2)/(2*Pi) dx, n >= 0. This representation is unique. - Karol A. Penson, Oct 11 2001
Narayana transform of [1, 2, 3, ...]. Let M = the Narayana triangle of A001263 as an infinite lower triangular matrix and V = the Vector [1, 2, 3, ...]. Then A001700 = M * V. - Gary W. Adamson, Apr 25 2006
a(n) = A122366(n,n). - Reinhard Zumkeller, Aug 30 2006
a(n) = C(2*n, n) + C(2*n, n-1) = A000984(n) + A001791(n). - Zerinvary Lajos, Jan 23 2007
a(n-1) = (n+1)*(n+2)*...*(2*n-1)/(n-1)! (product of n-1 consecutive integers, divided by (n-1)!). - Jonathan Vos Post, Apr 09 2007; [Corrected and shortened by Giovanni Ciriani, Mar 26 2019]
a(n-1) = (2*n - 1)!/(n!*(n - 1)!). - William A. Tedeschi, Feb 27 2008
a(n) = (2*n + 1)*A000108(n). - Paul Barry, Aug 21 2007
Binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double binomial transform of A001405. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A132813. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A134285. - Gary W. Adamson, Nov 19 2007
a(n) = 2*A000984(n) - A000108(n), that is, a(n) = 2*C(2*n, n) - n-th Catalan number. - Joseph Abate, Jun 11 2010
Conjectured: 4^n GaussHypergeometric(1/2,-n; 2; 1) -- Solution for the path which stays in the first and second quadrant. - Benjamin Phillabaum, Feb 20 2011
a(n)= Sum_{k=0..n} A038231(n,k) * (-1)^k * A000108(k). - Philippe Deléham, Nov 27 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1] = -1, A[i,j] = Catalan(j-i), (i <= j), and A[i,j] = 0, otherwise. Then, for n >= 1, a(n) = (-1)^n * charpoly(A,-2). - Milan Janjic, Jul 08 2010
a(n) is the upper left term of M^(n+1), where M is the infinite matrix in which a column of (1,2,3,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros, as follows:
1, 1, 0, 0, 0, ...
2, 1, 1, 0, 0, ...
3, 1, 1, 1, 0, ...
4, 1, 1, 1, 1, ...
...
Alternatively, a(n) is the upper left term of M^n where M is the infinite matrix:
3, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
...
- Gary W. Adamson, Jul 14 2011
a(n) = (n + 1)*hypergeom([-n, -n], [2], 1). - Peter Luschny, Oct 24 2011
a(n) = Pochhammer(n+1, n+1)/(n+1)!. - Peter Luschny, Nov 07 2011
E.g.f.: 1 + 6*x/(U(0) - 6*x); U(k) = k^2 + (4*x + 3)*k + 6*x + 2 - 2*x*(k + 1)*(k + 2)*(2*k + 5)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2011
a(n) = 2*A000984(n) - A000108(n). [Abate & Whitt]
a(n) = 2^(2*n+1)*binomial(n+1/2, -1/2). - Peter Luschny, May 06 2014
For n > 1: a(n-1) = A166454(2*n, n), central terms in A166454. - Reinhard Zumkeller, Mar 04 2015
a(n) = 2*4^n*Gamma(3/2 + n)/(sqrt(Pi)*Gamma(2+n)). - Peter Luschny, Dec 14 2015
a(n) ~ 2*4^n*(1 - (5/8)/n + (73/128)/n^2 - (575/1024)/n^3 + (18459/32768)/n^4)/sqrt(n*Pi). - Peter Luschny, Dec 16 2015
a(n) = (-1)^(n)*B(n, n+1, -n-1)/n!, where B(n,a,x) is a generalized Bernoulli polynomial. - Vladimir Kruchinin, Apr 06 2016
a(n) = Gamma(2 + 2*n)/(n!*Gamma(2 + n)). Andres Cicuttin, Apr 06 2016
a(n) = (n + (n + 1))!/(Gamma(n)*Gamma(1 + n)*A002378(n)), for n > 0. Andres Cicuttin, Apr 07 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
Sum_{n >= 0} 1/a(n) = 2*(9 + 2*sqrt(3)*Pi)/27 = A248179.
Sum_{n >= 0} (-1)^n/a(n) = 2*(5 + 4*sqrt(5)*arcsinh(1/2))/25 = 2*(5*A145433 - 1).
Sum_{n >= 0} (-1)^n*a(n)/n! = BesselI(2,2)*exp(-2) = A229020*A092553. (End)
Conjecture: a(n) = Sum_{k=2^n..2^(n+1)-1} A178244(k). - Mikhail Kurkov, Feb 20 2021
a(n-1) = 1 + (1/n)*Sum_{t=1..n/2} (2*cos((2*t-1)*Pi/(2*n)))^(2*n). - Greg Dresden, Oct 11 2022
a(n) = Product_{1 <= i <= j <= n} (i + j + 1)/(i + j - 1). Cf. A006013. - Peter Bala, Feb 21 2023
Sum_{n >= 0} a(n)*x^(n+1)/(n+1) = x + 3*x^2/2 + 10*x^3/3 + 35*x^4/4 + ... = the series reversion of exp(-x)*(1 - exp(-x)). - Peter Bala, Sep 06 2023

