cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 32 results. Next

A000984 Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.

Original entry on oeis.org

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100, 126410606437752, 495918532948104, 1946939425648112
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).
Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.
Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001
Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002
Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of directed, convex polyominoes having semiperimeter n+2.
Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002
The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003
Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003
Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003
Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003
Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003
a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004
The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2) * (sqrt(n)) * zeta(1/2). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006
Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007
So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013
The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008
Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008
Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008
Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009
Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009
Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.
Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011
Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012
From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.
Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1). Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012
a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 2013
Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments. For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013
The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014
From Fung Lam, May 19 2014: (Start)
Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)
a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048. The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014
The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(-1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k = -i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018
From Isaac Saffold, Dec 28 2018: (Start)
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019
a(n) is the number of walks of length 2n 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 from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:
(1) (2,1) (3,2)
(1,1,1) (1,2,2)
(2,2,1)
(1,1,2,1)
(2,1,1,1)
(1,1,1,1,1)
The following relate to these compositions:
- The unordered version is A000070.
- The alternating sum -1 version is counted by A001791, ranked by A345910/A345912.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
- Ranked by A345909 (reverse: A345911).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.
(End)
From Michael Wallner, Jan 25 2022: (Start)
a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(2)=6:
3 4 2 4 3 4 3|4 4|3 2|4
1|2, 1|3, 2|1, 1 2, 1 2, 1 3
a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.
Example for a(2)=6:
3|4 2|4 4|3 3|4 4|3 4|2
1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)
From Shel Kaphan, Jan 12 2023: (Start)
a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.
a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.
Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)
p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023
Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023
a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

Examples

			G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. - _Michael B. Porter_, Jul 06 2016

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.
  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 160.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
  • Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
  • Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
  • Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
  • Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
  • Leonard Lipshitz and A. van der Poorten, "Rational functions, diagonals, automata and arithmetic", in Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), 339-358.
  • 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).

Links

Crossrefs

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
Cf. A226078, A051924 (first differences).
Row sums of A059481, A008459, A152229, A158815, A205946.
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Cf. A000346, A001700, A001791, A008549, A032443, A073016, A097805, A126869.
Cf. A003418, A056040, A001316, A261130, A356637.

Programs

  • GAP
    List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018
  • Haskell
    a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011
  • Magma
    a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];
  • Maple
    A000984 := n-> binomial(2*n,n); seq(A000984(n), n=0..30);
with(combstruct); [seq(count([S,{S=Prod(Set(Z,card=i),Set(Z,card=i))}, labeled], size=(2*i)), i=0..20)];
with(combstruct); [seq(count([S,{S=Sequence(Union(Arch,Arch)), Arch=Prod(Epsilon, Sequence(Arch),Z)},unlabeled],size=i), i=0..25)];
with(combstruct):bin := {B=Union(Z,Prod(B,B))}: seq (count([B,bin,unlabeled],size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007
A000984List := proc(m) local A, P, n; A := [1,2]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
A := [op(A), 2*P[-1]] od; A end: A000984List(28); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
CoefficientList[Series[1/Sqrt[1-4x],{x,0,25}],x]  (* Harvey P. Dale, Mar 14 2011 *)
  • Maxima
    A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */
  • PARI
    A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=prodeuler(p=2,2*n,p^(fv(2*n,p)-2*fv(n,p))) \\ Charles R Greathouse IV, Aug 21 2013
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=my(s=1);forprime(p=2,2*n,s*=p^(fv(2*n,p)-2*fv(n,p)));s \\ Charles R Greathouse IV, Aug 21 2013
  • Python
    from _future_ import division
A000984_list, b = [1], 1
for n in range(10**3):
    b = b*(4*n+2)//(n+1)
    A000984_list.append(b) # Chai Wah Wu, Mar 04 2016

