cp's OEIS Frontend

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

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A376467 Triangular array read by rows: A063007 * A007318.

Original entry on oeis.org

1, 3, 2, 13, 18, 6, 63, 132, 90, 20, 321, 900, 930, 420, 70, 1683, 5910, 8190, 5600, 1890, 252, 8989, 37926, 65940, 60480, 30870, 8316, 924, 48639, 239624, 501228, 577080, 395010, 160776, 36036, 3432, 265729, 1497096, 3660300, 5072760, 4358970, 2378376, 804804, 154440, 12870, 1462563, 9274410, 25951860, 42060480, 43513470, 29801772, 13513500, 3912480, 656370, 48620
Offset: 0

Views

Author

Peter Bala, Sep 30 2024

Keywords

Comments

Note that the n-th row generating polynomial of A063007 is equal to P(n,2*x + 1), where P(n,x) denotes the n-th Legendre polynomial.
The matrix product A063007 * A007318^(-1) is equal to a signed version of A063007 and A007318^(-1) * A063007 = A115951.

Examples

			Triangle begins
 n\k|     0       1       2       3       4       5      6     7
 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |     1
  1 |     3       2
  2 |    13      18       6
  3 |    63     132      90      20
  4 |   321     900     930     420      70
  5 |  1683    5910    8190    5600    1890     252
  6 |  8989   37926   65940   60480   30870    8316    924
  7 | 48639  239624  501228  577080  395010  160776  36036  3432
  ...
		

Crossrefs

A000984 (main diagonal), A002457( (1/3)*first subdiagonal ), A001850 (Column 0), A002695 ( (1/2)*Column 1 ), A277660 ( (1/3)*Column 2 ), A006442 (row sums).

Programs

  • Maple
    A376467 := proc(n, k); add(binomial(n, j)*binomial(n+j, j)*binomial(j, k), j = k..n) end:
    seq(print(seq(A376467(n, k) , k = 0..n)), n = 0..10);

Formula

T(n, k) = Sum_{j = k..n} binomial(n, j)*binomial(n+j, j)*binomial(j, k).
(n - k)*T(n, k) = 3*(2*n - 1)*T(n-1, k) - (n + k - 1)*T(n-2, k).
T(n, k) = (1/k!) * (d/dx)^k (P(n, 2*x+1)) evaluated at x = 1, where P(n,x) denotes the n-th Legendre polynomial.
G.f. for triangle: 1/sqrt(1 - 6*t + t^2 - 4*t*x) = 1 + (3 + 2*x)*t + (13 + 18*x + 6*x^2)*t^2 + ....
G.f. for column k: binomial(2*k, k) * x^k/(1 - 6*x + x^2)^(k+1/2).
T(n, k) is divisible by binomial(2*k, k) and the array ( T(n, k)/binomial(2*k, k) )n,k >= 0 is the Riordan array (1/sqrt(1 - 6*x + x^2), x/(1 - 6*x + x^2)).
T(n, k) is divisible by binomial(n+k, k) and the array ( T(n, k)/binomial(n+k, k) )n,k >= 0 is the Riordan array A118384.
T(n, n) = binomial(2*n, n); T(n, n-1) = 3*(2*n-1)!/(n-1)!^2 = 3 * A002457(n-1) for n >= 1.
The n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*(1 + x)^k = P(n, 2*x+3) = hypergeom([-n, n+1], [1], -1-x).
Recurrence: n*R(n, x) = (2*x + 3)*(2*n - 1)*R(n-1, x) - (n - 1)*R(n-2, x) with R(0, x) = 1.
If we set R(-1,x) = 1, we can run the recurrence backwards to give R(-n, x) = Sum_{k = 0..n} binomial(-n, k)*binomial(-n+k, k)*(1 + x)^k = R(n-1, x).
R(n, x) = (-1)^n * R(n, -x-3).
R(n, x) = 1/n! * (d/dx)^n( ((1 + x)*(2 + x))^n ).
R(1, x) = 3 + 2*x divides R(2*n+1, x) in the polynomial ring Z[x].

A303987 Triangle read by rows: T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0, k = 0..n.