Extensions

Name corrected by Paul S. Coombes, Jan 11 2012
Name corrected by Robert Tanniru, Feb 01 2014

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

Original entry on oeis.org

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

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Keywords

Comments

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

Examples

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

References

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

Crossrefs

Other versions are in A090181 and A131198. - Philippe Deléham, Nov 18 2007
Cf. variants: A181143, A181144. - Paul D. Hanna, Oct 13 2010
Row sums give A000108 (Catalan numbers), n>0.
A008459 (h-vectors type B associahedra), A033282 (f-vectors type A associahedra), A145903 (h-vectors type D associahedra), A145904 (Hilbert transform). - Peter Bala, Oct 27 2008
Cf. A016098 and A189232 for numbers of crossing set partitions.
Cf. A243752.
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

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

Formula

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

Extensions

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

A002411 Pentagonal pyramidal numbers: a(n) = n^2*(n+1)/2.

Original entry on oeis.org

0, 1, 6, 18, 40, 75, 126, 196, 288, 405, 550, 726, 936, 1183, 1470, 1800, 2176, 2601, 3078, 3610, 4200, 4851, 5566, 6348, 7200, 8125, 9126, 10206, 11368, 12615, 13950, 15376, 16896, 18513, 20230, 22050, 23976, 26011, 28158, 30420, 32800, 35301, 37926, 40678
Offset: 0

Views

Author

Keywords

Comments

a(n) = n^2(n+1)/2 is half the number of colorings of three points on a line with n+1 colors. - R. H. Hardin, Feb 23 2002
Sum of n smallest multiples of n. - Amarnath Murthy, Sep 20 2002
a(n) = number of (n+6)-bit binary sequences with exactly 7 1's none of which is isolated. A 1 is isolated if its immediate neighbor(s) are 0. - David Callan, Jul 15 2004
Also as a(n) = (1/6)*(3*n^3+3*n^2), n > 0: structured trigonal prism numbers (cf. A100177 - structured prisms; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
If Y is a 3-subset of an n-set X then, for n >= 5, a(n-4) is the number of 5-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
a(n-1), n >= 2, is the number of ways to have n identical objects in m=2 of altogether n distinguishable boxes (n-2 boxes stay empty). - Wolfdieter Lang, Nov 13 2007
a(n+1) is the convolution of (n+1) and (3n+1). - Paul Barry, Sep 18 2008
The number of 3-character strings from an alphabet of n symbols, if a string and its reversal are considered to be the same.
Partial sums give A001296. - Jonathan Vos Post, Mar 26 2011
a(n-1):=N_1(n), n >= 1, is the number of edges 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
Partial sums of pentagonal numbers A000326. - Reinhard Zumkeller, Jul 07 2012
From Ant King, Oct 23 2012: (Start)
For n > 0, the digital roots of this sequence A010888(A002411(n)) form the purely periodic 9-cycle {1,6,9,4,3,9,7,9,9}.
For n > 0, the units' digits of this sequence A010879(A002411(n)) form the purely periodic 20-cycle {1,6,8,0,5,6,6,8,5,0,6,6,3,0,0,6,1,8,0,0}.
(End)
a(n) is the number of inequivalent ways to color a path graph having 3 nodes using at most n colors. Note, here there is no restriction on the color of adjacent nodes as in the above comment by R. H. Hardin (Feb 23 2002). Also, here the structures are counted up to graph isomorphism, where as in the above comment the "three points on a line" are considered to be embedded in the plane. - Geoffrey Critzer, Mar 20 2013
After 0, row sums of the triangle in A101468. - Bruno Berselli, Feb 10 2014
Latin Square Towers: Take a Latin square of order n, with symbols from 1 to n, and replace each symbol x with a tower of height x. Then the total number of unit cubes used is a(n). - Arun Giridhar, Mar 29 2015
This is the case k = n+4 of b(n,k) = n*((k-2)*n-(k-4))/2, which is the n-th k-gonal number. Therefore, this is the 3rd upper diagonal of the array in A139600. - Luciano Ancora, Apr 11 2015
For n > 0, a(n) is the number of compositions of n+7 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
Also the Wiener index of the n-antiprism graph. - Eric W. Weisstein, Sep 07 2017
For n > 0, a(2n+1) is the number of non-isomorphic 5C_m-snakes, where m = 2n+1 or m = 2n (for n >= 2). A kC_n-snake is a connected graph in which the k >= 2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 15 2019
For n >= 1, a(n-1) is the number of 0°- and 45°-tilted squares that can be drawn by joining points in an n X n lattice. - Paolo Xausa, Apr 13 2021
a(n) is the number of all possible products of n rolls of a six-sided die. This can be easily seen by the recursive formula a(n) = a(n - 1) + 2 * binomial(n, 2) + binomial(n + 1, 2). - Rafal Walczak, Jun 15 2024
a(n) is the number of all triples consisting of nonnegative integers smaller than n such that the sum of the first two integers is less than n. - Ruediger Jehn, Aug 17 2025