Formula

a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n) =A099976.
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1)=0.
a(n) = 2^n/n! * Product_{k=0..n-1} (2*k+1).
a(n) = a(n-1)*(4-2/n) = Product_{k=1..n} (4-2/k) = 4*a(n-1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n). - Henry Bottomley, Nov 10 2000
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
Integral representation as n-th moment of a positive function on the interval [0, 4]: a(n) = Integral_{x=0..4}(x^n*((x*(4-x))^(-1/2))/Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001
Sum_{n>=1} 1/a(n) = (2*Pi*sqrt(3) + 9)/27. [Lehmer 1985, eq. (15)] - Benoit Cloitre, May 01 2002 (= A073016. - Bernard Schott, Jul 20 2022)
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
a(n) = Sum_{k=0..n} binomial(n+k-1,k), row sums of A059481. - Vladeta Jovovic, Aug 28 2002
E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002
E.g.f.: I_0(2*x) = Sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09 2002
a(n) = Sum_{k=0..n} binomial(n, k)^2. - Benoit Cloitre, Jan 31 2003
Determinant of n X n matrix M(i, j) = binomial(n+i, j). - Benoit Cloitre, Aug 28 2003
Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -log(log(16)/(m^2*Pi)), we have f(m) = ceiling( (q + log(q)) / log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003
a(n) = 2*Sum_{k=0..(n-1)} a(k)*a(n-k+1)/(k+1). - Philippe Deléham, Jan 01 2004
a(n+1) = Sum_{j=n..n*2+1} binomial(j, n). E.g., a(4) = C(7,3) + C(6,3) + C(5,3) + C(4,3) + C(3,3) = 35 + 20 + 10 + 4 + 1 = 70. - Jon Perry, Jan 20 2004
a(n) = (-1)^(n)*Sum_{j=0..(2*n)} (-1)^j*binomial(2*n, j)^2. - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004
a(n) = Sum_{k=0..n} binomial(2n+1, k)*sin((2n-2k+1)*Pi/2). - Paul Barry, Nov 02 2004
a(n-1) = (1/2)*(-1)^n*Sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j). - Benoit Cloitre, Jun 18 2005
a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 2005
G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108. - Paul Barry, Feb 03 2006
a(n) = A006480(n) / A005809(n). - Zerinvary Lajos, Jun 28 2007
a(n) = Sum_{k=0..n} A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007
a(n) = Sum_{k>=0} A039599(n, k). a(n) = Sum_{k>=0} A050165(n, k). a(n) = Sum_{k>=0} A059365(n, k)*2^k, n>0. a(n+1) = Sum_{k>=0} A009766(n, k)*2^(n-k+1). - Philippe Deléham, Jan 01 2004
a(n) = 4^n*Sum_{k=0..n} C(n,k)(-4)^(-k)*A000108(n+k). - Paul Barry, Oct 18 2007
a(n) = Sum_{k=0..n} A039598(n,k)*A059841(k). - Philippe Deléham, Nov 12 2008
A007814(a(n)) = A000120(n). - Vladimir Shevelev, Jul 20 2009
From Paul Barry, Aug 05 2009: (Start)
G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);
G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
If n>=3 is prime, then a(n) == 2 (mod 2*n). - Vladimir Shevelev, Sep 05 2010
Let A(x) be the g.f. and B(x) = A(-x), then B(x) = sqrt(1-4*x*B(x)^2). - Vladimir Kruchinin, Jan 16 2011
a(n) = (-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)). - Gerry Martens, May 03 2011
a(n) = upper left term in M^n, M = the infinite square production matrix:
2, 2, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, 1, .... - Gary W. Adamson, Jul 14 2011
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
E.g.f.: hypergeometric([1/2],[1],4*x). - Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n-1} a(k)*A000108(n-k-1). - Alzhekeyev Ascar M, Mar 09 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2*H(k)/(2*H(n)-H(2*n)), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(1-4*x), where Q(k) = 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013
a(n) = 2^(4*n)/((2*n+1)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1)). - Mircea Merca, Nov 12 2013
a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
Sum_{n>=0} a(n)/n! = A234846. - Richard R. Forberg, Feb 10 2014
0 = a(n)*(16*a(n+1) - 6*a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
a(n+1) = 4*a(n) - 2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(1-1/(2*k)). - Stanislav Sykora, Aug 09 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n) = (-4)^n*binomial(-1/2,n). - Jean-François Alcover, Feb 10 2015
a(n) = 4^n*hypergeom([-n,1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi). - Peter Luschny, Oct 14 2015
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
a(n) = GegenbauerC(n,-n,-1). - Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2. - Andres Cicuttin, May 30 2016
Sum_{n>=0} (-1)^n/a(n) = 4*(5 - sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio. - Ilya Gutkovskiy, Jul 04 2016
From Peter Bala, Jul 22 2016: (Start)
This sequence occurs as the closed-form expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(n+k)*binomial(2*n - 1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n-1} binomial(n - 1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x - k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x - k,n)*binomial(y - k,n) appear to equal Kronecker's delta(n,0).
a(n) = (-1)^n*Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y - k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
From Ralf Steiner, Apr 07 2017: (Start)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for q in N, p in Z/{-4q< (some p) <-2}.
...
Sum_{k>=0} a(k)/(-4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for p>4q.(End)
Boas-Buck recurrence: a(n) = (2/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(2*n+1, k) for n in N. - Rene Adad, Sep 30 2017
a(n) = A034870(n,n). - Franck Maminirina Ramaharo, Nov 26 2018
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*(sqrt(4-x) + sqrt(x)*arcsin(sqrt(x)/2)) / (4-x)^(3/2).
E.g.f. for {1/a(n)}: 1 + exp(x/4)*sqrt(Pi*x)*erf(sqrt(x)/2)/2.
Sum_{n>=0} (-1)^n/a(n) = 4*(1/5 - arcsinh(1/2)/(5*sqrt(5))). (End)
From Peter Luschny, Sep 08 2022: (Start)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = A001316(n) * A356637(n) * A261130(n) for n >= 2. (End)
a(n) = 4^n*binomial(n-1/2,-1/2) = 4^n*GegenbauerC(n,1/4,1). - Gerry Martens, Oct 19 2022
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^2 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^2 * binomial(2*n, n+k)^3 = x*(x + 2*n)*a(n) (x arbitrary). Compare with the identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^2 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Mar 31 2024: (Start)
4^n*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
16^n = Sum_{k = 0..2*n} a(k)*a(2*n-k). (End)
From Gary Detlefs, May 28 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*binomial(2*k,k)*2^(n-2*k). (H. W. Gould) - Gary Detlefs, May 28 2024
a(n) = Sum_{k=0..2*n} (-1)^k*binomial(2n,k)*binomial(2*n+2*k,n+k)*3^(2*n-k). (H. W. Gould) (End)
a(n) = Product_{k>=n+1} k^2/(k^2 - n^2). - Antonio Graciá Llorente, Sep 08 2024
a(n) = Product_{k=1..n} A003418(floor(2*n/k))^((-1)^(k+1)) (Golomb, 2003). - Amiram Eldar, Aug 08 2025

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

N. J. A. Sloane

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).