Original entry on oeis.org

1, 1, 4, 1, 36, 36, 1, 144, 900, 400, 1, 400, 8100, 19600, 4900, 1, 900, 44100, 313600, 396900, 63504, 1, 1764, 176400, 2822400, 9922500, 7683984, 853776, 1, 3136, 571536, 17640000, 133402500, 276623424, 144288144, 11778624, 1, 5184, 1587600, 85377600, 1200622500, 5194373184, 7070119056, 2650190400, 165636900
Offset: 0

Views

Author

Wolfdieter Lang, May 14 2018

Keywords

Comments

The row sums of this triangle are b(n) = A005259(n), for n >= 0. This sequence b was used in R. Apéry's 1979 proof of the irrationality of Zeta(3). See A005259 for references and links.
Row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], x), hence b(n) = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], 1) (see the formula in A005259 given by K. A. Penson. This is the solution to Exercise 2.14 of the Koepf reference given there, p. 29).

Examples

			The triangle T begins:
n\k  0    1       2        3          4          5          6          7 ...
0:   1
1:   1    4
2:   1   36      36
3:   1  144     900      400
4:   1  400    8100    19600       4900
5:   1  900   44100   313600     396900      63504
6:   1 1764  176400  2822400    9922500    7683984     853776
7:   1 3136  571536 17640000  133402500  276623424  144288144   11778624
----------------------------------------------------------------------------
row n = 8:   1 5184 1587600 85377600 1200622500 5194373184 7070119056 2650190400 165636900,
row n = 9: 1 8100 3920400 341510400 8116208100 63631071504 176752976400 169612185600 47869064100 2363904400,
row n = 10: 1 12100 8820900 1177862400 44188244100 572679643536 2828047622400 5446435737600 3877394192100 853369488400 34134779536.
...
		

Crossrefs

The column sequences (without zeros) are A000012, A035287(n+1) = 4*A000217(n)^2, 36*A288876, 400*A000579(n+6)^2, 4900*A000581(n+8)^2, 63504*A001287(n+10)^2, ...

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->(Binomial(n,k)*Binomial(n+k,k))^2))); # Muniru A Asiru, May 15 2018
  • Mathematica
    T[n_, k_] := (Gamma[k + n + 1]/(Gamma[k + 1]^2*Gamma[-k + n + 1]))^2;
    Flatten[Table[T[n, k], {n, 0, 8}, {k, 0, n}]] (* Peter Luschny, May 14 2018 *)

Formula

T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0 and k = 0..n.
T(n, k) = (binomial(n+k, 2*k)*cbi(k))^2, with cbi(k) = A000984(k) = binomial(2*k, k), and cbi(k)^2 = A002894(k).
G.f. for column sequences (without leading zeros):
cbi(k)^2*P2(2*k, x)/(1 - x)^(4*k+1), with the row polynomials of A008459 (Pascal entries squared) P2(2*k, x) = Sum_{j=0..2*k} A008459(2*k, j)*x^j. For a proof see the general comment in A288876 on the diagonals and columns of A008459.

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

Original entry on oeis.org

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

Views

Author

Keywords

Comments

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

Examples

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

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 160.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
  • Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
  • Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
  • Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
  • Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
  • Leonard Lipshitz and A. van der Poorten, "Rational functions, diagonals, automata and arithmetic", in Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), 339-358.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
Cf. A226078, A051924 (first differences).
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.

Programs

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

Formula

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

A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

Original entry on oeis.org

1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, 212941, 616227, 1787607, 5196627, 15134931, 44152809, 128996853, 377379369, 1105350729, 3241135527, 9513228123, 27948336381, 82176836301, 241813226151, 712070156203, 2098240353907, 6186675630819
Offset: 0