Examples

			a(3)=18 because 4 identical balls can be put into m=2 of n=4 distinguishable boxes in binomial(4,2)*(2!/(1!*1!) + 2!/2!) = 6*(2+1) = 18 ways. The m=2 part partitions of 4, namely (1,3) and (2,2), specify the filling of each of the 6 possible two-box choices. - _Wolfdieter Lang_, Nov 13 2007
		

References

  • V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_1.
  • Christian Barrientos, Graceful labelings of cyclic snakes, Ars Combin., Vol. 60 (2001), pp. 85-96.
  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 166, Table 10.4/I/5).
  • 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. 2.
  • 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).

Crossrefs

A006002(n) = -a(-1-n).
a(n) = A093560(n+2, 3), (3, 1)-Pascal column.
A row or column of A132191.
Second column of triangle A103371.
Cf. similar sequences listed in A237616.

Programs

  • GAP
    List([0..45], n->n^2*(n+1)/2); # Muniru A Asiru, Feb 19 2018
  • Haskell
    a002411 n = n * a000217 n  -- Reinhard Zumkeller, Jul 07 2012
    
  • Magma
    [n^2*(n+1)/2: n in [0..40]]; // Wesley Ivan Hurt, May 25 2014
    
  • Maple
    seq(n^2*(n+1)/2, n=0..40);
  • Mathematica
    Table[n^2 (n + 1)/2, {n, 0, 40}]
    LinearRecurrence[{4, -6, 4, -1}, {0, 1, 6, 18}, 50] (* Harvey P. Dale, Oct 20 2011 *)
    Nest[Accumulate, Range[1, 140, 3], 2] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
    CoefficientList[Series[x (1 + 2 x) / (1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jan 08 2016 *)
  • PARI
    a(n)=n^2*(n+1)/2
    
  • PARI
    concat(0, Vec(x*(1+2*x)/(1-x)^4 + O(x^100))) \\ Altug Alkan, Jan 07 2016
    

Formula

Average of n^2 and n^3.
G.f.: x*(1+2*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
a(n) = n*Sum_{k=0..n} (n-k) = n*Sum_{k=0..n} k. - Paul Barry, Jul 21 2003
a(n) = n*A000217(n). - Xavier Acloque, Oct 27 2003
a(n) = (1/2)*Sum_{j=1..n} Sum_{i=1..n} (i+j) = (1/2)*(n^2+n^3) = (1/2)*A011379(n). - Alexander Adamchuk, Apr 13 2006
Row sums of triangle A127739, triangle A132118; and binomial transform of [1, 5, 7, 3, 0, 0, 0, ...] = (1, 6, 18, 40, 75, ...). - Gary W. Adamson, Aug 10 2007
G.f.: x*F(2,3;1;x). - Paul Barry, Sep 18 2008
Sum_{j>=1} 1/a(j) = hypergeom([1, 1, 1], [2, 3], 1) = -2 + 2*zeta(2) = A195055 - 2. - Stephen Crowley, Jun 28 2009
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=6, a(3)=18. - Harvey P. Dale, Oct 20 2011
From Ant King, Oct 23 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3.
a(n) = (n+1)*(2*A000326(n)+n)/6 = A000292(n) + 2*A000292(n-1).
a(n) = A000330(n)+A000292(n-1) = A000217(n) + 3*A000292(n-1).
a(n) = binomial(n+2,3) + 2*binomial(n+1,3).
(End)
a(n) = (A000330(n) + A002412(n))/2 = (A000292(n) + A002413(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = (24/(n+3)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+3). - Vladimir Kruchinin, Jun 04 2013
Sum_{n>=1} a(n)/n! = (7/2)*exp(1). - Richard R. Forberg, Jul 15 2013
E.g.f.: x*(2 + 4*x + x^2)*exp(x)/2. - Ilya Gutkovskiy, May 31 2016
From R. J. Mathar, Jul 28 2016: (Start)
a(n) = A057145(n+4,n).
a(n) = A080851(3,n-1). (End)
For n >= 1, a(n) = (Sum_{i=1..n} i^2) + Sum_{i=0..n-1} i^2*((i+n) mod 2). - Paolo Xausa, Apr 13 2021
a(n) = Sum_{k=1..n} GCD(k,n) * LCM(k,n). - Vaclav Kotesovec, May 22 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 2 + Pi^2/6 - 4*log(2). - Amiram Eldar, Jan 03 2022

A000894 a(n) = (2*n)!*(2*n+1)! /((n+1)! *n!^3).

