keyword:walk - cp's OEIS frontend

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

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

N. J. A. Sloane

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

			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

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CombOS - Combinatorial Object Server, Generate middle levels Gray codes
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Index entries for "core" sequences.

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.

List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018

a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011

a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];

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

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 *)

A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */

A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014

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

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

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

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

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

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

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

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

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Emanuel Sperner, Ein Satz über Untermengen einer endlichen Menge, Mathematische Zeitschrift (in German), Vol. 27, No. 1 (1928), pp. 544-548.
P. K. Stockmeyer, The charm bracelet problem and its applications, pp. 339-349 of Graphs and Combinatorics (Washington, Jun 1973), Ed. by R. A. Bari and F. Harary. Lect. Notes Math., Vol. 406. Springer-Verlag, 1974.
P. J. Stockmeyer, The charm bracelet problem and its applications, pp. 339-349 of Graphs and Combinatorics (Washington, Jun 1973), Ed. by R. A. Bari and F. Harary. Lect. Notes Math., Vol. 406. Springer-Verlag, 1974. [Scanned annotated and corrected copy]
I. Tasoulas, K. Manes, A. Sapounakis, and P. Tsikouras, Chains with Small Intervals in the Lattice of Binary Paths, arXiv:1911.10883 [math.CO], 2019.
C. G. Wagner, Letter to N. J. A. Sloane, Sep 30 1974.
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Quota System.
W. H. W. Wong and E. G. Tay, On Cross-intersecting Sperner Families, arXiv:2001.01910 [math.CO], 2020.
Index entries for sequences of k-nomial coefficients
Index entries for "core" sequences.

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

List([0..40],n->Binomial(n,Int(n/2))); # Muniru A Asiru, Apr 08 2018

a001405 n = a007318_row n !! (n `div` 2) -- Reinhard Zumkeller, Nov 09 2011

[Binomial(n, Floor(n/2)): n in [0..40]]; // Vincenzo Librandi, Nov 16 2014

A001405 := n->binomial(n, floor(n/2)): seq(A001405(n), n=0..33);

Table[Binomial[n, Floor[n/2]], {n, 0, 40}] (* Stefan Steinerberger, Apr 08 2006 *)
Table[DifferenceRoot[Function[{a,n},{-4 n a[n]-2 a[1+n]+(2+n) a[2+n] == 0,a[1] == 1,a[2] == 1}]][n], {n, 30}] (* Luciano Ancora, Jul 08 2015 *)
Array[Binomial[#,Floor[#/2]]&,40,0] (* Harvey P. Dale, Mar 05 2018 *)

A001405(n):=binomial(n,floor(n/2))$
makelist(A001405(n),n,0,30); /* Martin Ettl, Nov 01 2012 */

PARI
```
a(n) = binomial(n, n\2);
```

first(n) = x='x+O('x^n); Vec((-1+2*x+sqrt(1-4*x^2))/(2*x-4*x^2)) \\ Iain Fox, Dec 20 2017 (edited by Iain Fox, May 07 2018)

from math import comb
def A001405(n): return comb(n,n//2) # Chai Wah Wu, Jun 07 2022

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

A003946 Expansion of (1+x)/(1-3*x).

Original entry on oeis.org

1, 4, 12, 36, 108, 324, 972, 2916, 8748, 26244, 78732, 236196, 708588, 2125764, 6377292, 19131876, 57395628, 172186884, 516560652, 1549681956, 4649045868, 13947137604, 41841412812, 125524238436, 376572715308, 1129718145924

Offset: 0

N. J. A. Sloane

Coordination sequence for infinite tree with valency 4.
The n-th term of the coordination sequence of the infinite tree with valency 2m is the same as the number of reduced words of size n in the free group on m generators. In the five sequences A003946, A003948, A003950, A003952, A003954 m is 2, 3, 4, 5, 6. - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001 and Ola Veshta (olaveshta(AT)my-deja.com), Mar 30 2001
a(n) is the number of nonreversing random walks of the length of n edges on a two-dimensional square lattice, all beginning at a fixed point P. - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Apr 06 2005
Binomial transform of {1, 3, 5, 11, 21, 43, ...}, see A001045. Binomial transform is {1, 5, 21, 85, 341, 1365, ...}, see A002450. - Philippe Deléham, Jul 22 2005
For n >= 2, a(n) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3} such that for fixed, different x_1, x_2 in {1,2,...,n} and fixed y_1, y_2 in {1,2,3} we have f(x_1) <> y_1 and f(x_2) <> y_2. - Milan Janjic, Apr 19 2007
Equals row sums of triangle A143865. - Gary W. Adamson, Sep 04 2008
Equals INVERT transform of the odd integers = 1/(1 - x - 3x^2 - 5x^3 - ...). - Gary W. Adamson, Jul 27 2009
a(n) is the number of generalized compositions of n+1 when there are 2 *i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Aug 26 2010
Number of length-n strings of 4 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - Joerg Arndt, Oct 11 2012
The sequence is the INVERTi transform of A015448: (1, 5, 21, 89, 377, ...). - Gary W. Adamson, Aug 06 2016
Let D(m) = {d(m,i)}, i = 1..q, denote the set of the q divisors of a number m, and consider s1(m) and s2(m) the sums of the divisors that are congruent to 1 and 2 (mod 3) respectively. For n > 0, the sequence a(n) lists the numbers m such that s1(m) = 5 and s2(m) = 2. - Michel Lagneau, Feb 09 2017
a(n) is the number of quaternary sequences of length n such that no two consecutive terms have distance 2. - David Nacin, May 31 2017
Also the number of maximal cliques in the n-Sierpinski gasket graph. - Eric W. Weisstein, Dec 01 2017
Number of 3-permutations of n elements avoiding the patterns 231, 321. See Bonichon and Sun. - Michel Marcus, Aug 19 2022

			G.f. = 1 + 4*x + 12*x^2 + 36*x^3 + 108*x^4 + 324*x^5 + 972*x^6 + 2916*x^7 + ...

T. D. Noe, Table of n, a(n) for n = 0..200
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, (an + b)-color compositions, arXiv:1707.07798 [math.CO], 2017.
Nicolas Bonichon and Pierre-Jean Morel, Baxter d-permutations and other pattern avoiding classes, arXiv:2202.12677 [math.CO], 2022.
D. J. Broadhurst, On the enumeration of irreducible k-fold Euler sums and their roles in knot theory and field theory, arXiv:hep-th/9604128, 1996.
John Elias, Illustration: Sierpinski Hexagrams
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 305
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Amya Luo, Pattern Avoidance in Nonnesting Permutations, Undergraduate Thesis, Dartmouth College (2024). See p. 3.
A. M. Nemirovsky et al., Marriage of exact enumeration and 1/d expansion methods: lattice model of dilute polymers, J. Statist. Phys., 67 (1992), 1083-1108.
Nathan Sun, On d-permutations and Pattern Avoidance Classes, arXiv:2208.08506 [math.CO], 2022.
Eric Weisstein's World of Mathematics, Maximal Clique.
Eric Weisstein's World of Mathematics, Sierpiński Gasket Graph.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (3).
Index entries for sequences related to trees

Cf. A029653, A143865, column 4 in A265583, A015448.