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Comments

Number of ordered trees with n + 1 edges, having root of odd degree and nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 02 2002
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), running from (0,0) to (n,0) (i.e., grand Motzkin paths of length n). For example, a(3) = 7 because we have HHH, HUD, HDU, UDH, DUH, UHD and DHU. - Emeric Deutsch, May 31 2003
Number of lattice paths from (0,0) to (n,n) using steps (2,0), (0,2), (1,1). It appears that 1/sqrt((1 - x)^2 - 4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1). - Joerg Arndt, Jul 01 2011
Number of lattice paths from (0,0) to (n,n) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Binomial transform of A000984, with interpolated zeros. - Paul Barry, Jul 01 2003
Number of leaves in all 0-1-2 trees with n edges, n > 0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) - Emeric Deutsch, Nov 30 2003
a(n) is the number of UDU-free paths of n + 1 upsteps (U) and n downsteps (D) that start U. For example, a(2) = 3 counts UUUDD, UUDDU, UDDUU. - David Callan, Aug 18 2004
Diagonal sums of triangle A063007. - Paul Barry, Aug 31 2004
Number of ordered ballots from n voters that result in an equal number of votes for candidates A and B in a three candidate election. Ties are counted even when candidates A and B lose the election. For example, a(3) = 7 because ballots of the form (voter-1 choice, voter-2 choice, voter-3 choice) that result in equal votes for candidates A and B are the following: (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A) and (C,C,C). - Dennis P. Walsh, Oct 08 2004
a(n) is the number of weakly increasing sequences (a_1,a_2,...,a_n) with each a_i in [n]={1,2,...,n} and no element of [n] occurring more than twice. For n = 3, the sequences are 112, 113, 122, 123, 133, 223, 233. - David Callan, Oct 24 2004
Note that n divides a(n+1) - a(n). In fact, (a(n+1) - a(n))/n = A007971(n+1). - T. D. Noe, Mar 16 2005
Row sums of triangle A105868. - Paul Barry, Apr 23 2005
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having no H steps on the x-axis. Example: a(3) = 7 because we have UDU, UHD, UHH, UHU, UUD, UUH and UUU. - Emeric Deutsch, Oct 07 2007
Equals right border of triangle A152227; starting with offset 1, the row sums of triangle A152227. - Gary W. Adamson, Nov 29 2008
Starting with offset 1 = iterates of M * [1,1,1,...] where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
a(n) is prime for n = 2, 3 and 4, with no others for n <= 10^5 (E. W. Weisstein, Mar 14 2005). It has apparently not been proved that no [other] prime central trinomials exist. - Jonathan Vos Post, Mar 19 2010
a(n) is not divisible by 3 for n whose base-3 representation contains no 2 (A005836).
a(n) = number of (n-1)-lettered words in the alphabet {1,2,3} with as many occurrences of the substring (consecutive subword) [1,2] as those of [2,1]. See the papers by Ekhad-Zeilberger and Zeilberger. - N. J. A. Sloane, Jul 05 2012
a(n) = coefficient of x^n in (1 + x + x^2)^n. - L. Edson Jeffery, Mar 23 2013
a(n) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that (i.) A and B are disjoint and (ii.) A and B contain the same number of elements. For example, a(2) = 3 because we have: ({},{}) ; ({1},{2}) ; ({2},{1}). - Geoffrey Critzer, Sep 04 2013
Also central terms of A082601. - Reinhard Zumkeller, Apr 13 2014
a(n) is the number of n-tuples with entries 0, 1, or 2 and with the sum of entries equal to n. For n=3, the seven 3-tuples are (1,1,1), (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), and (2,1,0). - Dennis P. Walsh, May 08 2015
The series 2*a(n) + 3*a(n+1) + a(n+2) = 2*A245455(n+3) has Hankel transform of L(2n+1)*2^n, offset n = 1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
The series (2*a(n) + 3*a(n+1) + a(n+2))/2 = A245455(n+3) has Hankel transform of L(2n+1), offset n=1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
Conjecture: An integer n > 3 is prime if and only if a(n) == 1 (mod n^2). We have verified this for n up to 8*10^5, and proved that a(p) == 1 (mod p^2) for any prime p > 3 (cf. A277640). - Zhi-Wei Sun, Nov 30 2016
This is the analog for Coxeter type B of Motzkin numbers (A001006) for Coxeter type A. - F. Chapoton, Jul 19 2017
a(n) is also the number of solutions to the equation x(1) + x(2) + ... + x(n) = 0, where x(1), ..., x(n) are in the set {-1,0,1}. Indeed, the terms in (1 + x + x^2)^n that produce x^n are of the form x^i(1)*x^i(2)*...*x^i(n) where i(1), i(2), ..., i(n) are in {0,1,2} and i(1) + i(2) + ... + i(n) = n. By setting j(t) = i(t) - 1 we obtain that j(1), ..., j(n) satisfy j(1) + ... + j(n) =0 and j(t) in {-1,0,1} for all t = 1..n. - Lucien Haddad, Mar 10 2018
If n is a prime greater than 3 then a(n)-1 is divisible by n^2. - Ira M. Gessel, Aug 08 2021
Let f(m) = ceiling((q+log(q))/log(9)), where q = -log(log(27)/(2*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024
Diagonal of the rational function 1 / (1 - x^2 - y^2 - x*y). - Ilya Gutkovskiy, Apr 23 2025

Examples

			For n = 2, (x^2 + x + 1)^2 = x^4 + 2*x^3 + 3*x^2 + 2*x + 1, so a(2) = 3. - _Michael B. Porter_, Sep 06 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 78 and 163, #19.
  • L. Euler, Exemplum Memorabile Inductionis Fallacis, Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 15, p. 59.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 575.
  • P. Henrici, Applied and Computational Complex Analysis. Wiley, NY, 3 vols., 1974-1986. (Vol. 1, p. 42.)
  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 579.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 74.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 6.3.8.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 22.
  • Lin Yang and S.-L. Yang, The parametric Pascal rhombus. Fib. Q., 57:4 (2019), 337-346. See p. 341.