Original entry on oeis.org

1, 6, 60, 700, 8820, 116424, 1585584, 22084920, 312869700, 4491418360, 65166397296, 953799087696, 14062422446800, 208618354980000, 3111393751416000, 46619049708716400, 701342468412012900
Offset: 0

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Keywords

Comments

This sequence is one half of the odd part of the bisection of A241530. The even part is given in A002894. - Wolfdieter Lang, Sep 06 2016

Examples

			G.f. = 1 + 6*x + 60*x^2 + 700*x^3 + 8820*x^4 + 116424*x^5 + ...
		

References

  • E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.

Crossrefs

Programs

  • Haskell
    a000894 n = a132813 (2 * n) n  -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    [Factorial(2*n)*Factorial(2*n+1) /(Factorial(n+1)* Factorial(n)^3): n in [0..20]]; // Vincenzo Librandi, Oct 25 2011
    
  • Magma
    A000894:= func< n | Binomial(2*n+2,2)*Catalan(n)^2 >;
    [A000894(n): n in [0..40]]; // G. C. Greubel, Mar 12 2025
    
  • Maple
    seq(binomial(2*n+1,n)*binomial(2*n,n), n=0..16); # Zerinvary Lajos, Jan 23 2007
  • Mathematica
    a[ n_] := Binomial[2 n + 1, n] Binomial[2 n, n]; (* Michael Somos, May 28 2014 *)
    a[ n_] := SeriesCoefficient[ (EllipticK[ 16 x] - EllipticE[ 16 x]) / (4 x Pi), {x, 0, n}]; (* Michael Somos, May 28 2014 *)
    Table[(2 n)!*(2 n + 1)!/((n + 1)!*n!^3), {n, 0, 16}] (* Michael De Vlieger, Sep 06 2016 *)
  • PARI
    {a(n) =  binomial( 2*n + 1, n) * binomial( 2*n, n)}; /* Michael Somos, May 28 2014 */
    
  • SageMath
    def A000894(n): return binomial(2*n+2,2)*catalan_number(n)^2
    print([A000894(n) for n in range(41)]) # G. C. Greubel, Mar 12 2025

Formula

From Zerinvary Lajos, Jan 23 2007: (Start)
a(n) = C(2*n+1,n)*C(2*n,n) = A001700(n)*A000984(n).
a(n) = A000984(n)*A000984(n+1)/2, n>=0. (End)
G.f.: (EllipticK(4*sqrt(x)) - EllipticE(4*sqrt(x)))/(4*Pi*x). - Mark van Hoeij, Oct 24 2011
n*(n+1)*a(n) = 4*(2*n-1)*(2*n+1)*a(n-1). - R. J. Mathar, Sep 08 2013
a(n) = A103371(2*n,n) = A132813(2*n,n). - Reinhard Zumkeller, Apr 04 2014
0 = a(n)*(+65536*a(n+2) - 23040*a(n+3) + 1400*a(n+4)) + a(n+1)*(-1536*a(n+2) + 1184*a(n+3) - 90*a(n+4)) + a(n+2)*(-24*a(n+2) - 6*a(n+3) + a(n+4)) for all n in Z. - Michael Somos, May 28 2014
0 = a(n+1)^3 * (+256*a(n) - 6*a(n+1) + a(n+2)) + a(n) * a(n+1) * a(n+
2) * (-768*a(n) - 20*a(n+1) - 3*a(n+2)) + 90*a(n)^2*a(n+2)^2 for all n in Z. - Michael Somos, Sep 17 2014
a(n) = (n+1) * A000891(n) = A248045(n+1) / A000142(n). - Reinhard Zumkeller, Sep 30 2014
a(n) = A241530(2n+1)/2, n >= 0. - Wolfdieter Lang, Sep 06 2016
a(n) ~ 2^(4*n+1)/(Pi*n). - Ilya Gutkovskiy, Sep 06 2016
a(n) = A000217(n+1)*A000108(n)*A000108(n+1) = A000217(2*n+1)*A000108(n)^2. - G. C. Greubel, Mar 12 2025

A035206 Number of multisets associated with least integer of each prime signature.