[1] cat [4*3^(n-1): n in [1..25]]; // Vincenzo Librandi, Dec 11 2012

Maple
```
if n = 0 then 1 else 4*3^(n-1); fi;
```

Join[{1}, 4 3^Range[0, 30]] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2009 *)
Join[{1}, NestList[3# &, 4, 30]] (* Harvey P. Dale, Nov 30 2011 *)
CoefficientList[Series[(1 + x)/(1 - 3 x), {x, 0, 30}], x] (* Vincenzo Librandi, Dec 11 2012 *)
Join[{1}, LinearRecurrence[{3}, {4}, 20]] (* Eric W. Weisstein, Dec 01 2017 *)

A003946[n]:=if n<1 then 1 else 4*3^(n-1)$
makelist(A003946[n],n,0,30); /* Martin Ettl, Oct 29 2012 */

{a(n) = if( n<1, n==0, 4 * 3^(n-1))}; /* Michael Somos, Jun 18 2002 */

Vec((1+x)/(1-3*x) + O(x^100)) \\ Altug Alkan, Dec 07 2015

a(n) = floor(4*3^(n-1)). - Michael Somos, Jun 18 2002
a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 2. - Philippe Deléham, Jul 10 2005
The Hankel transform of this sequence is [1,-4,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
a(n + 1) = (((1 + sqrt(-11))/2)^n + ((1 - sqrt(-11))/2)^n)^2 - (((1 + sqrt(-11))/2)^n - ((1 - sqrt(-11))/2)^n)^2. - Raphie Frank, Dec 07 2015
From Mario C. Enriquez, Apr 01 2017: (Start)
(L(a(n+k)) - 1)/a(n) reduces to the form C/a(n-1), where n > 1, k >= 0, L(a(n)) is the a(n)-th Lucas number and C = (L(a(n+k)) - 1)/3.
(L(a(n+k)) - 1)/3 mod (L(a(n)) - 1)/3 = (L(a(n)) - 1)/3 - 1, where n >= 1, k >= 0 and L(a(n)) is the a(n)-th Lucas number. (End)
E.g.f.: (4*exp(3*x) - 1)/3. - Stefano Spezia, Jan 31 2025

Additional comments from Michael Somos, Jun 18 2002

Edited by N. J. A. Sloane, Dec 04 2009

A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, 44287, 132860, 398581, 1195742, 3587227, 10761680, 32285041, 96855122, 290565367, 871696100, 2615088301, 7845264902, 23535794707, 70607384120, 211822152361, 635466457082

Offset: 0

Olivier Gérard

Number of walks of length n between any two distinct vertices of the complete graph K_4. - Paul Barry and Emeric Deutsch, Apr 01 2004
For n >= 1, a(n) is the number of integers k, 1 <= k <= 3^(n-1), whose ternary representation ends in an even number of zeros (see A007417). - Philippe Deléham, Mar 31 2004
Form the digraph with matrix A=[0,1,1,1;1,0,1,1;1,1,0,1;1,0,1,1]. A015518(n) corresponds to the (1,3) term of A^n. - Paul Barry, Oct 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
(A046717(n))^2 + (2*a(n))^2 = A046717(2n). E.g., A046717(3) = 13, 2*a(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
For n >= 2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and three types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are three types of twins; namely, both F, both M, or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=7, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F twins; (6) M,M twins; or (7) M,F twins (emphasizing that birth order is irrelevant here when both/all children are the same gender and when two children are within the same pair of twins). - Rick L. Shepherd, Sep 18 2004
a(n) is prime for n = {2, 3, 5, 7, 13, 23, 43, 281, 359, ...}, where only a(2) = 2 corresponds to a prime of the form (3^k - 1)/4. All prime terms, except a(2) = 2, are the primes of the form (3^k + 1)/4. Numbers k such that (3^k + 1)/4 is prime are listed in A007658. Note that all prime terms have prime indices. Prime terms are listed in A111010. - Alexander Adamchuk, Nov 19 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 26 2010
Select an odd size subset S from {1,2,...,n}, then select an even size subset from S. - Geoffrey Critzer, Mar 02 2010
a(n) is the number of ternary sequences of length n where the numbers of (0's, 1's) are (even, odd) respectively, and, by symmetry, the number of such sequences where those numbers are (odd, even) respectively. A122983 covers (even, even), and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
An elephant sequence, see A175654. For the corner squares just one A[5] vector, with decimal value 341, leads to this sequence (without the leading 0). For the central square this vector leads to the companion sequence A046717 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Let R be the commutative algebra resulting from adjoining the elements of the Klein four-group to the integers (equivalently, K = Z[x,y,z]/{x*y - z, y*z - x, x*z - y, x^2 - 1, y^2 - 1, z^2 - 1}). Then a(n) is equal to the coefficients of x, y, and z in the expansion of (x + y + z)^n. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 06 2010
Pisano period lengths:  1, 2, 2, 4, 4, 2, 6, 8, 2, 4, 10, 4, 6, 6, 4, 16, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is a divisibility sequence, also the values of Chebyshev polynomials, and also the number of ways of packing a 2 X n-1 rectangle with dominoes and unit squares. - R. K. Guy, Dec 16 2016
For n>0, gcd(a(n),a(n+1))=1. - Kengbo Lu, Jul 02 2020

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, and Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
Katerina Böhmová, Cristina Dalfó, and Clemens Huemer, On cyclic Kautz digraphs, Research Report, 2015.
Garry Bowlin and Matthew G. Brin, Coloring Planar Graphs via Colored Paths in the Associahedra, arXiv preprint arXiv:1301.3984 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 12 2013
AJ Bu and Doron Zeilberger, Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas, arXiv:2305.09030 [math.CO], 2023.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Sergio Falcón, Binomial Transform of the Generalized k-Fibonacci Numbers, Communications in Mathematics and Applications (2019) Vol. 10, No. 3, 643-651.
Dale Gerdemann, Fractal generated from (2,3) recursion, YouTube Video, Dec 05 2014.
Toufik Mansour and Mark Shattuck, Pattern avoidance in flattened derangements, Disc. Math. Lett. (2025) Vol. 15, 67-74. See p. 74.
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 2.
Felix Pozon Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014.
Index entries for linear recurrences with constant coefficients, signature (2,3).
Index entries for sequences related to Chebyshev polynomials.

a(n) = A080926(n-1) + 1 = (1/3)*A054878(n+1) = (1/3)*abs(A084567(n+1)).
First differences of A033113 and A039300.
Partial sums of A046717.
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Cf. A046717.
Cf. A007658, A111010.
Cf. A001045, A078008, A097073, A115341. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Cf. A007482, A059260.

[Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)

a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */

PARI
```
a(n)=round(3^n/4)
```

for n in range(0, 20): print(int((3**n-(-1)**n)/4), end=', ') # Stefano Spezia, Nov 30 2018

Sage
```
[round(3^n/4) for n in range(0,27)]
```

G.f.: x/((1+x)*(1-3*x)).
a(n) = (3^n - (-1)^n)/4 = floor(3^n/4 + 1/2).
a(n) = 3^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
E.g.f.: (exp(3*x) - exp(-x))/4. Second inverse binomial transform of (5^n-1)/4, A003463. Inverse binomial transform for powers of 4, A000302 (when preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)*2^(2k). - Paul Barry, May 14 2003
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*4^(k-1). - Paul Barry, Apr 02 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2*k)*3^k. - Paul Barry, Jul 13 2004
a(n) = U(n-1, i/sqrt(3))(-i*sqrt(3))^(n-1), i^2=-1. - Paul Barry, Nov 17 2003
G.f.: x*(1+x)^2/(1 - 6*x^2 - 8*x^3 - 3*x^4) = x(1+x)^2/characteristic polynomial(x^4*adj(K_4)(1/x)). - Paul Barry, Feb 03 2004
a(n) = sum_{k=0..3^(n-1)} A014578(k) = -(-1)^n*A014983(n) = A051068(3^(n-1)), for n > 0. - Philippe Deléham, Mar 31 2004
E.g.f.: exp(x)*sinh(2*x)/2. - Paul Barry, Oct 02 2004
a(2*n+1) = A054880(n) + 1. - M. F. Hasler, Mar 20 2008
2*a(n) + (-1)^n = A046717(n). - M. F. Hasler, Mar 20 2008
a(n) = ((1+sqrt(4))^n - (1-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = abs(A014983(n)). - Zerinvary Lajos, May 28 2009
a(n) = round(3^n/4). - Mircea Merca, Dec 28 2010
a(n) = Sum_{k=1,3,5,...} binomial(n,k)*2^(k-1). - Geoffrey Critzer, Mar 02 2010
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
G.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - 1/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + 1/G(k+1)))))); (continued fraction).
E.g.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + (2*k+2)/G(k+1)))))); (continued fraction).  (End)
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n+1) = Sum_{k = 0..n} A238801(n,k)*2^k. - Philippe Deléham, Mar 07 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-4)^k = (-1)^(n-1)*Sum_{k=0..n-1} (-3)^k. Equals (-1)^(n-1)*Phi(n,-3), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n) = 2*A006342(n-1) - n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Nov 30 2018
a(n) = 2*A033113(n-2) + n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Aug 16 2019
a(2*k) = 2*A002452(k), a(2*k+1) = A066443(k). - Yuchun Ji, Aug 14 2019
a(n+1) = 2*Sum_{k=0..n} a(k) if n odd, and 1 + 2*Sum_{k=0..n} a(k) if n even. - Kengbo Lu, May 30 2020
a(n) = F(n) + Sum_{k=1..(n-1)} a(k)*L(n-k), for F(n) and L(n) the Fibonacci and Lucas numbers. - Kengbo Lu and Greg Dresden, Jun 05 2020
From Kengbo Lu, Jun 11 2020: (Start)
a(n) = A002605(n) + Sum_{k = 1..n-2} a(k)*A002605(n-k-1).
a(n) = A006130(n-1) + Sum_{k = 1..n-1} a(k)*A006130(n-k-1). (End)
a(2n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)* 2^(2n-2i-2j-1)* 3^(i+j). - Kengbo Lu, Jul 02 2020
a(n) = 3*a(n-1) - (-1)^n. - Dimitri Papadopoulos, Nov 28 2023
G.f.: x/((1 + x)*(1 - 3*x)) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 3*x + 1)(1 + k*x) (a telescoping series). Cf. A007482. - Peter Bala, May 08 2024
From Peter Bala, Jun 29 2025: (Start)
For n >= 1, a(n+1) = 2^n * hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -3).
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) =  x + 2*x^2 + 7*x^3 + 20*x^4 + ....
sqrt(A(x)/x) is the g.f. of A002426.
The following series telescope:
Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)) = -1; Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = -1/98.
In general, for k >= 0, Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*...*a(n+2*k+1)) = -1/((a(1)*a(2)*...*a(2*k+1))*a(2*k+1)).
Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)) = 1/4; Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)* a(n+3)*a(n+4)) = 1/5600.
In general, for k >= 1, Sum_{n >= 1} 3^n/(a(n)*a(n+1)*...*a(n+2*k)) = 1/((a(1)*a(2)*...*a(2*k))*a(2*k)). (End)

More terms from Emeric Deutsch, Apr 01 2004

Edited by Ralf Stephan, Aug 30 2004

A002893 a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(2*k,k).

Original entry on oeis.org

1, 3, 15, 93, 639, 4653, 35169, 272835, 2157759, 17319837, 140668065, 1153462995, 9533639025, 79326566595, 663835030335, 5582724468093, 47152425626559, 399769750195965, 3400775573443089, 29016970072920387, 248256043372999089

Offset: 0

N. J. A. Sloane

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
a(n) is the 2n-th moment of the distance from the origin of a 3-step random walk in the plane. - Peter M. W. Gill (peter.gill(AT)nott.ac.uk), Feb 27 2004
a(n) is the number of Abelian squares of length 2n over a 3-letter alphabet. - Jeffrey Shallit, Aug 17 2010
Consider 2D simple random walk on honeycomb lattice. a(n) gives number of paths of length 2n ending at origin. - Sergey Perepechko, Feb 16 2011
Row sums of A318397 the square of A008459. - Peter Bala, Mar 05 2013
Conjecture: For each n=1,2,3,... the polynomial g_n(x) = Sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
a(n) is the sum of the squares of the coefficients of (x + y + z)^n. - Michael Somos, Aug 25 2018
a(n) is the constant term in the expansion of (1 + (1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 28 2019

			G.f.: A(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 639*x^4 + 4653*x^5 + 35169*x^6 + ...
G.f.: A(x) = 1/(1-3*x) + 6*x^2*(1-x)/(1-3*x)^4 + 90*x^4*(1-x)^2/(1-3*x)^7 + 1680*x^6*(1-x)^3/(1-3*x)^10 + 34650*x^8*(1-x)^4/(1-3*x)^13 + ... - _Paul D. Hanna_, Feb 26 2012

Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
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).