Crossrefs

INVERT transform is A007971. Partial sums are A097893. Squares are A168597.
Main column of A027907. Column k=2 of A305161. Column k=0 of A328347. Column 1 of A201552(?).
Cf. A001006, A002878, A005043, A005717, A082758 (bisection), A273055 (bisection), A102445, A113302, A113303, A113304, A113305 (divisibility of central trinomial coefficients), A152227, A277640.

Programs

  • Haskell
    a002426 n = a027907 n n  -- Reinhard Zumkeller, Jan 22 2013
    
  • Magma
    P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2)^n)): n in [0..26]]; // Bruno Berselli, Jul 05 2011
    
  • Maple
    A002426 := proc(n) local k;
        sum(binomial(n, k)*binomial(n-k, k), k=0..floor(n/2));
    end proc: # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    # Alternatively:
    a := n -> simplify(GegenbauerC(n,-n,-1/2)):
    seq(a(n), n=0..29); # Peter Luschny, May 07 2016
  • Mathematica
    Table[ CoefficientList[ Series[(1 + x + x^2)^n, {x, 0, n}], x][[ -1]], {n, 0, 27}] (* Robert G. Wilson v *)
    a=b=1; Join[{a,b}, Table[c=((2n-1)b + 3(n-1)a)/n; a=b; b=c; c, {n,2,100}]]; Table[Sqrt[-3]^n LegendreP[n,1/Sqrt[-3]],{n,0,26}] (* Wouter Meeussen, Feb 16 2013 *)
    a[ n_] := If[ n < 0, 0, 3^n Hypergeometric2F1[ 1/2, -n, 1, 4/3]]; (* Michael Somos, Jul 08 2014 *)
    Table[4^n *JacobiP[n,-n-1/2,-n-1/2,-1/2], {n,0,29}] (* Peter Luschny, May 13 2016 *)
    a[n_] := a[n] = Sum[n!/((n - 2*i)!*(i!)^2), {i, 0, n/2}]; Table[a[n], {n, 0, 29}] (* Shara Lalo and Zagros Lalo, Oct 03 2018 *)
  • Maxima
    trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
    makelist(trinomial(n,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */
    
  • Maxima
    makelist(ultraspherical(n,-n,-1/2),n,0,12); /* Emanuele Munarini, Dec 20 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, n))};
    
  • PARI
    /* as lattice paths: same as in A092566 but use */
    steps=[[2, 0], [0, 2], [1, 1]];
    /* Joerg Arndt, Jul 01 2011 */
    
  • PARI
    a(n)=polcoeff(sum(m=0, n, (2*m)!/m!^2 * x^(2*m) / (1-x+x*O(x^n))^(2*m+1)), n) \\ Paul D. Hanna, Sep 21 2013
    
  • Python
    from math import comb
    def A002426(n): return sum(comb(n,k)*comb(k,n-k) for k in range(n+1)) # Chai Wah Wu, Nov 15 2022
  • Sage
    A002426 = lambda n: hypergeometric([-n/2, (1-n)/2], [1], 4)
    [simplify(A002426(n)) for n in (0..29)]
    # Peter Luschny, Sep 17 2014
    
  • Sage
    def A():
        a, b, n = 1, 1, 1
        yield a
        while True:
            yield b
            n += 1
            a, b = b, ((3 * (n - 1)) * a + (2 * n - 1) * b) // n
    A002426 = A()
    print([next(A002426) for  in range(30)])  # _Peter Luschny, May 16 2016
    