Links

Crossrefs

Cf. A000110, A007318, A030662, A046097, A060897-A060900, A049027, A076025, A076026, A060150, A001263, A005773, A001405, A132813, A134285.
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
Cf. A005043, A091867, A001792, A001813, A166454.

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

A051924 a(n) = binomial(2*n,n) - binomial(2*n-2,n-1); or (3n-2)*C(n-1), where C = Catalan numbers (A000108).

Original entry on oeis.org

1, 4, 14, 50, 182, 672, 2508, 9438, 35750, 136136, 520676, 1998724, 7696444, 29716000, 115000920, 445962870, 1732525830, 6741529080, 26270128500, 102501265020, 400411345620, 1565841089280, 6129331763880, 24014172955500, 94163002754652, 369507926510352
Offset: 1

Views

Author

Barry E. Williams, Dec 19 1999

Keywords

Comments

Number of partitions with Ferrers plots that fit inside an n X n box, but not in an n-1 X n-1 box. - Wouter Meeussen, Dec 10 2001
From Benoit Cloitre, Jan 29 2002: (Start)
Let m(1,j)=j, m(i,1)=i and m(i,j) = m(i-1,j) + m(i,j-1); then a(n) = m(n,n):
1 2 3 4 ...
2 4 7 11 ...
3 7 14 25 ...
4 11 25 50 ... (End)
This sequence also gives the number of clusters and non-crossing partitions of type D_n. - F. Chapoton, Jan 31 2005
If Y is a 2-subset of a 2n-set X then a(n) is the number of (n+1)-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
Prefaced with a 1: (1, 1, 4, 14, 50, ...) and convolved with the Catalan sequence = A097613: (1, 2, 7, 25, 91, ...). - Gary W. Adamson, May 15 2009
Total number of up steps before the second return in all Dyck n-paths. - David Scambler, Aug 21 2012
Conjecture: a(n) mod n^2 = n+2 iff n is an odd prime. - Gary Detlefs, Feb 19 2013
First differences of A000984 and A030662. - J. M. Bergot, Jun 22 2013
From R. J. Mathar, Jun 30 2013: (Start)
Equivalent to the Meeussen comment and the Bergot comment: The array view of A007318 is
1, 1, 1, 1, 1, 1,
1, 2, 3, 4, 5, 6,
1, 3, 6, 10, 15, 21,
1, 4, 10, 20, 35, 56,
1, 5, 15, 35, 70, 126,
1, 6, 21, 56, 126, 252,
and a(n) are the hook sums Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). (End)
From Gus Wiseman, Apr 12 2019: (Start)
Equivalent to Wouter Meeussen's comment, a(n) is the number of integer partitions (of any positive integer) such that the maximum of the length and the largest part is n. For example, the a(1) = 1 through a(3) = 14 partitions are:
(1) (2) (3)
(11) (31)
(21) (32)
(22) (33)
(111)
(211)
(221)
(222)
(311)
(321)
(322)
(331)
(332)
(333)
(End)
Coxeter-Catalan numbers for Coxeter groups of type D_n [Armstrong]. - N. J. A. Sloane, Mar 09 2022
a(n+1) is the number of ways that a best of n pairs contest with early termination can go. For example, the first stage of an association football (soccer) penalty-kick shoot out has n=5 pairs of shots and there are a(6)=672 distinct ways it can go. For n=2 pairs, writing G for goal and M for miss, and listing the up-to-four shots in chronological order with teams alternating shots, the n(3)=14 possibilities are MMMM, MMMG, MMGM, MMGG, MGM, MGGM, MGGG, GMMM, GMMG, GMG, GGMM, GGMG, GGGM, and GGGG. Not all four shots are taken in two cases because it becomes impossible for one team to overcome the lead of the other team. - Lee A. Newberg, Jul 20 2024

Examples

			Sums of {1}, {2, 1, 1}, {2, 2, 3, 3, 2, 1, 1}, {2, 2, 4, 5, 7, 6, 7, 5, 5, 3, 2, 1, 1}, ...

References

  • Drew Armstrong, Generalized Noncrossing Partitions and Combinatorics of Coxeter Groups, Mem. Amer. Math. Soc. 202 (2009), no. 949, x+159. MR 2561274 16; See Table 2.8.

Links

Crossrefs

Left-central elements of the (1, 2)-Pascal triangle A029635.
Column sums of A096771.
Cf. A000108, A024482 (diagonal from 2), A076540 (diagonal from 3), A000124 (row from 2), A004006 (row from 3), A006522 (row from 4).
Cf. A128064; first differences of A000984.
Cf. A097613.
Cf. A115720, A252464, A257990, A263297, A325189, A325192, A325193.