Seiichi Manyama, Table of n, a(n) for n = 0..1051 (terms 0..100 from T. D. Noe)
B. Adamczewski, J. P. Bell, and E. Delaygue, Algebraic independence of G-functions and congruences à la Lucas", arXiv preprint arXiv:1603.04187 [math.NT], 2016.
Rostislav Akhmechet, Mikhail Khovanov, and Melissa Zhang, Annular SL(2) and SL(3) web algebras, arXiv:2508.05605 [math.GT], 2025. See p. 48.
Gert Almkvist, The art of finding Calabi-Yau differential equations, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
David H. Bailey, Jonathan M. Borwein, David Broadhurst and M. L. Glasser, Elliptic integral evaluations of Bessel moments, arXiv:0801.0891 [hep-th], 2008.
P. Barrucand, A combinatorial identity, Problem 75-4, SIAM Rev., 17 (1975), 168. Solution by D. R. Breach, D. McCarthy, D. Monk and P. E. O'Neil, SIAM Rev. 18 (1976), 303.
P. Barrucand, Problem 75-4, A Combinatorial Identity, SIAM Rev., 17 (1975), 168. [Annotated scanned copy of statement of problem]
Arnaud Beauville, Les familles stables de courbes sur P_1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, no. 294, May 24 1982.
Frits Beukers and Jan Stienstra, On the Picard-Fuchs Equation and the Formal Brauer Group of Certain Elliptic K3-Surfaces, Mathematische Annalen (1985), Vol. 271, pp. 269-304 (see Part III).
Artur Bille, Victor Buchstaber, Simon Coste, Satoshi Kuriki, and Evgeny Spodarev, Random eigenvalues of graphenes and the triangulation of plane, arXiv:2306.01462 [math.SP], 2023.
Jonathan M. Borwein, A short walk can be beautiful, 2015.
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, 2010.
Jonathan M. Borwein and Armin Straub, Mahler measures, short walks and log-sine integrals.
Jonathan M. Borwein, Armin Straub and Christophe Vignat, Densities of short uniform random walks, Part II: Higher dimensions, Preprint, 2015
Jonathan M. Borwein, Armin Straub and James Wan, Three-Step and Four-Step Random Walk Integrals, Exper. Math., 22 (2013), 1-14.
Charles Burnette and Chung Wong, Abelian Squares and Their Progenies, arXiv:1609.05580 [math.CO], 2016.
David Callan, A combinatorial interpretation for an identity of Barrucand, JIS 11 (2008) 08.3.4.
Shaun Cooper, Apéry-like sequences defined by four-term recurrence relations, arXiv:2302.00757 [math.NT], 2023.
M. Coster, Email, Nov 1990
Eric Delaygue, Arithmetic properties of Apery-like numbers, arXiv preprint arXiv:1310.4131 [math.NT], 2013.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Jeffrey S. Geronimo, Hugo J. Woerdeman, and Chung Y. Wong, The autoregressive filter problem for multivariable degree one symmetric polynomials, arXiv:2101.00525 [math.CA], 2021.
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See C p. 2.
Victor J. W. Guo, Proof of two conjectures of Z.-W. Sun on congruences for Franel numbers, arXiv preprint arXiv:1201.0617 [math.NT], 2012.
Victor J. W. Guo, Guo-Shuai Mao and Hao Pan, Proof of a conjecture involving Sun polynomials, arXiv preprint arXiv:1511.04005 [math.NT], 2015.
E. Hallouin and M. Perret, A Graph Aided Strategy to Produce Good Recursive Towers over Finite Fields, arXiv preprint arXiv:1503.06591 [math.NT], 2015.
J. A. Hendrickson, Jr., On the enumeration of rectangular (0,1)-matrices, Journal of Statistical Computation and Simulation, 51 (1995), 291-313.
S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
Pakawut Jiradilok and Elchanan Mossel, Gaussian Broadcast on Grids, arXiv:2402.11990 [cs.IT], 2024. See p. 27.
Davidson Noby Joseph and Igor Boettcher, Walking on Archimedean Lattices: Insights from Bloch Band Theory, arXiv:2507.12662 [cond-mat.stat-mech], 2025. See p. 18.
Tanya Khovanova and Konstantin Knop, Coins of three different weights, arXiv:1409.0250 [math.HO], 2014.
Murray S. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 148-149.
Bradley Klee, Checking Weierstrass data, 2023.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5
Mathematics Stack Exchange, sum involving the product of binomial coefficients, Nov 10 2016.
L. B. Richmond and Jeffrey Shallit, Counting abelian squares, Electronic J. Combinatorics 16 (1), #R72, June 2009. [From _Jeffrey Shallit_, Aug 17 2010]
Armin Straub, Arithmetic aspects of random walks and methods in definite integration, Ph. D. Dissertation, School Of Science And Engineering, Tulane University, 2012. - From _N. J. A. Sloane_, Dec 16 2012
Zhi-Hong Sun, Congruences for Apéry-like numbers, arXiv:1803.10051 [math.NT], 2018.
Zhi-Hong Sun, New congruences involving Apéry-like numbers, arXiv:2004.07172 [math.NT], 2020.
Zhi-Wei Sun, Connections between p = x^2+3y^2 and Franel numbers, J. Number Theory 133(2013), 2919-2928.
Zhi-Wei Sun, Congruences involving g_n(x) = Sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k, Ramanujan J., in press. Doi: 10.1007/s11139-015-9727-3.
Brani Vidakovic, All roads lead to Rome--even in the honeycomb world, Amer. Statist., 48 (1994) no. 3, 234-236.
Yi Wang and Bao-Xuan Zhu, Proofs of some conjectures on monotonicity of number-theoretic and combinatorial sequences, arXiv preprint arXiv:1303.5595 [math.CO], 2013.
D. Zagier, Integral solutions of Apery-like recurrence equations. See line C in sporadic solutions table of page 5.