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 12, 6, 12, 1, 5, 20, 20, 30, 30, 20, 1, 6, 30, 30, 15, 60, 120, 20, 60, 90, 30, 1, 7, 42, 42, 42, 105, 210, 105, 105, 140, 420, 140, 105, 210, 42, 1, 8, 56, 56, 56, 28, 168, 336, 336, 168, 168, 280, 840, 420, 840, 70, 280, 1120, 560, 168, 420, 56, 1, 9, 72
Offset: 0

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Author

Keywords

Comments

a(n,k) multiplied by A036038(n,k) yields A049009(n,k).
a(n,k) enumerates distributions of n identical objects (balls) into m of altogether n distinguishable boxes. The k-th partition of n, taken in the Abramowitz-Stegun (A-St) order, specifies the occupation of the m =m(n,k)= A036043(n,k) boxes. m = m(n,k) is the number of parts of the k-th partition of n. For the A-St ordering see pp.831-2 of the reference given in A117506. - Wolfdieter Lang, Nov 13 2007
The sequence of row lengths is p(n)= A000041(n) (partition numbers).
For the A-St order of partitions see the Abramowitz-Stegun reference given in A117506.
The corresponding triangle with summed row entries which belong to partitions of the same number of parts k is A103371. [Wolfdieter Lang, Jul 11 2012]

Examples

			n\k 1  2  3  4   5   6   7   8   9  10  11  12  13 14 15
0   1
1   1
2   2  1
3   3  6  1
4   4 12  6 12   1
5   5 20 20 30  30  20   1
6   6 30 30 15  60 120  20  60  90  30   1
7   7 42 42 42 105 210 105 105 140 420 140 105 210 42  1
...
Row No. 8:  8  56 56 56 28 168 336 336 168 168 280  840 420 840 70 280 1120 560 168 420 56 1
Row No. 9: 9 72 72 72 72 252 504 504 252 252 504 84 504 1512 1512 1512 1512 504 630 2520 1260 3780 630 504 2520 1680 252 756 72 1
[rewritten and extended table by _Wolfdieter Lang_, Jul 11 2012]
a(5,5) relates to the partition (1,2^2) of n=5. Here m=3 and 5 indistinguishable (identical) balls are put into boxes b1,...,b5 with m=3 boxes occupied; one with one ball and two with two balls.
Therefore a(5,5) = binomial(5,3)*3!/(1!*2!) = 10*3 = 30. _Wolfdieter Lang_, Nov 13 2007
		

Crossrefs

Cf. A001700 (row sums).
Cf. A103371(n-1, m-1) (triangle obtained after summing in every row the numbers with like part numbers m).

Programs

  • PARI
    C(sig)={my(S=Set(sig)); binomial(vecsum(sig), #sig)*(#sig)!/prod(k=1, #S, (#select(t->t==S[k], sig))!)}
    Row(n)={apply(C, [Vecrev(p) | p<-partitions(n)])}
    { for(n=0, 7, print(Row(n))) } \\ Andrew Howroyd, Oct 18 2020

Formula

a(n,k) = A048996(n,k)*binomial(n,m(n,k)),n>=1, k=1,...,p(n) and m(n,k):=A036043(n,k) gives the number of parts of the k-th partition of n.

Extensions

More terms from Joshua Zucker, Jul 27 2006
a(0)=1 prepended by Andrew Howroyd, Oct 18 2020

A132813 Triangle read by rows: A001263 * A127648 as infinite lower triangular matrices.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 12, 18, 4, 1, 20, 60, 40, 5, 1, 30, 150, 200, 75, 6, 1, 42, 315, 700, 525, 126, 7, 1, 56, 588, 1960, 2450, 1176, 196, 8, 1, 72, 1008, 4704, 8820, 7056, 2352, 288, 9, 1, 90, 1620, 10080, 26460, 31752, 17640, 4320, 405, 10
Offset: 0

Views

Author

Gary W. Adamson, Sep 01 2007

Keywords

Comments

Also T(n,k) = binomial(n-1, k-1)*binomial(n, k-1), related to Narayana polynomials (see Sulanke reference). - Roger L. Bagula, Apr 09 2008
h-vector for cluster complex associated to the root system B_n. See p. 8, Athanasiadis and C. Savvidou. - Tom Copeland, Oct 19 2014

Examples

			First few rows of the triangle are:
  1;
  1,  2;
  1,  6,   3;
  1, 12,  18,   4;
  1, 20,  60,  40,   5;
  1, 30, 150, 200,  75,   6;
  1, 42, 315, 700, 525, 126, 7;
  ...
		