Programs

  • Haskell
    a051924 n = a051924_list !! (n-1)
a051924_list = zipWith (-) (tail a000984_list) a000984_list
-- Reinhard Zumkeller, May 25 2013
  • Magma
    [Binomial(2*n, n)-Binomial(2*n-2, n-1): n in [1..28]]; // Vincenzo Librandi, Dec 21 2016
  • Maple
    C:= n-> binomial(2*n, n)/(n+1): seq((n+1)*C(n)-n*C(n-1), n=1..25); # Emeric Deutsch, Jan 08 2008
Z:=(1-z-sqrt(1-4*z))/sqrt(1-4*z): Zser:=series(Z, z=0, 32): seq(coeff(Zser, z, n), n=1..24); # Zerinvary Lajos, Jan 01 2007
a := n -> 2^(-2+2*n)*GAMMA(-1/2+n)*(3*n-2)/(sqrt(Pi)*GAMMA(1+n)):
seq(simplify(a(n)), n=1..24); # Peter Luschny, Dec 14 2015
  • Mathematica
    Table[Binomial[2n,n]-Binomial[2n-2,n-1],{n,30}] (* Harvey P. Dale, Jan 15 2012 *)
  • PARI
    a(n)=binomial(2*n,n)-binomial(2*n-2,n-1) \\ Charles R Greathouse IV, Jun 25 2013
  • PARI
    {a(n)=polcoeff((1-x) / sqrt(1-4*x +x*O(x^n)) - 1,n)}
for(n=1,30,print1(a(n),", ")) \\ Paul D. Hanna, Nov 08 2014
  • PARI
    {a(n)=polcoeff( sum(m=1, n, x^m * sum(k=0, m, binomial(m, k)^2 * x^k) / (1-x +x*O(x^n))^(2*m)), n)}
for(n=1, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Nov 08 2014
  • Sage
    a = lambda n: 2^(-2+2*n)*gamma(n-1/2)*(3*n-2)/(sqrt(pi)*gamma(1+n))
[a(n) for n in (1..120)] # Peter Luschny, Dec 14 2015

Formula

G.f.: (1-x) / sqrt(1-4*x) - 1. - Paul D. Hanna, Nov 08 2014
G.f.: Sum_{n>=1} x^n/(1-x)^(2*n) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n+1) = binomial(2*n, n) + 2*Sum_{i=0..n-1} binomial(n+i, i) (V's in Pascal's Triangle). - Jon Perry Apr 13 2004
a(n) = n*C(n-1) - (n-1)*C(n-2), where C(n) = A000108(n) = Catalan(n). For example, a(5) = 50 = 5*C(4) - 4*C(3) - 5*14 - 3*5 = 70 - 20. Triangle A128064 as an infinite lower triangular matrix * A000108 = A051924 prefaced with a 1: (1, 1, 4, 14, 50, 182, ...). - Gary W. Adamson, May 15 2009
Sum of 3 central terms of Pascal's triangle: 2*C(2+2*n, n)+C(2+2*n, 1+n). - Zerinvary Lajos, Dec 20 2005
a(n+1) = A051597(2n,n). - Philippe Deléham, Nov 26 2006
The sequence 1,1,4,... has a(n) = C(2*n,n)-C(2*(n-1),n-1) = 0^n+Sum_{k=0..n} C(n-1,k-1)*A002426(k), and g.f. given by (1-x)/(1-2*x-2*x^2/(1-2*x-x^2/(1-2*x-x^2/(1-2*x-x^2/(1-.... (continued fraction). - Paul Barry, Oct 17 2009
a(n) = (3*n-2)*(2*n-2)!/(n*(n-1)!^2) = A001700(n) + A001791(n-1). - David Scambler, Aug 21 2012
D-finite with recurrence: a(n) = 2*(3*n-2)*(2*n-3)*a(n-1)/(n*(3*n-5)). - Alois P. Heinz, Apr 25 2014
a(n) = 2^(-2+2*n)*Gamma(-1/2+n)*(3*n-2)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
a(n) ~ (3/4)*4^n*(1-(7/24)/n-(7/128)/n^2-(85/3072)/n^3-(581/32768)/n^4-(2611/262144)/n^5)/sqrt(n*Pi). - Peter Luschny, Dec 16 2015
E.g.f.: ((1 - x)*BesselI(0,2*x) + x*BesselI(1,2*x))*exp(2*x) - 1. - Ilya Gutkovskiy, Dec 20 2016
a(n) = 2 * A097613(n) for n > 1. - Bruce J. Nicholson, Jan 06 2019
Sum_{n>=1} a(n)/8^n = 7/(4*sqrt(2)) - 1. - Amiram Eldar, May 06 2023

Extensions

Edited by N. J. A. Sloane, May 03 2008, at the suggestion of R. J. Mathar

A047909 Array read by antidiagonals upwards: h(n,k) = number of sequences with n copies each of 1,2,...,k and longest increasing subsequence of length k (n>=1, k>=1).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 19, 47, 1, 1, 69, 1306, 641, 1, 1, 251, 31451, 195709, 11389, 1, 1, 923, 729811, 46922017, 50775091, 248749, 1, 1, 3431, 16928840, 10258694241, 162588279629, 20117051281, 6439075, 1, 1, 12869, 397222288, 2176464012941, 449363984934526, 1077273394836829, 11260558754404, 192621953, 1
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Old name was: Triangle of numbers arising from problem of complete increasing subsequences.
Table h_p(k) on page 80 in the Horton & Kurn reference has two typos. - Alois P. Heinz, Feb 05 2016
Conjecture: Column k > 1 is asymptotic to k^(k*n + 1/2) / (2*Pi*n)^((k-1)/2). - Vaclav Kotesovec, Feb 21 2016
Conjecture: Row k > 1 is asymptotic to sqrt(k) * (k^k/(k-1)!)^n * n^((k-1)*n) / exp((k-1)*(n+1)). - Vaclav Kotesovec, Feb 21 2016

Examples

			First few antidiagonals are:
  1;
  1,   1;
  1,   5,      1;
  1,  19,     47,        1;
  1,  69,   1306,      641,        1;
  1, 251,  31451,   195709,    11389,      1;
  1, 923, 729811, 46922017, 50775091, 248749,   1;
  ...
First few rows are:
  1,   1,        1,             1,                   1, ...
  1,   5,       47,           641,               11389, ...
  1,  19,     1306,        195709,            50775091, ...
  1,  69,    31451,      46922017,        162588279629, ...
  1, 251,   729811,   10258694241,     449363984934526, ...
  1, 923, 16928840, 2176464012941, 1162145520205261219, ...
  ...