Cf. A000172, A002895, A000984, A006480, A087457, A274600, A318397, A244973.
Cf. A169714, A169715.
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.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

[&+[Binomial(n, k)^2 * Binomial(2*k, k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 26 2018

series(1/GaussAGM(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3)), x=0, 42) # Gheorghe Coserea, Aug 17 2016
A002893 := n -> hypergeom([1/2, -n, -n], [1, 1], 4):
seq(simplify(A002893(n)), n=0..20); # Peter Luschny, May 23 2017

Table[Sum[Binomial[n,k]^2 Binomial[2k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Aug 19 2011 *)
a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {1/2, -n, -n}, {1, 1}, 4]]; (* Michael Somos, Oct 16 2013 *)
a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^3, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 30 2013 *)
a[ n_] := If[ n < 0, 0, Block[ {x, y, z},  Expand[(x + y + z)^n] /. {t_Integer -> t^2, x -> 1, y -> 1, z -> 1}]]; (* Michael Somos, Aug 25 2018 *)

{a(n) = if( n<0, 0, n!^2 * polcoeff( besseli(0, 2*x + O(x^(2*n+1)))^3, 2*n))};

{a(n) = sum(k=0, n, binomial(n, k)^2 * binomial(2*k, k))}; /* Michael Somos, Jul 25 2007 */

{a(n)=polcoeff(sum(m=0,n, (3*m)!/m!^3 * x^(2*m)*(1-x)^m / (1-3*x+x*O(x^n))^(3*m+1)),n)} \\ Paul D. Hanna, Feb 26 2012

N = 42; x='x + O('x^N); v = Vec(1/agm(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3))); vector((#v+1)\2, k, v[2*k-1])  \\ Gheorghe Coserea, Aug 17 2016

def A002893(n): return simplify(hypergeometric([1/2,-n,-n], [1,1], 4))
[A002893(n) for n in range(31)] # G. C. Greubel, Jan 21 2023

a(n) = Sum_{m=0..n} binomial(n, m) * A000172(m). [Barrucand]
D-finite with recurrence: (n+1)^2 a(n+1) = (10*n^2+10*n+3) * a(n) - 9*n^2 * a(n-1). - Matthijs Coster, Apr 28 2004
Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0, 2*sqrt(x))^3. - Vladeta Jovovic, Mar 11 2003
a(n) = Sum_{p+q+r=n} (n!/(p!*q!*r!))^2 with p, q, r >= 0. - Michael Somos, Jul 25 2007
a(n) = 3*A087457(n) for n>0. - Philippe Deléham, Sep 14 2008
a(n) = hypergeom([1/2, -n, -n], [1, 1], 4). - Mark van Hoeij, Jun 02 2010
G.f.: 2*sqrt(2)/Pi/sqrt(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z))) *  EllipticK(8*z^(3/2)/(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z)))). - Sergey Perepechko, Feb 16 2011
G.f.: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)*(1-x)^n / (1-3*x)^(3*n+1). - Paul D. Hanna, Feb 26 2012
Asymptotic: a(n) ~ 3^(2*n+3/2)/(4*Pi*n). - Vaclav Kotesovec, Sep 11 2012
G.f.: 1/(1-3*x)*(1-6*x^2*(1-x)/(Q(0)+6*x^2*(1-x))), where Q(k) = (54*x^3 - 54*x^2 + 9*x -1)*k^2 + (81*x^3 - 81*x^2 + 18*x -2)*k + 33*x^3 - 33*x^2 +9*x - 1 - 3*x^2*(1-x)*(1-3*x)^3*(k+1)^2*(3*k+4)*(3*k+5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/(2*(1-9*x)^(2/3)), where G(k) = 1 + 1/(1 - 3*(3*k+1)^2*x*(1-x)^2/(3*(3*k+1)^2*x*(1-x)^2 - (k+1)^2*(1-9*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013
a(n) = [x^(2n)] 1/agm(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3)). - Gheorghe Coserea, Aug 17 2016
0 = +a(n)*(+a(n+1)*(+729*a(n+2) -1539*a(n+3) +243*a(n+4)) +a(n+2)*(-567*a(n+2) +1665*a(n+3) -297*a(n+4)) +a(n+3)*(-117*a(n+3) +27*a(n+4))) +a(n+1)*(+a(n+1)*(-324*a(n+2) +720*a(n+3) -117*a(n+4)) +a(n+2)*(+315*a(n+2) -1000*a(n+3) +185*a(n+4)) +a(n+3)*(+80*a(n+3) -19*a(n+4))) +a(n+2)*(+a(n+2)*(-9*a(n+2) +35*a(n+3) -7*a(n+4)) +a(n+3)*(-4*a(n+3) +a(n+4))) for all n in Z. - Michael Somos, Oct 30 2017
G.f. y=A(x) satisfies: 0 = x*(x - 1)*(9*x - 1)*y'' + (27*x^2 - 20*x + 1)*y' + 3*(3*x - 1)*y. - Gheorghe Coserea, Jul 01 2018
Sum_{k>=0} binomial(2*k,k) * a(k) / 6^(2*k) = A086231 = (sqrt(3)-1) * (Gamma(1/24) * Gamma(11/24))^2 / (32*Pi^3). - Vaclav Kotesovec, Apr 23 2023
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=3*(-1 + 3*x)*T(x) + (1 - 20*x + 27*x^2)*T'(x) + x*(-1 + x)*(-1 + 9*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (3/64)*(1 + 3*x)*(1 - 15*x + 75*x^2 + 3*x^3);
g3 = -(1/512)*(-1 + 6*x + 3*x^2)*(1 - 12*x + 30*x^2 - 540*x^3 + 9*x^4);
which determine an elliptic surface with four singular fibers. (End)
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(3*k, 2*n) (Almkvist, p. 16). - Peter Bala, May 22 2025

A001411 Number of n-step self-avoiding walks on square lattice.

Original entry on oeis.org

1, 4, 12, 36, 100, 284, 780, 2172, 5916, 16268, 44100, 120292, 324932, 881500, 2374444, 6416596, 17245332, 46466676, 124658732, 335116620, 897697164, 2408806028, 6444560484, 17266613812, 46146397316, 123481354908, 329712786220, 881317491628

Offset: 0

N. J. A. Sloane, Anthony Guttmann

B. Bollobas and O. Riordan, Percolation, Cambridge, 2006, see p. 15.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, section 5.10, pp. 331-338.
B. D. Hughes, Random Walks and Random Environments, Oxford 1995, vol. 1, p. 461.
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).