Crossrefs

Family of polynomials (see A062145): A008459 (c=1), this sequence (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Columns: A000012 (k=0), A002378 (k=1), A006011 (k=2), 4*A006542 (k=3), 5*A006857 (k=4), 6*A108679 (k=5), 7*A134288 (k=6), 8*A134289 (k=7), 9*A134290 (k=8), 10*A134291 (k=9).
Diagonals: A000027 (k=n), A002411 (k=n-1), A004302 (k=n-2), A108647 (k=n-3), A134287 (k=n-4).
Main diagonal: A000894.
Sums: (-1)^floor((n+1)/2)*A001405 (signed row), A001700 (row), A203611 (diagonal).
Cf. A103371 (mirrored).

Programs

  • GAP
    Flat(List([0..10],n->List([0..n], k->(k+1)*Binomial(n+1,k+1)*Binomial(n+1,k)/(n+1)))); # Muniru A Asiru, Feb 26 2019
    
  • Haskell
    a132813 n k = a132813_tabl !! n !! k
    a132813_row n = a132813_tabl !! n
    a132813_tabl = zipWith (zipWith (*)) a007318_tabl $ tail a007318_tabl
    -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    /* triangle */ [[(k+1)*Binomial(n+1,k+1)*Binomial(n+1,k)/(n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 19 2014
    
  • Maple
    P := (n, x) -> hypergeom([1-n, -n], [1], x): for n from 1 to 9 do PolynomialTools:-CoefficientList(simplify(P(n,x)),x) od; # Peter Luschny, Nov 26 2014
  • Mathematica
    T[n_,k_]=Binomial[n-1,k-1]*Binomial[n,k-1]; Table[Table[T[n,k],{k,1,n}],{n,1,11}]; Flatten[%] (* Roger L. Bagula, Apr 09 2008 *)
    P[n_, x_] := HypergeometricPFQ[{1-n, -n}, {1}, x]; Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten (* Jean-François Alcover, Nov 27 2014, after Peter Luschny *)
  • PARI
    tabl(nn) = {for (n = 1, nn, for (k = 1, n, print1(binomial(n-1, k-1)*binomial(n, k-1) , ", ");););} \\ Michel Marcus, Feb 12 2014
    
  • SageMath
    def A132813(n,k): return binomial(n,k)*binomial(n+1,k)
    print(flatten([[A132813(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 12 2025

Formula

T(n,k) = (k+1)*binomial(n+1,k+1)*binomial(n+1,k)/(n+1), n >= k >= 0.
From Roger L. Bagula, May 14 2010: (Start)
T(n, m) = coefficients(p(x,n)), where
p(x,n) = (1-x)^(2*n)*Sum_{k >= 0} binomial(k+n-1, k)*binomial(n+k, k)*x^k,
or p(x,n) = (1-x)^(2*n)*Hypergeometric2F1([n, n+1], [1], x). (End)
T(n,k) = binomial(n,k) * binomial(n+1,k). - Reinhard Zumkeller, Apr 04 2014
These are the coefficients of the polynomials Hypergeometric2F1([1-n,-n], [1], x). - Peter Luschny, Nov 26 2014
G.f.: A(x,y) = A281260(x,y)/(1-A281260(x,y))/x. - Vladimir Kruchinin, Oct 10 2020

A004302 a(n) = n^2*(n+1)^2*(n+2)/12.

Original entry on oeis.org

0, 1, 12, 60, 200, 525, 1176, 2352, 4320, 7425, 12100, 18876, 28392, 41405, 58800, 81600, 110976, 148257, 194940, 252700, 323400, 409101, 512072, 634800, 780000, 950625, 1149876, 1381212, 1648360, 1955325, 2306400, 2706176, 3159552, 3671745, 4248300, 4895100
Offset: 0

Views

Author

Keywords

Comments

Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 19 2005
a(n-2), n>=3, is the number of ways to have n identical objects in m=3 of altogether n distinguishable boxes (n-3 boxes stay empty). - Wolfdieter Lang, Nov 13 2007
Starting with offset 1 = row sums of triangle A096948 and binomial transform of [1, 11, 37, 55, 38, 10, 0, 0, 0, ...]. - Gary W. Adamson, Aug 08 2008

Examples

			a(3)=60 because n=5 identical balls can be put into m=3 of n=5 distinguishable boxes in binomial(5,3)*(3!/(2!*1!)+ 3!/(1!*2!) ) = 10*(3+3) = 60 ways. The m=3 part partitions of 5, namely (1^2,3) and (1,2^2) specify the filling of each of the 10 possible three-box choices. - _Wolfdieter Lang_, Nov 13 2007
		

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 233, # 11).
  • T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Crossrefs

Third column of triangle A103371.
Main diagonal of A103252.