Links

Crossrefs

Columns k=1-10 give: A000012, A030662, A047911, A268840, A268841, A268842, A268843, A268844, A268845, A268846.
Rows n=1-10 give: A000012, A006902, A047910, A268847, A268848, A268849, A268850, A268851, A268852, A268853.
Main diagonal gives A268485.

Programs

  • Maple
    g:= proc(l) option remember; (n-> f(l[1..nops(l)-1])*
      binomial(n-1, l[-1]-1)+ add(f(sort(subsop(j=l[j]
      -1, l))), j=1..nops(l)-1))(add(i, i=l))
    end:
f:= l-> (n-> `if`(n<2 or l[-1]=1, 1, `if`(l[1]=0, 0, `if`(
         n=2, binomial(l[1]+l[2], l[1])-1, g(l)))))(nops(l)):
h:= (n, k)-> f([n$k]):
seq(seq(h(1+d-k, k), k=1..d), d=1..10);  # Alois P. Heinz, Feb 11 2016
# second Maple program:
b:= proc(k, p, j, l, t) option remember;
      `if`(k=0, (-1)^t/l!, `if`(p<0, 0, add(b(k-i, p-1,
       j+1, l+i*j, irem(t+i*j, 2))/(i!*p!^i), i=0..k)))
    end:
h:= (p, k)-> k!*(p*k)!*b(k, p-1, 1, 0, irem(k, 2)):
seq(seq(h(1+d-k, k), k=1..d), d=1..10);  # Alois P. Heinz, Mar 03 2016
  • Mathematica
    g[l_] := g[l] = Function[n, f[l[[1 ;; -2]]]*Binomial[n-1, l[[-1]]-1] + Sum[ f[Sort[ReplacePart[l, j -> l[[j]] - 1]]], {j, 1, Length[l] -1}]][ Total[ l]]; f[l_] := Function [n, If[n<2 || l[[-1]]==1, 1, If[l[[1]]==0, 0, If[n == 2, Binomial[l[[1]] + l[[2]], l[[1]] ] - 1, g[l]]]]][Length[l]]; h[n_, k_] := f[Array[n&, k]]; Table[Table[h[1+d-k, k], {k, 1, d}], {d, 1, 10}] // Flatten (* Jean-François Alcover, Feb 19 2016, after Alois P. Heinz *)

Formula

Reference gives explicit formula.

Extensions

New name, two terms corrected and more terms from Alois P. Heinz, Feb 08 2016

A109128 Triangle read by rows: T(n,k) = T(n-1,k-1) + T(n-1,k) + 1 for 0

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 11, 7, 1, 1, 9, 19, 19, 9, 1, 1, 11, 29, 39, 29, 11, 1, 1, 13, 41, 69, 69, 41, 13, 1, 1, 15, 55, 111, 139, 111, 55, 15, 1, 1, 17, 71, 167, 251, 251, 167, 71, 17, 1, 1, 19, 89, 239, 419, 503, 419, 239, 89, 19, 1, 1, 21, 109, 329, 659, 923, 923, 659, 329, 109, 21, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Jun 20 2005

Keywords

Comments

Eigensequence of the triangle = A001861. - Gary W. Adamson, Apr 17 2009

Examples

			Triangle begins as:
  1;
  1   1;
  1   3   1;
  1   5   5   1;
  1   7  11   7   1;
  1   9  19  19   9   1;
  1  11  29  39  29  11   1;
  1  13  41  69  69  41  13   1;
  1  15  55 111 139 111  55  15   1;
  1  17  71 167 251 251 167  71  17   1;
  1  19  89 239 419 503 419 239  89  19   1;

Links

Crossrefs

Cf. A000035, A000295, A001861, A005408, A014473, A028387, A030662.
Cf. A134760, A281362, A327767.
Cf. A000325 (row sums).
Sequence m*binomial(n,k) - (m-1): A007318 (m=1), this sequence (m=2), A131060 (m=3), A131061 (m=4), A131063 (m=5), A131065 (m=6), A131067 (m=7), A168625 (m=8).

Programs

  • Haskell
    a109128 n k = a109128_tabl !! n !! k
a109128_row n = a109128_tabl !! n
a109128_tabl = iterate (\row -> zipWith (+)
   ([0] ++ row) (1 : (map (+ 1) $ tail row) ++ [0])) [1]
-- Reinhard Zumkeller, Apr 10 2012
  • Magma
    [2*Binomial(n,k) -1: k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 12 2020
  • Maple
    A109128 := proc(n,k)
    2*binomial(n,k)-1 ;
end proc: # R. J. Mathar, Jul 12 2016
  • Mathematica
    Table[2*Binomial[n,k] -1, {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 12 2020 *)
  • Sage
    [[2*binomial(n,k) -1 for k in (0..n)] for n in (0..12)] # G. C. Greubel, Mar 12 2020