Formula

G.f.: 1/sqrt(1 - 2*x - 3*x^2).
E.g.f.: exp(x)*I_0(2x), where I_0 is a Bessel function. - Michael Somos, Sep 09 2002
a(n) = 2*A027914(n) - 3^n. - Benoit Cloitre, Sep 28 2002
a(n) is asymptotic to d*3^n/sqrt(n) with d around 0.5.. - Benoit Cloitre, Nov 02 2002, d = sqrt(3/Pi)/2 = 0.4886025119... - Alec Mihailovs (alec(AT)mihailovs.com), Feb 24 2005
D-finite with recurrence: a(n) = ((2*n - 1)*a(n-1) + 3*(n - 1)*a(n-2))/n; a(0) = a(1) = 1; see paper by Barcucci, Pinzani and Sprugnoli.
Inverse binomial transform of A000984. - Vladeta Jovovic, Apr 28 2003
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, k/2)*(1 + (-1)^k)/2; a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*binomial(2*k, k). - Paul Barry, Jul 01 2003
a(n) = Sum_{k>=0} binomial(n, 2*k)*binomial(2*k, k). - Philippe Deléham, Dec 31 2003
a(n) = Sum_{i+j=n, 0<=j<=i<=n} binomial(n, i)*binomial(i, j). - Benoit Cloitre, Jun 06 2004
a(n) = 3*a(n-1) - 2*A005043(n). - Joost Vermeij (joost_vermeij(AT)hotmail.com), Feb 10 2005
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, n-k). - Paul Barry, Apr 23 2005
a(n) = (-1/4)^n*Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-2*k, n-k)*(-3)^k. - Philippe Deléham, Aug 17 2005
a(n) = A111808(n,n). - Reinhard Zumkeller, Aug 17 2005
a(n) = Sum_{k=0..n} (((1 + (-1)^k)/2)*Sum_{i=0..floor((n-k)/2)} binomial(n, i)*binomial(n-i, i+k)*((k + 1)/(i + k + 1))). - Paul Barry, Sep 23 2005
a(n) = 3^n*Sum_{j=0..n} (-1/3)^j*C(n, j)*C(2*j, j); follows from (a) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/2)^n*Sum_{j=0..n} 3^j*binomial(n, j)*binomial(2*n-2*j, n) = (3/2)^n*Sum_{j=0..n} (1/3)^j*binomial(n, j)*binomial(2*j, n); follows from (c) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/Pi)*Integral_{x=-1..3} x^n/sqrt((3 - x)*(1 + x)) is moment representation. - Paul Barry, Sep 10 2007
G.f.: 1/(1 - x - 2x^2/(1 - x - x^2/(1 - x - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 05 2009
a(n) = sqrt(-1/3)*(-1)^n*hypergeometric([1/2, n+1], [1], 4/3). - Mark van Hoeij, Nov 12 2009
a(n) = (1/Pi)*Integral_{x=-1..1} (1 + 2*x)^n/sqrt(1 - x^2) = (1/Pi)*Integral_{t=0..Pi} (1 + 2*cos(t))^n. - Eli Wolfhagen, Feb 01 2011
In general, g.f.: 1/sqrt(1 - 2*a*x + x^2*(a^2 - 4*b)) = 1/(1 - a*x)*(1 - 2*x^2*b/(G(0)*(a*x - 1) + 2*x^2*b)); G(k) = 1 - a*x - x^2*b/G(k+1); for g.f.: 1/sqrt(1 - 2*x - 3*x^2) = 1/(1 - x)*(1 - 2*x^2/(G(0)*(x - 1) + 2*x^2)); G(k) = 1 - x - x^2/G(k+1), a = 1, b = 1; (continued fraction). - Sergei N. Gladkovskii, Dec 08 2011
a(n) = Sum_{k=0..floor(n/3)} (-1)^k*binomial(2*n-3*k-1, n-3*k)*binomial(n, k). - Gopinath A. R., Feb 10 2012
G.f.: A(x) = x*B'(x)/B(x) where B(x) satisfies B(x) = x*(1 + B(x) + B(x)^2). - Vladimir Kruchinin, Feb 03 2013 (B(x) = x*A001006(x) - Michael Somos, Jul 08 2014)
G.f.: G(0), where G(k) = 1 + x*(2 + 3*x)*(4*k + 1)/(4*k + 2 - x*(2 + 3*x)*(4*k + 2)*(4*k + 3)/(x*(2 + 3*x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
E.g.f.: exp(x) * Sum_{k>=0} (x^k/k!)^2. - Geoffrey Critzer, Sep 04 2013
G.f.: Sum_{n>=0} (2*n)!/n!^2*(x^(2*n)/(1 - x)^(2*n+1)). - Paul D. Hanna, Sep 21 2013
0 = a(n)*(9*a(n+1) + 9*a(n+2) - 6*a(n+3)) + a(n+1)*(3*a(n+1) + 4*a(n+2) - 3*a(n+3)) + a(n+2)*(-a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = hypergeometric([-n/2, (1-n)/2], [1], 4). - Peter Luschny, Sep 17 2014
a(n) = A132885(n,0), that is, a(n) = A132885(A002620(n+1)). - Altug Alkan, Nov 29 2015
a(n) = GegenbauerC(n,-n,-1/2). - Peter Luschny, May 07 2016
a(n) = 4^n*JacobiP[n,-n-1/2,-n-1/2,-1/2]. - Peter Luschny, May 13 2016
From Alexander Burstein, Oct 03 2017: (Start)
G.f.: A(4*x) = B(-x)*B(3*x), where B(x) is the g.f. of A000984.
G.f.: A(2*x)*A(-2*x) = B(x^2)*B(9*x^2).
G.f.: A(x) = 1 + x*M'(x)/M(x), where M(x) is the g.f. of A001006. (End)
a(n) = Sum_{i=0..n/2} n!/((n - 2*i)!*(i!)^2). [Cf. Lalo and Lalo link. It is Luschny's terminating hypergeometric sum.] - Shara Lalo and Zagros Lalo, Oct 03 2018
From Peter Bala, Feb 07 2022: (Start)
a(n)^2 = Sum_{k = 0..n} (-3)^(n-k)*binomial(2*k,k)^2*binomial(n+k,n-k) and has g.f. Sum_{n >= 0} binomial(2*n,n)^2*x^n/(1 + 3*x)^(2*n+1). Compare with the g.f. for a(n) given above by Hanna.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all prime p and positive integers n and k.
Conjecture: The stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all prime p >= 5 and positive integers n and k. (End)
a(n) = A005043(n) + A005717(n) for n >= 1. - Amiram Eldar, May 17 2024
For even n, a(n) = (n-1)!!* 2^{n/2}/ (n/2)!* 2F1(-n/2,-n/2;1/2;1/4). For odd n, a(n) = n!! *2^(n/2-1/2) / (n/2-1/2)! * 2F1(1/2-n/2,1/2-n/2;3/2;1/4). - R. J. Mathar, Mar 19 2025