Programs

  • Haskell
    a004302 0 = 0
    a004302 n = a103371 (n + 1) 2 -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    [n^2*(n+1)^2*(n+2)/12: n in [0..40]]; // Vincenzo Librandi, May 22 2011
    
  • Maple
    a:=n->n^2*(n+1)^2*(n+2)/12: seq(a(n),n=0..33); # Emeric Deutsch, Jun 19 2005
  • Mathematica
    Table[n^2 (n+1)^2 (n+2)/12,{n,0,30}] (* or *) LinearRecurrence[{6,-15,20, -15,6,-1}, {0,1,12,60,200,525}, 30] (* Harvey P. Dale, Oct 19 2014 *)
  • PARI
    a(n)=n^2*(n+1)^2*(n+2)/12 \\ Charles R Greathouse IV, Oct 07 2015
    
  • SageMath
    def A004302(n): return 3*binomial(n+2,3)^2//(n+2)
    print([A004302(n) for n in range(41)]) # G. C. Greubel, Mar 12 2025

Formula

From Paul Barry, Feb 03 2005: (Start)
G.f.: x*(1 + 6*x + 3*x^2)/(1 - x)^6.
a(n) = C(n, 2)*C(n+1, 3). (End)
a(n) = 3*C(n+2,3)^2/(n+2). - Zerinvary Lajos, May 09 2008
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5. - Harvey P. Dale, Oct 19 2014
a(n) = A000217(n)*A000292(n). - Bruno Berselli, Jan 13 2015
a(n) = Sum_{k=0..n} Sum_{i=0..n} i*C(k+1,k-1). - Wesley Ivan Hurt, Sep 21 2017
a(n) = Sum_{i=0..n} (n+2)*(n-i)^3/3. - Bruno Berselli, Oct 31 2017
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=1} 1/a(n) = 3*Pi^2 - 57/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 45/2 - Pi^2/2 - 24*log(2). (End)
E.g.f.: exp(x)*x*(12 + 60*x + 54*x^2 + 14*x^3 + x^4)/12. - Stefano Spezia, May 22 2023

A108647 a(n) = (n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144.

Original entry on oeis.org

1, 20, 150, 700, 2450, 7056, 17640, 39600, 81675, 157300, 286286, 496860, 828100, 1332800, 2080800, 3162816, 4694805, 6822900, 9728950, 13636700, 18818646, 25603600, 34385000, 45630000, 59889375, 77808276, 100137870, 127747900
Offset: 0

Views

Author

Emeric Deutsch, Jun 13 2005

Keywords

Comments

Kekulé numbers for certain benzenoids.
a(n-4), n>=4, is the number of ways to have n identical objects in m=4 of altogether n distinguishable boxes (n-4 boxes stay empty). - Wolfdieter Lang, Nov 13 2007

Examples

			a(2) = 150 because n=6 identical balls can be put into m=4 of n=6 distinguishable boxes in binomial(6,4)*(4!/(3!*1!)+ 4!/(2!*2!)) = 15*(4 + 6) = 150 ways. The m=4 part partitions of 6, namely (1^3,3) and (1^2,2^2) specify the filling of each of the 15 possible four box choices. - _Wolfdieter Lang_, Nov 13 2007
		

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, no. 23).

Crossrefs

Fourth column of triangle A103371.

Programs

  • Haskell
    a108647 = flip a103371 3 . (+ 3)  -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    [4*Binomial(n+4, 4)^2/(n+4): n in [0..30]]; // G. C. Greubel, Oct 28 2022
    
  • Maple
    a:=(n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144: seq(a(n),n=0..30);
  • Mathematica
    Array[Binomial[# + 4, 4] Binomial[# + 3, 2] (# + 1)/3 &, 28, 0] (* or *)
    CoefficientList[Series[(1 + 12 x + 18 x^2 + 4 x^3)/(1 - x)^8, {x, 0, 27}], x] (* Michael De Vlieger, Dec 17 2017 *)
  • MuPAD
    4*binomial(n,4)^2/n $ n = 4..35; // Zerinvary Lajos, May 09 2008
    
  • SageMath
    [4*binomial(n+4, 4)^2/(n+4) for n in (0..30)] # G. C. Greubel, Oct 28 2022

Formula

a(n) = C(n+4,4)*C(n+3,2)(n+1)/3. - Paul Barry, May 13 2006
G.f.: (1+12*x+18*x^2+4*x^3)/(1-x)^8.
a(n) = 4*C(n,4)^2/n, n >= 4. - Zerinvary Lajos, May 09 2008
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 20*Pi^2 - 589/3.
Sum_{n>=0} (-1)^n/a(n) = 64*log(2) - 2*Pi^2 - 71/3. (End)
E.g.f.: (144 + 2736*x + 7992*x^2 + 7416*x^3 + 2826*x^4 + 486*x^5 + 37*x^6 + x^7)*exp(x)/144. - G. C. Greubel, Oct 28 2022

