

A001405


a(n) = binomial(n, floor(n/2)).
(Formerly M0769 N0294)


407



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

0,3


COMMENTS

Sperner's theorem says that this is the maximal number of subsets of an nset such that no one contains another.
When computed from index 1, [seq(binomial(n,floor(n/2)), n = 1..30)]; > [1,1,1,2,3,6,10,20,35,70,126,...] and convolved with aerated Catalan numbers [seq(((n+1) mod 2)*binomial(n,n/2)/((n/2)+1), n = 0..30)]; > [1,0,1,0,2,0,5,0,14,0,42,0,132,0,...] shifts left by one: [1,1,2,3,6,10,20,35,70,126,252,...] and if again convolved with aerated Catalan numbers, gives A037952 apart from the initial term.  Antti Karttunen, Jun 05 2001 [This is correct because the g.f.'s satisfy (1+x*g001405(x))*g126120(x) = g001405(x) and g001405(x)*g126120(x) = g037952(x)/x.  R. J. Mathar, Sep 23 2021]
Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 2.  Emeric Deutsch, Aug 02 2002
Gives for n >= 1 the maximum absolute column sum norm of the inverse of the Vandermonde matrix (a_ij) i=0..n1, j=0..n1 with a_00=1 and a_ij=i^j for (i,j) != (0,0).  Torsten Muetze, Feb 06 2004
Number of left factors of Dyck paths, consisting of n steps. Example: a(4)=6 because we have UDUD, UDUU, UUDD, UUDU, UUUD and UUUU, where U=(1,1) and D=(1,1).  Emeric Deutsch, Apr 23 2005
Number of dispersed Dyck paths of length n; they are defined as concatenations of Dyck paths and (1,0)steps on the xaxis; equivalently, Motzkin paths with no (1,0)steps at positive height. Example: a(4)=6 because we have HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD, where U=(1,1), H=(1,0), and D=(1,1).  Emeric Deutsch, Jun 04 2011
a(n) is odd iff n=2^k1.  Jon Perry, May 05 2005
An inverse Chebyshev transform of binomial(1,n)=(1,1,0,0,0,...) where g(x)>(1/sqrt(14*x^2))*g(x*c(x^2)), with c(x) the g.f. of A000108.  Paul Barry, May 13 2005
In a random walk on the number line, starting at 0 and with 0 absorbing after the first step, number of ways of ending up at a positive integer after n steps.  Joshua Zucker, Jul 31 2005
Maximum number of sums of the form Sum_{i=1..n} e(i)*a(i) that are congruent to 0 mod q, where e_i=0 or 1 and gcd(a_i,q)=1, provided that q > ceiling(n/2).  Ralf Stephan, Apr 27 2003
Also the number of standard tableaux of height <= 2.  Mike Zabrocki, Mar 24 2007
A001263 * [1, 2, 3, 4, 5, ...] = [1, 1, 2, 3, 6, 10, 20, 35, 70, 126, ...].  Gary W. Adamson, Jan 02 2008
a(n) is also the number of distinct strings of length n, each of which is a prefix of a string of balanced parentheses; see example.  Lee A. Newberg, Apr 26 2010
Number of symmetric balanced strings of n pairs of parentheses; see example.  Joerg Arndt, Jul 25 2011
a(n) is the number of permutation patterns modulo 2.  Olivier Gérard, Feb 25 2011
For n >= 2, a(n1) is the number of incongruent twocolor bracelets of 2*n1 beads, n of which are black (A007123), having a diameter of symmetry.  Vladimir Shevelev, May 03 2011
The number of permutations of n elements where p(k2) < p(k) for all k.  Joerg Arndt, Jul 23 2011
Also size of the equivalence class of S_{n+1} containing the identity permutation under transformations of positionally adjacent elements of the form abc <> cba where a < b < c, cf. A210668.  Tom Roby, May 15 2012
a(n) is the number of symmetric Dyck paths of length 2n.  Matt Watson, Sep 26 2012
a(n) is the number of permutations of length n avoiding both 213 and 231 in the classical sense which are breadthfirst search reading words of increasing unarybinary trees. For more details, see the entry for permutations avoiding 231 at A245898.  Manda Riehl, Aug 05 2014
Number of symmetric standard Young tableaux of shape (n,n).  Ran Pan, Apr 10 2015
Also "stepped path" in the array formed by partial sums of the all 1's sequence (or a Pascal's triangle displayed as a square). Example:
[1], [1], 1, 1, 1, 1, 1, ... A000012
1, [2], [3], 4, 5, 6, 7, ...
1, 3, [6], [10], 15, 21, 28, ...
1, 4, 10, [20], [35], 56, 84, ...
1, 5, 15, 35, [70], [126], 210, ...