A001850 Central Delannoy numbers: a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k).

Original entry on oeis.org

1, 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823, 252055236609, 1425834724419, 8079317057869, 45849429914943, 260543813797441, 1482376214227923, 8443414161166173, 48141245001931263
Offset: 0

Views

Author

Keywords

Comments

Number of paths from (0,0) to (n,n) in an n X n grid using only steps north, northeast and east (i.e., steps (1,0), (1,1), and (0,1)).
Also the number of ways of aligning two sequences (e.g., of nucleotides or amino acids) of length n, with at most 2*n gaps (-) inserted, so that while unnecessary gappings: - -a a- - are forbidden, both b- and -b are allowed. (If only other of the latter is allowed, then the sequence A000984 gives the number of alignments.) There is an easy bijection from grid walks given by Dickau to such set of alignments (e.g., the straight diagonal corresponds to the perfect alignment with no gaps). - Antti Karttunen, Oct 10 2001
Also main diagonal of array A008288 defined by m(i,1) = m(1,j) = 1, m(i,j) = m(i-1,j-1) + m(i-1,j) + m(i,j-1). - Benoit Cloitre, May 03 2002
So, as a special case of Dmitry Zaitsev's Dec 10 2015 comment on A008288, a(n) is the number of points in Z^n that are L1 (Manhattan) distance <= n from any given point. These terms occur in the crystal ball sequences: a(n) here is the n-th term in the sequence for the n-dimensional cubic lattice. See A008288 for a list of crystal ball sequences (rows or columns of A008288). - Shel Kaphan, Dec 26 2022
a(n) is the number of n-matchings of a comb-like graph with 2*n teeth. Example: a(2) = 13 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has 13 2-matchings: any of the six possible pairs of teeth and {Aa, BC}, {Aa, CD}, {Bb, CD}, {Cc, AB}, {Dd, AB}, {Dd, BC}, {AB, CD}. - Emeric Deutsch, Jul 02 2002
Number of ordered trees with 2*n+1 edges, having root of odd degree, nonroot nodes of outdegree at most 2 and branches of odd length. - Emeric Deutsch, Aug 02 2002
The sum of the first n coefficients of ((1 - x) / (1 - 2*x))^n is a(n-1). - Michael Somos, Sep 28 2003
Row sums of A063007 and A105870. - Paul Barry, Apr 23 2005
The Hankel transform (see A001906 for definition) of this sequence is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
Also number of paths from (0,0) to (n,0) using only steps U = (1,1), H = (1,0) and D =(1,-1), U can have 2 colors and H can have 3 colors. - N-E. Fahssi, Jan 27 2008
Equals row sums of triangle A152250 and INVERT transform of A109980: (1, 2, 8, 36, 172, 852, ...). - Gary W. Adamson, Nov 30 2008
Number of overpartitions in the n X n box (treat a walk of the type in the first comment as an overpartition, by interpreting a NE step as N, E with the part thus created being overlined). - William J. Keith, May 19 2017
Diagonal of rational functions 1/(1 - x - y - x*y), 1/(1 - x - y*z - x*y*z). - Gheorghe Coserea, Jul 03 2018
Dimensions of endomorphism algebras End(R^{(n)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023
a(n) is the number of ways to tile a strip of length n with white squares, black squares, and red dominos, where we must have an equal number of white and black squares. - Greg Dresden and Leo Zhang, Jul 11 2025