Formula

T(n,k) = T(n-1,k-1) + T(n-1,k) + 1 with T(n,0) = T(n,n) = 1.
Sum_{k=0..n} T(n, k) = A000325(n+1) (row sums).
T(n, k) = 2*binomial(n,k) - 1. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Sep 30 2007
T(n, 1) = 2*n - 1 = A005408(n+1) for n>0.
T(n, 2) = n^2 + n - 1 = A028387(n-2) for n>1.
T(n, k) = Sum_{j=0..n-k} C(n-k,j)*C(k,j)*(2 - 0^j) for k <= n. - Paul Barry, Apr 27 2006
T(n,k) = A014473(n,k) + A007318(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 10 2012
From G. C. Greubel, Apr 06 2024: (Start)
T(n, n-k) = T(n, k).
T(2*n, n) = A134760(n).
T(2*n-1, n) = A030662(n), for n >= 1.
Sum_{k=0..n-1} T(n, k) = A000295(n+1), for n >= 1.
Sum_{k=0..n} (-1)^k*T(n, k) = 2*[n=0] - A000035(n+1).
Sum_{k=0..n-1} (-1)^k*T(n, k) = A327767(n), for n >= 1.
Sum_{k=0..floor(n/2)} T(n-k, k) = A281362(n).
Sum_{k=0..floor((n-1)/2)} T(n-k, k) = A281362(n-1) - (1+(-1)^n)/2 for n >= 1.
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = b(n), where b(n) is the repeating pattern {1,1,0,-2,-3,-1,2,2,-1,-3,-2,0} with b(n) = b(n-12). (End)

Extensions

Offset corrected by Reinhard Zumkeller, Apr 10 2012

A014473 Pascal's triangle - 1: Triangle read by rows: T(n, k) = A007318(n, k) - 1.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 5, 3, 0, 0, 4, 9, 9, 4, 0, 0, 5, 14, 19, 14, 5, 0, 0, 6, 20, 34, 34, 20, 6, 0, 0, 7, 27, 55, 69, 55, 27, 7, 0, 0, 8, 35, 83, 125, 125, 83, 35, 8, 0, 0, 9, 44, 119, 209, 251, 209, 119, 44, 9, 0, 0, 10, 54, 164, 329, 461, 461, 329, 164, 54, 10, 0
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Indexed as a square array A(n,k): If X is an (n+k)-set and Y a fixed k-subset of X then A(n,k) is equal to the number of n-subsets of X intersecting Y. - Peter Luschny, Apr 20 2012

Examples

			Triangle begins:
   0;
   0, 0;
   0, 1,  0;
   0, 2,  2,  0;
   0, 3,  5,  3,  0;
   0, 4,  9,  9,  4,  0;
   0, 5, 14, 19, 14,  5, 0;
   0, 6, 20, 34, 34, 20, 6, 0;
   ...
Seen as a square array read by antidiagonals:
  [0] 0, 0,  0,  0,   0,   0,   0,    0,    0,    0,    0,     0, ... A000004
  [1] 0, 1,  2,  3,   4,   5,   6,    7,    8,    9,   10,    11, ... A001477
  [2] 0, 2,  5,  9,  14,  20,  27,   35,   44,   54,   65,    77, ... A000096
  [3] 0, 3,  9, 19,  34,  55,  83,  119,  164,  219,  285,   363, ... A062748
  [4] 0, 4, 14, 34,  69, 125, 209,  329,  494,  714, 1000,  1364, ... A063258
  [5] 0, 5, 20, 55, 125, 251, 461,  791, 1286, 2001, 3002,  4367, ... A062988
  [6] 0, 6, 27, 83, 209, 461, 923, 1715, 3002, 5004, 8007, 12375, ... A124089

Links

Crossrefs

Triangle without zeros: A014430.
Related: A323211 (A007318(n, k) + 1).
A000295 (row sums), A059841 (alternating row sums), A030662(n-1) (central terms).
Columns include A000096, A062748, A062988, A063258.
Diagonals of A(n, n+d): A030662 (d=0), A010763 (d=1), A322938 (d=2).
Cf. A000004, A001477, A007318, A030662, A059841, A109128, A124089, A129696.

Programs

  • Haskell
    a014473 n k = a014473_tabl !! n !! k
a014473_row n = a014473_tabl !! n
a014473_tabl = map (map (subtract 1)) a007318_tabl
-- Reinhard Zumkeller, Apr 10 2012
  • Magma
    [Binomial(n,k)-1: k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 08 2024
  • Maple
    with(combstruct): for n from 0 to 11 do seq(-1+count(Combination(n), size=m), m = 0 .. n) od; # Zerinvary Lajos, Apr 09 2008
# The rows of the square array:
Arow := proc(n, len) local gf, ser;
gf := (x - 1)^(-n - 1) + (-1)^(n + 1)/(x*(x - 1));
ser := series(gf, x, len+2): seq((-1)^(n+1)*coeff(ser, x, j), j=0..len) end:
for n from 0 to 9 do lprint([n], Arow(n, 12)) od; # Peter Luschny, Feb 13 2019
  • Mathematica
    Table[Binomial[n,k] -1, {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 08 2024 *)
  • SageMath
    flatten([[binomial(n,k)-1 for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Apr 08 2024

Formula

G.f.: x^2*y/((1 - x)*(1 - x*y)*(1 - x*(1 + y))). - Ralf Stephan, Jan 24 2005
T(n,k) = A109128(n,k) - A007318(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 10 2012
T(n, k) = T(n-1, k-1) + T(n-1, k) + 1, 0 < k < n with T(n, 0) = T(n, n) = 0. - Reinhard Zumkeller, Jul 18 2015
If seen as a square array read by antidiagonals the generating function of row n is: G(n) = (x - 1)^(-n - 1) + (-1)^(n + 1)/(x*(x - 1)). - Peter Luschny, Feb 13 2019
From G. C. Greubel, Apr 08 2024: (Start)
T(n, n-k) = T(n, k).
Sum_{k=0..floor(n/2)} T(n-k, k) = A129696(n-2).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = b(n-1), where b(n) is the repeating pattern {0, 0, -1, -2, -1, 1, 1, -1, -2, -1, 0, 0}_{n=0..11}, with b(n) = b(n-12). (End)

Extensions

More terms from Erich Friedman

A064315 Triangle of number of permutations by length of shortest ascending run.