Sequences in second formula are the mixed diagonals shown in this array. (End)
Number of meanders (walks starting at the origin and ending at any altitude >= 0 that may touch but never go below the xaxis) with n steps from {1,1}.  David Nguyen, Dec 20 2016
a(n) is also the number of paths of n steps (either up or down by 1) that end at the maximal value achieved along the path.  Winston Luo, Jun 01 2017
Number of binary ntuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions.  Juan A. Olmos, Dec 21 2017
Equivalently, a(n) is the number of subsets of {1,...,n} containing as many even numbers as odd numbers.  Gus Wiseman, Mar 17 2018
a(n) is the number of Dyck paths with semilength = n+1, returns to the xaxis = floor((n+3)/2) and up movements in odd positions = floor((n+3)/2). Example: a(4)=6, U=up movement in odd position, u=up movement in even position, d=down movement, =return to xaxis: UududdUdUd, UdUuddUudd, UuddUuddUd, UdUdUududd, UuddUdUudd, UdUududdUd.  Roger Ford, Dec 29 2017
Let C_n(R, H) denote the transition matrix from the ribbon basis to the homogeneous basis of the graded component of the algebra of noncommutative symmetric functions of order n. Letting I(2^(n1)) denote the identity matrix of order 2^(n1), it has been conjectured that the dimension of the kernel of C_n(R, H)  I(2^(n1)) is always equal to a(n1).  John M. Campbell, Mar 30 2018
The number of Uequivalence classes of Łukasiewicz paths. Łukasiewicz paths are Uequivalent iff the positions of pattern U are identical in these paths.  Sergey Kirgizov, Apr 2018
All binary selfdual codes of length 2n, for n > 0, must contain at least a(n) codewords of weight n. More to the point, there will always be at least one, perhaps unique, binary selfdual code of length 2n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (n). This code can be constructed by direct summing the unique binary selfdual code of length 2 (up to permutation equivalence) to itself n times. A permutation equivalent code can be constructed by augmenting two identity matrices of length n together.  Nathan J. Russell, Nov 25 2018
The sequence starting (1, 2, 3, 6, ...) is the invert transform of A097331: (1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, ...).  Gary W. Adamson, Feb 22 2020
The sequence is the culminating limit of an infinite set of sequences with convergents of 2*cos(Pi/N), N = (3, 5, 7, 9, ...).
The first few such sequences are:
N = 3: (1, 1, 1, 1, 1, 1, 1, 1, ...)
N = 5: (1, 1, 2, 3, 5, 8, 13, 21, ...) = A000045
N = 7: (1, 1, 2, 3, 6, 10, 19, 33, ...) = A028495, a(n)/a(n1) tends to 1.801937...
N = 9 (1, 1, 2, 3, 6, 10, 20, 35, ...) = A061551, a(n)/a(n_1) tends to 1.879385...
...
In the limit one gets the current sequence with ratio 2. (End)
a(n) is also the number of monotone lattice paths from (0,0) to (floor(n/2),ceiling(n/2)). These are the number of Grand Dyck paths when n is even.  Nachum Dershowitz, Aug 12 2020
The maximum number of preimages that a permutation of length n+1 can have under the consecutive132avoiding stacksorting map.  Colin Defant, Aug 28 2020
Counts faro permutations of length n. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one.  Sergey Kirgizov, Jan 12 2021
Per "Sperner's Theorem", the largest possible familes of finite sets none of which contain any other sets in the family.  Renzo Benedetti, May 26 2021
a(n1) are the incomplete, primitive Dyck paths of n steps without a first return: paths of U and D steps starting at the origin, never touching the horizontal axis later on, and ending above the horizontal axis. n=1: {U}, n=2: {UU}, n=3: {UUU, UUD}, n=4: {UUUU, UUUD, UUDU}, n=5: {UUUUU, UUUUD, UUUDD, UUDUU, UUUDU, UUDUD}. For comparison: A037952 counts incomplete Dyck paths with n steps with any number of intermediate returns to the horizontal axis, ending above the horizontal axis.  R. J. Mathar, Sep 24 2021
a(n) is the number of noncrossing partitions of [n] whose nontrivial blocks are of type {a,b}, with a <= n/2, b > n/2.  Francesca Aicardi, May 29 2022