N. J. A. Sloane, Hugo Pfoertner, Table of n, a(n) for n = 0..79 (from the Jensen links below)
H. Bottomley, Illustration of initial terms
A. R. Conway et al., Algebraic techniques for enumerating self-avoiding walks on the square lattice, J. Phys A 26 (1993) 1519-1534.
A. R. Conway et al., Algebraic techniques for enumerating self-avoiding walks on the square lattice, arXiv:hep-lat/9211062, 1992.
Steven R. Finch, Self-Avoiding-Walk Connective Constants
M. E. Fisher and M. F. Sykes, Excluded-volume problem and the Ising model of ferromagnetism, Phys. Rev. 114 (1959), 45-58.
A. J. Guttmann, On the critical behavior of self-avoiding walks, J. Phys. A 20 (1987), 1839-1854.
A. J. Guttmann and A. R. Conway, Self-Avoiding Walks and Polygons, Annals of Combinatorics 5 (2001) 319-345.
B. J. Hiley and M. F. Sykes, Probability of initial ring closure in the restricted random-walk model of a macromolecule, J. Chem. Phys., 34 (1961), 1531-1537.
I. Jensen, Series Expansions for Self-Avoiding Walks
Iwan Jensen, A new transfer-matrix algorithm for exact enumerations: self-avoiding walks on the square lattice, arXiv:1309.6709 [math-ph], 26 Sep 2013.
D. Randall, Counting in Lattices: Combinatorial Problems from Statistical Mechanics, PhD Thesis.
D. Randall, Counting in Lattices: Combinatorial Problems from Statistical Mechanics, PhD Thesis, 1994.
G. Slade, Self-avoiding walks, Math. Intelligencer, 16 (No. 1, 1994), 29-35.
M. F. Sykes, Some counting theorems in the theory of the Ising problem and the excluded volume problem, J. Math. Phys., 2 (1961), 52-62.
M. F. Sykes et al., The asymptotic behavior of selfavoiding walks and returns on a lattice, J. Phys. A 5 (1972), 653-660.

Twice A002900.

noloop:=X->evalb(nops(X)=nops({op(X)})):
extend:=proc(L) local L1,U,X,res:
   U:=[[1,0],[0,1],[-1,0],[0,-1]]:
   res:=NULL:for X in U do L1:=[op(L),L[nops(L)]+X]:
   if noloop(L1) then res:=res,L1 fi od:
   return(res) end:
walks:={[[0,0]]}: A001411:=1:
to 12 do walks:=map(x->extend(x),walks): A001411:=A001411,nops(walks) od:
[A001411];
# Robert FERREOL, Mar 29 2019

mo=Tuples[{-1,1}, 2]; a[0]=1; a[tg_, p_:{{0,0}}] := Block[{e, mv = Complement[Last[p]+# & /@ mo, p]}, If[tg == 1, Length@mv, Sum[a[tg-1, Append[p, e]], {e, mv}]]]; a /@ Range[0, 10] (* Giovanni Resta, May 06 2016 *)

def add(L,x):
    M=[y for y in L];M.append(x)
    return(M)
plus=lambda L,M : [x+y for x,y in zip(L,M)]
mo=[[1,0],[0,1],[-1,0],[0,-1]]
def a(n,P=[[0,0]]):
    if n==0: return(1)
    mv1=[plus(P[-1],x) for x in mo]
    mv2=[x for x in mv1 if x not in P]
    if n==1: return(len(mv2))
    else: return(sum(a(n-1,add(P,x)) for x in mv2))
[a(n) for n in range(11)]
# Robert FERREOL, Nov 30 2018; after the Mathematica program.

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

1, 1, 3, 4, 11, 16, 42, 64, 163, 256, 638, 1024, 2510, 4096, 9908, 16384, 39203, 65536, 155382, 262144, 616666, 1048576, 2449868, 4194304, 9740686, 16777216, 38754732, 67108864, 154276028, 268435456, 614429672, 1073741824, 2448023843

Offset: 0

Clark Kimberling

Inverse binomial transform of A027914. Hankel transform (see A001906 for definition) is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 21 2005
Number of walks of length n on a line that starts at the origin and ends at or above 0. - Benjamin Phillabaum, Mar 05 2011
Number of binary integers (i.e., with a leading 1 bit) of length n+1 which have a majority of 1-bits. E.g., for n+1=4: (1011, 1101, 1110, 1111) a(3)=4. - Toby Gottfried, Dec 11 2011
Number of distinct symmetric staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum > 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A345917. For example, the a(0) = 1 through a(4) = 11 compositions are:
  (1)  (2)  (3)    (4)    (5)
            (21)   (31)   (32)
            (111)  (112)  (41)
                   (211)  (113)
                          (122)
                          (212)
                          (221)
                          (311)
                          (1121)
                          (2111)
                          (11111)
The following relate to these compositions:
- The unordered version is A027193.
- The complement is counted by A058622.
- The reverse unordered version is A086543.
- The version for alternating sum >= 0 is A116406.
- The version for alternating sum < 0 is A294175.
- Ranked by A345917. (End)
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

			From _Gus Wiseman_, Aug 20 2021: (Start)
The a(0) = 1 through a(4) = 11 binary numbers with a majority of 1-bits (Gottfried's comment) are:
  1   11   101   1011   10011
           110   1101   10101
           111   1110   10110
                 1111   10111
                        11001
                        11010
                        11011
                        11100
                        11101
                        11110
                        11111
The version allowing an initial zero is A058622.
(End)

A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.1.6)

Alois P. Heinz, Table of n, a(n) for n = 0..1000
F. Disanto, A. Frosini, and S. Rinaldi, Square involutions, J. Int. Seq. 14 (2011) # 11.3.5.
Zachary Hamaker and Eric Marberg, Atoms for signed permutations, arXiv:1802.09805 [math.CO], 2018.
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.

a(n) = Sum{(k+1)T(n, m-k)}, 0<=k<=[ (n+1)/2 ], T given by A008315.
Column k=2 of A226873. - Alois P. Heinz, Jun 21 2013
Cf. A008949, A248574, A001405, A246437.
The even bisection is A000302.
The odd bisection appears to be A032443.
Cf. A000984, A001700, A001791, A008549, A011782, A088218, A097805, A163493, A182616, A345197.