Examples

			G.f. = 1 + 3*x + 13*x^2 + 63*x^3 + 321*x^4 + 1683*x^5 + 8989*x^6 + ...
		

References

  • Frits Beukers, Arithmetic properties of Picard-Fuchs equations, Séminaire de Théorie des nombres de Paris, 1982-83, Birkhäuser Boston, Inc.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 593.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • L. Moser and W. Zayachkowski, Lattice paths with diagonal steps, Scripta Math., 26 (1961), 223-229.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 2, 1999; see Example 6.3.8 and Problem 6.49.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 28.

Crossrefs

Main diagonal of A064861.
Column k=2 of A262809 and A263159.

Programs

  • Maple
    seq(add(multinomial(n+k,n-k,k,k),k=0..n),n=0..20); # Zerinvary Lajos, Oct 18 2006
    seq(orthopoly[P](n,3), n=0..100); # Robert Israel, Nov 03 2015
  • Mathematica
    f[n_] := Sum[ Binomial[n, k] Binomial[n + k, k], {k, 0, n}]; Array[f, 21, 0] (* Or *)
    a[0] = 1; a[1] = 3; a[n_] := a[n] = (3(2 n - 1)a[n - 1] - (n - 1)a[n - 2])/n; Array[a, 21, 0] (* Or *)
    CoefficientList[ Series[1/Sqrt[1 - 6x + x^2], {x, 0, 20}], x] (* Robert G. Wilson v *)
    Table[LegendreP[n, 3], {n, 0, 22}] (* Jean-François Alcover, Jul 16 2012, from first formula *)
    a[n_] := Hypergeometric2F1[-n, n+1, 1, -1]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Feb 26 2013 *)
    a[ n_] := With[ {m = If[n < 0, -1 - n, n]}, SeriesCoefficient[ (1 - 6 x + x^2)^(-1/2), {x, 0, m}]]; (* Michael Somos, Jun 10 2015 *)
  • Maxima
    a(n):=coeff(expand((1+3*x+2*x^2)^n),x,n);
    makelist(a(n),n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • PARI
    {a(n) = if( n<0, n = -1 - n); polcoeff( 1 / sqrt(1 - 6*x + x^2 + x * O(x^n)), n)}; /* Michael Somos, Sep 23 2006 */
    
  • PARI
    {a(n) = if( n<0, n = -1 - n); subst( pollegendre(n), x, 3)}; /* Michael Somos, Sep 23 2006 */
    
  • PARI
    {a(n) = if( n<0, n = -1 - n); n++; subst( Pol(((1 - x) / (1 - 2*x) + O(x^n))^n), x, 1);} /* Michael Somos, Sep 23 2006 */
    
  • PARI
    a(n)=if(n<0, 0, polcoeff((1+3*x+2*x^2)^n, n)) \\ Paul Barry, Aug 22 2007
    
  • PARI
    /* same as in A092566 but use */
    steps=[[1,0], [0,1], [1,1]]; /* Joerg Arndt, Jun 30 2011 */
    
  • PARI
    a(n)=sum(k=0,n,binomial(n,k)*binomial(n+k,k)); \\ Joerg Arndt, May 11 2013
    
  • PARI
    my(x='x+O('x^30)); Vec(1/sqrt(1 - 6*x + x^2)) \\ Altug Alkan, Oct 17 2015
    