REFERENCES

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
M. Aigner and G. M. Ziegler, Proofs from The Book, SpringerVerlag, Berlin, 1999; see p. 135.
K. Engel, Sperner Theory, Camb. Univ. Press, 1997; Theorem 1.1.1.
P. Frankl, Extremal sets systems, Chap. 24 of R. L. Graham et al., eds, Handbook of Combinatorics, NorthHolland.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.16(b), p. 452.


LINKS

F. Disanto, A. Frosini, and S. Rinaldi, Square involutions, J. Int. Seq. 14 (2011) # 11.3.5.
Justine Falque, JeanChristophe Novelli, and JeanYves Thibon, Pinnacle sets revisited, arXiv:2106.05248 [math.CO], 2021.
F. Harary and R. W. Robinson, The number of achiral trees, Jnl. Reine Angewandte Mathematik, Vol. 278 (1975), pp. 322335. (Annotated scanned copy)
M. A. Narcowich, Problem 736, SIAM Review, Vol. 16, No. 1, 1974, p. 97.
P. J. Stockmeyer, The charm bracelet problem and its applications, pp. 339349 of Graphs and Combinatorics (Washington, Jun 1973), Ed. by R. A. Bari and F. Harary. Lect. Notes Math., Vol. 406. SpringerVerlag, 1974. [Scanned annotated and corrected copy]