List([0..35],n->Sum([0..Int(n/2)],k->Binomial(n,k))); # Muniru A Asiru, Nov 27 2018

a027306 n = a008949 n (n `div` 2)  -- Reinhard Zumkeller, Nov 14 2014

[2^(n-1)+(1+(-1)^n)/4*Binomial(n, n div 2): n in [0..40]]; // Vincenzo Librandi, Jun 19 2016

a:= proc(n) add(binomial(n, j), j=0..n/2) end:
seq(a(n), n=0..32); # Zerinvary Lajos, Mar 29 2009

Table[Sum[Binomial[n, k], {k, 0, Floor[n/2]}], {n, 1, 35}]
(* Second program: *)
a[0] = a[1] = 1; a[2] = 3; a[n_] := a[n] = (2(n-1)(2a[n-2] + a[n-1]) - 8(n-2) a[n-3])/n; Array[a, 33, 0] (* Jean-François Alcover, Sep 04 2016 *)

a(n)=if(n<0,0,(2^n+if(n%2,0,binomial(n, n/2)))/2)

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k).
Odd terms are 2^(n-1). Also a(2n) - 2^(2n-1) is given by A001700. a(n) = 2^n + (n mod 2)*binomial(n, (n-1)/2).
E.g.f.: (exp(2x) + I_0(2x))/2.
O.g.f.: 2*x/(1-2*x)/(1+2*x-((1+2*x)*(1-2*x))^(1/2)). - Vladeta Jovovic, Apr 27 2003
a(n) = A008949(n, floor(n/2)); a(n) + a(n-1) = A248574(n), n > 0. - Reinhard Zumkeller, Nov 14 2014
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 2*x - 1/(1 - x) )^n.
O.g.f.: (1/2)*( 1/sqrt(1 - 4*x^2) + 1/(1 - 2*x) ).
Inverse binomial transform is (-1)^n*A246437(n).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + ... is the o.g.f. for A001405. (End)
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n-1,(2n+1-(-1)^n)/4 -k). - Anthony Browne, Jun 18 2016
D-finite with recurrence: n*a(n) + 2*(-n+1)*a(n-1) + 4*(-n+1)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Aug 09 2017

Better description from Robert G. Wilson v, Aug 30 2000 and from Yong Kong (ykong(AT)curagen.com), Dec 28 2000

A002895 Domb numbers: number of 2n-step polygons on diamond lattice.

Original entry on oeis.org

1, 4, 28, 256, 2716, 31504, 387136, 4951552, 65218204, 878536624, 12046924528, 167595457792, 2359613230144, 33557651538688, 481365424895488, 6956365106016256, 101181938814289564, 1480129751586116848, 21761706991570726096, 321401321741959062016

Offset: 0

N. J. A. Sloane

a(n) is the (2n)th moment of the distance from the origin of a 4-step random walk in the plane. - Peter M.W. Gill (peter.gill(AT)nott.ac.uk), Mar 03 2004
Row sums of the cube of A008459. - Peter Bala, Mar 05 2013
Conjecture: Let D(n) be the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to a(i+j) for all i,j = 0..n. Then the number D(n)/12^n is always a positive odd integer. - Zhi-Wei Sun, Aug 14 2013
It appears that the expansions exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 4*x + 22*x^2 + 152*x^3 + 1241*x^4 + ... and exp( Sum_{n >= 1} 1/4*a(n)*x^n/n ) = 1 + x + 4*x^2 + 25*x^3 + 199*x^4 + ... have integer coefficients. See A267219. - Peter Bala, Jan 12 2016
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Named after the British-Israeli theoretical physicist Cyril Domb (1920-2012). - Amiram Eldar, Mar 20 2021

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).

Indranil Ghosh, Table of n, a(n) for n = 0..832 (terms 0..100 from T. D. Noe)
B. Adamczewski, Jason P. Bell and E. Delaygue, Algebraic independence of G-functions and congruences "a la Lucas", arXiv preprint arXiv:1603.04187 [math.NT], 2016.
David H. Bailey, Jonathan M. Borwein, David Broadhurst and M. L. Glasser, Elliptic integral evaluations of Bessel moments and applications, Journal of Physics A: Mathematical and Theoretical, Vol. 41, No. 20 (2008), 205203; arXiv preprint, arXiv:0801.0891 [hep-th], 2008.
Nikolai Beluhov, Powers of 2 in High-Dimensional Lattice Walks, arXiv:2506.12789 [math.CO], 2025. See p. 20.
Jonathan M. Borwein, A short walk can be beautiful, Journal of Humanistic Mathematics, Vol. 6, No. 1 (2016), pp. 86-109; preprint, 2015.
Jonathan M. Borwein, Adventures with the OEIS: Five sequences Tony may like, Guttmann 70th [Birthday] Meeting, 2015, revised May 2016.
Jonathan M. Borwein, Adventures with the OEIS: Five sequences Tony may like, Guttmann 70th [Birthday] Meeting, 2015, revised May 2016. [Cached copy, with permission]
Jonathan M. Borwein and Armin Straub, Mahler measures, short walks and log-sine integrals, Theoretical Computer Science, Vol. 479 (2013), pp. 4-21.
Jonathan M. Borwein, Armin Straub and Christophe Vignat, Densities of short uniform random walks, Part II: Higher dimensions, Preprint, 2015.
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, 2010.
Alin Bostan, Andrew Elvey Price, Anthony John Guttmann and Jean-Marie Maillard, Stieltjes moment sequences for pattern-avoiding permutations, arXiv:2001.00393 [math.CO], 2020.
H. Huat Chan, Song Heng Chan and Zhiguo Liu, Domb's numbers and Ramanujan-Sato type series for 1/pi, Adv. Math., Vol. 186, No. 2 (2004), pp. 396-410.
Shaun Cooper, Apéry-like sequences defined by four-term recurrence relations, arXiv:2302.00757 [math.NT], 2023. See Table 2 p. 7.
Shaun Cooper, James G. Wan and Wadim Zudilin, Holonomic Alchemy and Series for 1/pi, in: G. Andrews and F. Garvan (eds.) Analytic Number Theory, Modular Forms and q-Hypergeometric Series, ALLADI60 2016, Springer Proceedings in Mathematics & Statistics, Vol 221. Springer, Cham, 2016; arXiv preprint, arXiv:1512.04608 [math.NT], 2015.
Eric Delaygue, Arithmetic properties of Apéry-like numbers, Compositio Mathematica, Vol. 154, No. 2 (2018), pp. 249-274; arXiv preprint, arXiv:1310.4131 [math.NT], 2013-2015.
Cyril Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., Vol. 9 (1960), pp. 149-361.
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See alpha p. 3.
John A. Hendrickson, Jr., On the enumeration of rectangular (0,1)-matrices, Journal of Statistical Computation and Simulation, Vol. 51 (1995), pp. 291-313.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Pakawut Jiradilok and Elchanan Mossel, Gaussian Broadcast on Grids, arXiv:2402.11990 [cs.IT], 2024. See p. 27.
Ji-Cai Liu, Supercongruences for sums involving Domb numbers, arXiv:2008.02647 [math.NT], 2020.
Rui-Li Liu and Feng-Zhen Zhao, New Sufficient Conditions for Log-Balancedness, With Applications to Combinatorial Sequences, J. Int. Seq., Vol. 21 (2018), Article 18.5.7.
Yen Lee Loh, A general method for calculating lattice green functions on the branch cut, Journal of Physics A: Mathematical and Theoretical, Vol. 50, No. 40 (2017), 405203; arXiv preprint, arXiv:1706.03083 [math-ph], 2017.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5.
Guo-Shuai Mao and Yan Liu, Proof of some conjectural congruences involving Domb numbers, arXiv:2112.00511 [math.NT], 2021.
Guo-Shuai Mao and Michael J. Schlosser, Supercongruences involving Domb numbers and binary quadratic forms, arXiv:2112.12732 [math.NT], 2021.
Robert Osburn and Brundaban Sahu, A supercongruence for generalized Domb numbers, Functiones et Approximatio Commentarii Mathematici, Vol. 48, No. 1 (2013), pp. 29-36; preprint.
L. B. Richmond and Jeffrey Shallit, Counting Abelian Squares, The Electronic Journal of Combinatorics, Vol. 16, No. 1 (2009), Article R72; arXiv preprint, arXiv:0807.5028 [math.CO], 2008.
Armin Straub, Arithmetic aspects of random walks and methods in definite integration, Ph. D. Dissertation, School Of Science And Engineering, Tulane University, 2012.
Zhi-Hong Sun, Super congruences concerning binomial coefficients and Apéry-like numbers, arXiv:2002.12072 [math.NT], 2020.
Zhi-Hong Sun, New congruences involving Apéry-like numbers, arXiv:2004.07172 [math.NT], 2020.
Zhi-Wei Sun, Conjectures involving arithmetical sequences, in: S. Kanemitsu, H. Li and J. Liu (eds.), Number Theory: Arithmetic in Shangri-La, Proc. the 6th China-Japan Sem. Number Theory (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258; alternative link.
H. A. Verrill, Sums of squares of binomial coefficients, with applications to Picard-Fuchs equations, arXiv:math/0407327 [math.CO], 2004.
Chen Wang, Supercongruences and hypergeometric transformations, arXiv:2003.09888 [math.NT], 2020.
Yi Wang and BaoXuan Zhu, Proofs of some conjectures on monotonicity of number-theoretic and combinatorial sequences, Science China Mathematics, Vol. 57, No. 11 (2014), pp. 2429-2435; arXiv preprint, arXiv:1303.5595 [math.CO], 2013.
Bao-Xuan Zhu, Higher order log-monotonicity of combinatorial sequences, arXiv preprint, arXiv:1309.6025 [math.CO], 2013.