Original entry on oeis.org

1, 1, 1, 5, 0, 1, 18, 5, 0, 1, 101, 18, 0, 0, 1, 611, 89, 19, 0, 0, 1, 4452, 519, 68, 0, 0, 0, 1, 36287, 3853, 110, 69, 0, 0, 0, 1, 333395, 27555, 1679, 250, 0, 0, 0, 0, 1, 3382758, 233431, 11941, 418, 251, 0, 0, 0, 0, 1, 37688597, 2167152, 59470, 658, 922, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

David W. Wilson, Sep 07 2001

Keywords

Examples

			Sequence (1, 3, 2, 5, 4) has ascending runs (1, 3), (2, 5), (4), the shortest is length 1. Of all permutations of (1, 2, 3, 4, 5), T(5,1) = 101 have shortest ascending run of length 1.
Triangle T(n,k) begins:
      1;
      1,    1;
      5,    0,   1;
     18,    5,   0,  1;
    101,   18,   0,  0,  1;
    611,   89,  19,  0,  0, 1;
   4452,  519,  68,  0,  0, 0, 1,
  36287, 3853, 110, 69,  0, 0, 0, 1;
  ...

Links

Crossrefs

Row sums give: A000142.
Columns k=1-10 give: A228614, A185652, A228670, A228671, A228672, A228673, A228674, A228675, A228676, A228677.
Cf. A030662, A064316.

Programs

  • Maple
    A:= proc(n, k) option remember; local b; b:=
      proc(u, o, t) option remember; `if`(t+o<=k, (u+o)!,
        add(b(u+i-1, o-i, min(k, t)+1), i=1..o)+
        `if`(t<=k, u*(u+o-1)!, add(b(u-i, o+i-1, 1), i=1..u)))
      end: forget(b):
      add(b(j-1, n-j, 1), j=1..n)
    end:
T:= (n, k)-> A(n, k) -A(n, k-1):
seq(seq(T(n, k), k=1..n), n=1..12); # Alois P. Heinz, Aug 29 2013
  • Mathematica
    A[n_, k_] := A[n, k] = Module[{b}, b[u_, o_, t_] := b[u, o, t] = If[t+o <= k, (u+o)!, Sum[b[u+i-1, o-i, Min[k, t]+1], {i, 1, o}] + If[t <= k, u*(u+o-1)!, Sum[ b[u-i, o+i-1, 1], {i, 1, u}]]]; Sum[b[j-1, n-j, 1], {j, 1, n}]]; T[n_, k_] := A[n, k] - A[n, k-1]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Jan 28 2015, after Alois P. Heinz *)

Formula

T(2*n,n) = binomial(2*n,n)-1 = A030662(n).
Sum_{k=1..n} k * T(n,k) = A064316(n).

A050157 T(n, k) = S(2n, n, k) for 0<=k<=n and n>=0, where S(p, q, r) is the number of upright paths from (0, 0) to (p, p-q) that do not rise above the line y = x-r.

Original entry on oeis.org

1, 1, 2, 2, 5, 6, 5, 14, 19, 20, 14, 42, 62, 69, 70, 42, 132, 207, 242, 251, 252, 132, 429, 704, 858, 912, 923, 924, 429, 1430, 2431, 3068, 3341, 3418, 3431, 3432, 1430, 4862, 8502, 11050, 12310, 12750, 12854, 12869, 12870
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

Let V = (e(1),...,e(n)) consist of q 1's and p-q 0's; let V(h) = (e(1),...,e(h)) and m(h) = (#1's in V(h)) - (#0's in V(h)) for h=1,...,n. Then S(p,q,r) is the number of V having r >= max{m(h)}.

Examples

			The triangle starts:
                                1
                              1, 2
                            2, 5, 6
                         5, 14, 19, 20
                       14, 42, 62, 69, 70
                  42, 132, 207, 242, 251, 252
               132, 429, 704, 858, 912, 923, 924

Crossrefs

T(n, 0) = A000108(n).
T(n, 1) = A000108(n+1).
T(n, n) = A000984(n).
T(n, n-1) = A030662(n).
Row sums are A296771.
Cf. A039599, A050163.

Programs

  • Maple
    A050157 := (n, k) -> binomial(2*n, n) - binomial(2*n, n+k+1):
seq(seq(A050157(n,k), k=0..n), n=0..10); # Peter Luschny, Dec 21 2017

Formula

T(n, k) = Sum_{0<=j<=k} t(n, j), array t as in A039599.
T(n, k) = binomial(2*n, n) - binomial(2*n, n+k+1). - Peter Luschny, Dec 21 2017

A144660 a(n) = Sum_{i=0..n} Sum_{j=0..n} Sum_{k=0..n} (i+j+k)!/(i!*j!*k!).