FORMULA

a(n) = max_{k=0..n} binomial(n, k).
By symmetry, a(n) = binomial(n, ceiling(n/2)).  Labos Elemer, Mar 20 2003
Precursive with recurrence: a(0) = 1, a(1) = 1, and for n >= 2, (n+1)*a(n) = 2*a(n1) + 4*(n1)*a(n2).  Peter Bala, Feb 28 2011
G.f.: (1+x*c(x^2))/sqrt(14*x^2) = 1/(1  x  x^2*c(x^2)); where c(x) = g.f. for Catalan numbers A000108.
G.f.: (1 + 2*x + sqrt(14*x^2))/(2*x  4*x^2).  Lee A. Newberg, Apr 26 2010
G.f.: 1/(1  x  x^2/(1  x^2/(1  x^2/(1  x^2/(1  ... (continued fraction).  Paul Barry, Aug 12 2009
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = Sum_{k = 0..2*m} (1)^k*a(k)*a(2*mk).  Len Smiley, Dec 09 2001
The o.g.f. A(x) satisfies A(x) + x*A^2(x) = 1/(12*x).  Peter Bala, Feb 28 2011
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = 2*a(2*m)  c(m), where c(m)=A000108(m) are the Catalan numbers.  Christopher Hanusa (chanusa(AT)washington.edu), Nov 25 2003
a(n) = Sum_{k=0..n} (1)^k*2^(nk)*binomial(n, k)*A000108(k).  Paul Barry, Jan 27 2005
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(1, n2*k).  Paul Barry, May 13 2005
a(n) = Sum_{k=0..floor((n+1)/2)} (binomial(n+1, k)*(cos((n2*k+1)*Pi/2) + sin((n2*k+1)*Pi/2))).
a(n) = Sum_{k=0..n+1}, (binomial(n+1, (nk+1)/2)*(1(1)^(nk))*(cos(k*Pi/2) + sin(k*Pi))/2). (End)
a(n) = Sum_{k=floor(n/2)..n} (binomial(n,nk)  binomial(n,nk1)).  Paul Barry, Sep 06 2007
Inverse binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double inverse binomial transform of A001700. Row sums of triangle A132815.  Gary W. Adamson, Aug 31 2007
a(n) = Sum_{k = 0..floor(n/2)} (binomial(n,k)  binomial(n,k1)).  Nishant Doshi (doshinikki2004(AT)gmail.com), Apr 06 2009
Sum_{n>=0} a(n)/10^(n+1) = 0.1123724... = (sqrt(3)sqrt(2))/(2*sqrt(2)); Sum_{n>=0} a(n)/100^(n+1) = 0.0101020306102035... = (sqrt(51)sqrt(49))/(2*sqrt(49)).  Mark Dols, Jul 15 2010
Conjectured: a(n) = 2^n*2F1(1/2,n;2;2), useful for number of paths in 1d for which the coordinate is never negative.  Benjamin Phillabaum, Feb 20 2011
a(2*m+1) = (2*m+1)*a(2*m)/(m+1), e.g., a(7) = (7/4)*a(6) = (7/4)*20 = 35.  Jon Perry, Jan 20 2011
Let F(x) be the logarithmic derivative of the o.g.f. A(x). Then 1+x*F(x) is the o.g.f. for A027306.
Let G(x) be the logarithmic derivative of 1+x*A(x). Then x*G(x) is the o.g.f. for A058622. (End)
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal; and V = the vector [1,0,0,0,...]. a(n) = M^n*V, leftmost term.  Gary W. Adamson, Jun 13 2011
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal. a(n) = M^n_{1,1}.  Corrected by Gary W. Adamson, Jan 30 2012
a(n+2*p2) = Sum_{k=0..floor(n/2)} A009766(nk+p1, k+p1) + binomial(n+2*p2, p2), for p >= 1.  Johannes W. Meijer, Aug 02 2013
O.g.f.: (1x*c(x^2))/(12*x), with the o.g.f. c(x) of Catalan numbers A000108. See the rewritten formula given by Lee A. Newberg above. This is the o.g.f. for the row sums the Riordan triangle A053121.  Wolfdieter Lang, Sep 22 2013
a(2*k) = Sum_{i=0..k} binomial(k, i)*binomial(k, i), a(2*k+1) = Sum_{i=0..k} binomial(k+1, i)*binomial(k, i).  Juan A. Olmos, Dec 21 2017
a(0) = 1, a(n) = 2 * a(n1) for even n, a(n) = (2*n/(n+1)) * a(n1) for odd n.  James East, Sep 25 2019
Sum_{n>=0} 1/a(n) = 2*Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (1)^n/a(n) = 2/3  2*Pi/(9*sqrt(3)). (End)
For k>2, Sum_{n>=0} a(n)/k^n = (sqrt((k+2)/(k2))  1)*k/2.  Vaclav Kotesovec, May 13 2022
a(n) = Sum_{k = 0..n+1} (1)^(k+binomial(n+2,2)) * k/(n+1) * binomial(n+1,k)^2.
(n + 1)*(2*n  1)*a(n) = (1)^(n+1)*2*a(n1) + 4*(n  1)*(2*n + 1)*a(n2) with a(0) = a(1) = 1. (End)


EXAMPLE

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


MAPLE



MATHEMATICA

Table[DifferenceRoot[Function[{a, n}, {4 n a[n]2 a[1+n]+(2+n) a[2+n] == 0, a[1] == 1, a[2] == 1}]][n], {n, 30}] (* Luciano Ancora, Jul 08 2015 *)
Array[Binomial[#, Floor[#/2]]&, 40, 0] (* Harvey P. Dale, Mar 05 2018 *)


PROG

(PARI) a(n) = binomial(n, n\2);
(PARI) first(n) = x='x+O('x^n); Vec((1+2*x+sqrt(14*x^2))/(2*x4*x^2)) \\ Iain Fox, Dec 20 2017 (edited by Iain Fox, May 07 2018)
(Haskell)
(Maxima) A001405(n):=binomial(n, floor(n/2))$
(GAP) List([0..40], n>Binomial(n, Int(n/2))); # Muniru A Asiru, Apr 08 2018
(Python)
from math import comb


CROSSREFS

Row sums of Catalan triangle A053121 and of symmetric Dyck paths A088855.
Apparently a(n) = lim_{k>infinity} A094718(k, n).
Cf. A000984 gives the oddindexed terms of this sequence.
Cf. A000712, A001006, A001700, A005773, A005817, A007578, A007579, A022916, A022917 (permutation patterns mod k), A049401, A051920, A063886, A130820, A132815, A153585, A239241, A265848.


KEYWORD

nonn,easy,nice,core,walk


AUTHOR



STATUS

approved