Cf. A002893, A008459, A169714, A169715, A228289, A267219.
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.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

A002895 := n -> add(binomial(n,k)^2*binomial(2*n-2*k,n-k)*binomial(2*k,k), k=0..n): seq(A002895(n), n=0..25); # Wesley Ivan Hurt, Dec 20 2015
A002895 := n -> binomial(2*n,n)*hypergeom([1/2, -n, -n, -n],[1, 1, 1/2 - n], 1):
seq(simplify(A002895(n)), n=0..19); # Peter Luschny, May 23 2017

Table[Sum[Binomial[n,k]^2 Binomial[2n-2k,n-k]Binomial[2k,k],{k,0,n}], {n,0,30}] (* Harvey P. Dale, Aug 15 2011 *)
a[n_] = Binomial[2*n, n]*HypergeometricPFQ[{1/2, -n, -n, -n}, {1, 1, 1/2-n}, 1]; (* or *) a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^4, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Dec 30 2013, after Vladeta Jovovic *)
max = 19; Total /@ MatrixPower[Table[Binomial[n, k]^2, {n, 0, max}, {k, 0, max}], 3] (* Jean-François Alcover, Mar 24 2015, after Peter Bala *)

C=binomial;
a(n) = sum(k=0,n, C(n,k)^2 * C(2*n-2*k,n-k) * C(2*k,k) );
/* Joerg Arndt, Apr 19 2013 */

from math import comb
def A002895(n): return (sum(comb(n,k)**2*comb(n-k<<1,n-k)*comb(m:=k<<1,k) for k in range(n+1>>1))<<1) + (0 if n&1 else comb(n,n>>1)**4) # Chai Wah Wu, Jun 17 2025

a(n) = Sum_{k=0..n} binomial(n, k)^2 * binomial(2n-2k, n-k) * binomial(2k, k).
D-finite with recurrence: n^3*a(n) = 2*(2*n-1)*(5*n^2-5*n+2)*a(n-1) - 64*(n-1)^3*a(n-2). - Vladeta Jovovic, Jul 16 2004
Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0, 2*sqrt(x))^4. - Vladeta Jovovic, Aug 01 2006
G.f.: hypergeom([1/6, 1/3],[1],108*x^2/(1-4*x)^3)^2/(1-4*x). - Mark van Hoeij, Oct 29 2011
From Zhi-Wei Sun, Mar 20 2013: (Start)
Via the Zeilberger algorithm, Zhi-Wei Sun proved that:
(1) 4^n*a(n) = Sum_{k = 0..n} (binomial(2k,k)*binomial(2(n-k),n-k))^3/ binomial(n,k)^2,
(2) a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*binomial(2k,n)*binomial(2k,k)* binomial(2(n-k),n-k). (End)
a(n) ~ 2^(4*n+1)/((Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 20 2013
G.f. y=A(x) satisfies: 0 = x^2*(4*x - 1)*(16*x - 1)*y''' + 3*x*(128*x^2 - 30*x + 1)*y'' + (448*x^2 - 68*x + 1)*y' + 4*(16*x - 1)*y. - Gheorghe Coserea, Jun 26 2018
a(n) = Sum_{p+q+r+s=n} (n!/(p!*q!*r!*s!))^2 with p,q,r,s >= 0. See Verrill, p. 5. - Peter Bala, Jan 06 2020
From Peter Bala, Jul 25 2024: (Start)
a(n) = 2*Sum_{k = 1..n} (k/n)*binomial(n, k)^2*binomial(2*n-2*k, n-k)* binomial(2*k, k) for n >= 1.
a(n-1) = (1/2)*Sum_{k = 1..n} (k/n)^3*binomial(n, k)^2*binomial(2*n-2*k, n-k)* binomial(2*k, k) for n >= 1. Cf. A081085. (End)

More terms from Vladeta Jovovic, Mar 11 2003

cp's OEIS Frontend

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

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

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A003946 Expansion of (1+x)/(1-3*x).

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A015518 a(n) = 2*a(n-1) + 3*a(n-2), with a(0)=0, a(1)=1.

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A002893 a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(2*k,k).

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

A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.