A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*3 and containing (k+1)*3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 3, 1, 7, 14, 1, 13, 81, 39, 1, 21, 304, 456, 84, 1, 31, 875, 3000, 1750, 155, 1, 43, 2106, 13875, 18500, 5265, 258, 1, 57, 4459, 50421, 128625, 84035, 13377, 399, 1, 73, 8576, 153664, 669536, 836920, 307328, 30016, 584, 1, 91, 15309, 409536, 2815344, 6001128, 4223016, 955584, 61236, 819, 1
Offset: 0

Views

Author

Susanne Wienand, Oct 10 2011

Keywords

Comments

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated
a value of dir,
(c) consecutive L's increment the index of dir,
(d) consecutive R's decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 3.
The values in the triangle are proved by brute force for 0 <= n <= 11. The formulas are not yet proved in general. - Susanne Wienand
Let S(N,n,k) = C(n,k)^(N+1)*Sum_{j=0..N} Sum_{i=0..N} (-1)^(N-j+i)*C(N-i,j)*((n+1)/(k+1))^j. Then S(0,n,k) = A007318(n,k), S(1,n,k) = A103371(n,k), S(2,n,k) = T(n,k), S(3,n,k) = A197653(n,k), S(4,n,k) = A197654(n,k), S(5,n,k) = A197655(n,k). - Peter Luschny, Oct 21 2011
The number triangle can be calculated recursively by the number triangles A103371 and A007318. The first column of the triangle contains the central polygonal numbers A002061. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027444. Row sums are in A197657. - Susanne Wienand, Nov 24 2011
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 62. - Susanne Wienand, Jun 24 2015

Examples

			For n = 4 and k = 2, T(3,4,2) = 456.
Recursive example:
T(1,4,0) = 1
T(1,4,1) = 4
T(1,4,2) = 6
T(1,4,3) = 4
T(1,4,4) = 1
T(2,4,0) = 5
T(2,4,1) = 40
T(2,4,2) = 60
T(2,4,3) = 20
T(2,4,4) = 1
T(3,4,0) = T(1,4,0)^3 + T(1,4,0)*T(2,4,4-1-0) = 1^3 + 1*20 = 21
T(3,4,1) = T(1,4,1)^3 + T(1,4,1)*T(2,4,4-1-1) = 4^3 + 4*60 = 304
T(3,4,2) = T(1,4,2)^3 + T(1,4,2)*T(2,4,4-1-2) = 6^3 + 6*40 = 456
T(3,4,3) = T(1,4,3)^3 + T(1,4,3)*T(2,4,4-1-3) = 4^3 + 4*5  = 84
T(3,4,4) = 1.
Example for closed formula:
T(4,2) = (C(4,2))^3 + C(4,2) * C(4,3) * C(5,3) = 6^3 + 6 * 4 * 10 = 456.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*3 = 15 and S contains (2+1)*3 = 9 L's.
  S: L,L,L,L,L,L,L,L,L,R,R,R,R,R,R
dir: 1,2,0,1,2,0,1,2,0,0,2,1,0,2,1
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L
dir: 1,2,2,2,0,1,2,2,1,1,1,0,0,0,0
  S: L,R,R,R,L,L,L,L,R,R,L,L,L,R,L
dir: 1,1,0,2,2,0,1,2,2,1,1,2,0,0,0
Each value of dir occurs 15/3 = 5 times.
		

Crossrefs

Programs

  • Maple
    A194595 := (n,k)->binomial(n,k)^3*(k^2+k+1+n^2+n-k*n)/((k+1)^2);
    seq(print(seq(A194595(n,k),k=0..n)),n=0..7); # Peter Luschny, Oct 14 2011
  • Mathematica
    T[n_, k_] := Binomial[n, k]^3*(k^2 + k + 1 + n^2 + n - k*n)/((k + 1)^2);
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)
  • PARI
    A194595(n,k) = {if(n == 1+2*k,3,(1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n))*binomial(n,k)^3} \\ Peter Luschny, Nov 24 2011

Formula

Recursive formula (conjectured):
T(n,k) = T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k), 0 <= k < n
T(3,n,n) = 1, k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k), 0 <= k < n
T(2,n,n) = 1, k = n
T(2,n,k) = A103371,
T(1,n,k) = A007318 (Pascal's Triangle).
Closed formula (conjectured): T(n,k) = (C(n,k))^3 + C(n,k) * C(n,k+1) * C(n+1,k+1). - Susanne Wienand
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,2). - Peter Luschny, Oct 20 2011
T(n,k) = A073254(n+1,k+1)C(n,k)^3/(k+1)^2. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^3, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 3. - Peter Luschny, Nov 24 2011
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