Original entry on oeis.org

1, 16, 271, 5248, 110251, 2435200, 55621567, 1301226496, 30992872483, 748574130016, 18283414868863, 450657134765056, 11192820128307871, 279787295456009728, 7032532242167190271, 177611430242835570688, 4504491083159761986451, 114662734697313744041248
Offset: 0

Views

Author

N. J. A. Sloane, Jan 31 2009, Feb 01 2009

Keywords

Links

Crossrefs

Cf. A030662, A144661, A307318. This sum is very close to that in A144511.

Programs

  • Maple
    f:=n->add( add( add( (i+j+k)!/(i!*j!*k!), i=0..n),j=0..n),k=0..n); [seq(f(n),n=0..20)];
  • Mathematica
    Table[Sum[(i + j + k)!/(i!*j!*k!), {i, 0, n}, {j, 0, n}, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 02 2019 *)
Table[Sum[(1 + k + 2*n)! * HypergeometricPFQ[{1, -1 - k - n, -n}, {-1 - k - 2*n, -k - n}, 1] / ((1 + k + n)*k!*n!^2), {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 04 2019 *)
  • PARI
    {a(n) = sum(i=0, n, sum(j=0, n, sum(k=0, n, (i+j+k)!/(i!*j!*k!))))} \\ Seiichi Manyama, Apr 02 2019

Formula

From Vaclav Kotesovec, Apr 02 2019: (Start)
Recurrence: n^2*(2*n + 1)*(91*n^4 - 478*n^3 + 917*n^2 - 755*n + 222)*a(n) = 3*(2*n - 3)*(3*n - 5)*(3*n - 4)*(91*n^4 - 114*n^3 + 29*n^2 + 9*n - 3)*a(n-1) + n^2*(2*n + 1)*(91*n^4 - 478*n^3 + 917*n^2 - 755*n + 222)*a(n-2) - 3*(2*n - 3)*(3*n - 5)*(3*n - 4)*(91*n^4 - 114*n^3 + 29*n^2 + 9*n - 3)*a(n-3).
a(n) ~ 3^(3*n + 7/2) / (16*Pi*n). (End)

A181985 Generalized Euler numbers. Square array A(n,k), n >= 1, k >= 0, read by antidiagonals. A(n,k) = n-alternating permutations of length n*k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 19, 61, 1, 1, 1, 69, 1513, 1385, 1, 1, 1, 251, 33661, 315523, 50521, 1, 1, 1, 923, 750751, 60376809, 136085041, 2702765, 1, 1, 1, 3431, 17116009, 11593285251, 288294050521, 105261234643, 199360981, 1
Offset: 1

Views

Author

Peter Luschny, Apr 04 2012

Keywords

Comments

For an integer n > 0, a permutation s = s_1...s_k is an n-alternating permutation if it has the property that s_i < s_{i+1} if and only if n divides i.
The classical Euler numbers count 2-alternating permutations of length 2n.
Ludwig Seidel gave in 1877 an efficient algorithm to compute the coefficients of sec which carries immediately over to the computation of the generalized Euler numbers (see the Maple script).

Examples

			n\k [0][1]  [2]       [3]            [4]                 [5]
[1]  1, 1,   1,        1,             1,                   1
[2]  1, 1,   5,       61,          1385,               50521  [A000364]
[3]  1, 1,  19,     1513,        315523,           136085041  [A002115]
[4]  1, 1,  69,    33661,      60376809,        288294050521  [A211212]
[5]  1, 1, 251,   750751,   11593285251,     613498040952501
[6]  1, 1, 923, 17116009, 2301250545971, 1364944703949044401
       [A030662][A211213]   [A181991]
The (n,n)-diagonal is A181992.

Links

Crossrefs

Cf. A181937, A000364, A002115, A030662, A211212, A211213, A181991, A181992.

Programs

  • Maple
    A181985_list := proc(n, len) local E, dim, i, k;
dim := n*(len-1); E := array(0..dim, 0..dim); E[0, 0] := 1;
for i from 1 to dim do
   if i mod n = 0 then E[i, 0] := 0 ;
      for k from i-1 by -1 to 0 do E[k, i-k] := E[k+1, i-k-1] + E[k, i-k-1] od;
   else E[0, i] := 0;
      for k from 1 by 1 to i do E[k, i-k] := E[k-1, i-k+1] + E[k-1, i-k] od;
   fi od;
seq(E[0, n*k], k=0..len-1) end:
for n from 1 to 6 do print(A181985_list(n, 6)) od;
  • Mathematica
    nmax = 9; A181985[n_, len_] := Module[{e, dim = n*(len - 1)}, e[0, 0] = 1; For[i = 1, i <= dim, i++, If[Mod[i, n] == 0 , e[i, 0] = 0 ; For[k = i-1, k >= 0, k--, e[k, i-k] = e[k+1, i-k-1] + e[k, i-k-1] ], e[0, i] = 0; For[k = 1, k <= i, k++, e[k, i-k] = e[k-1, i-k+1] + e[k-1, i-k] ]; ]]; Table[e[0, n*k], { k, 0, len-1}]]; t = Table[A181985[n, nmax], {n, 1, nmax}]; a[n_, k_] := t[[n, k+1]]; Table[a[n-k, k], {n, 1, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Jun 27 2013, translated and adapted from Maple *)
  • Sage
    def A181985(m, n):
    shapes = ([x*m for x in p] for p in Partitions(n))
    return (-1)^n*sum((-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in shapes)
for m in (1..6): print([A181985(m, n) for n in (0..7)]) # Peter Luschny, Aug 10 2015
Showing 1-10 of 32 results. Next

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

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