  • Python
    # from Nick Hobson.
    def f(a, b):
        if a == 0 or b == 0:
            return 1
        return f(a, b - 1) + f(a - 1, b) + f(a - 1, b - 1)
    [f(n, n) for n in range(7)]
    
  • Python
    from gmpy2 import divexact
    A001850 = [1, 3]
    for n in range(2,10**3):
        A001850.append(divexact(A001850[-1]*(6*n-3)-(n-1)*A001850[-2],n))
    # Chai Wah Wu, Sep 01 2014
    
  • Sage
    a = lambda n: hypergeometric([-n, -n], [1], 2)
    [simplify(a(n)) for n in range(23)] # Peter Luschny, Nov 19 2014

Formula

a(n) = P_n(3), where P_n is n-th Legendre polynomial.
G.f.: 1 / sqrt(1 - 6*x + x^2).
a(n) = a(n-1) + 2*A002002(n) = Sum_{j} A063007(n, j). - Henry Bottomley, Jul 02 2001
Dominant term in asymptotic expansion is binomial(2*n, n)/2^(1/4)*((sqrt(2) + 1)/2)^(2*n + 1)*(1 + c_1/n + c_2/n^2 + ...). - Michael David Hirschhorn
a(n) = Sum_{i=0..n} (A000079(i)*A008459(n, i)) = Sum_{i=0..n} (2^i * C(n, i)^2). - Antti Karttunen, Oct 10 2001
a(n) = Sum_{k=0..n} C(n+k, n-k)*C(2*k, k). - Benoit Cloitre, Feb 13 2003
a(n) = Sum_{k=0..n} C(n, k)^2 * 2^k. - Michael Somos, Oct 08 2003
a(n - 1) = coefficient of x^n in A120588(x)^n if n>=0. - Michael Somos, Apr 11 2012
G.f. of a(n-1) = 1 / (1 - x / (1 - 2*x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / ...)))))). - Michael Somos, May 11 2012
INVERT transform is A109980. BINOMIAL transform is A080609. BINOMIAL transform of A006139. PSUM transform is A089165. PSUMSIGN transform is A026933. First backward difference is A110170. - Michael Somos, May 11 2012
E.g.f.: exp(3*x)*BesselI(0, 2*sqrt(2)*x). - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..n} C(2*n-k, n)*C(n, k). - Paul Barry, Apr 23 2005
a(n) = Sum_{k>=n} binomial(k, n)^2/2^(k+1). - Vladeta Jovovic, Aug 25 2006
a(n) = a(-1 - n) for all n in Z. - Michael Somos, Sep 23 2006
D-finite with recurrence: a(-1) = a(0) = 1; n*a(n) = 3*(2*n-1)*a(n-1) - (n-1)*a(n-2). Eq (4) in T. D. Noe's article in JIS 9 (2006) #06.2.7.
Define general Delannoy numbers by (i,j > 0): d(i,0) = d(0,j) = 1 =: d(0,0) and d(i,j) = d(i-1,j-1) + d(i-2,j-1) + d(i-1,j). Then a(k) = Sum_{j >= 0} d(k,j)^2 + d(k-1,j)^2 = A026933(n)+A026933(n-1). This is a special case of the following formula for general Delannoy numbers: d(k,j) = Sum_{i >= 0, p=0..n} d(p, i) * d(n-p, j-i) + d(p-1, i) * d(n-p-1, j-i-1). - Peter E John, Oct 19 2006
Coefficient of x^n in (1 + 3*x + 2*x^2)^n. - N-E. Fahssi, Jan 11 2008
a(n) = A008288(A046092(n)). - Philippe Deléham, Apr 08 2009
G.f.: 1/(1 - x - 2*x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ... (continued fraction). - Paul Barry, May 28 2009
G.f.: d/dx log(1/(1 - x*A001003(x))). - Vladimir Kruchinin, Apr 19 2011
G.f.: 1/(2*Q(0) + x - 1) where Q(k) = 1 + k*(1-x) - x - x*(k + 1)*(k + 2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n) = Sum_{k=0..n} C(n,k) * C(n+k,k). - Joerg Arndt, May 11 2013
G.f.: G(0), where G(k) = 1 + x*(6 - x)*(4*k + 1)/(4*k + 2 - 2*x*(6-x)*(2*k + 1)*(4*k + 3)/(x*(6 - x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k - 1)/(x*(6 - x)*(2*k - 1) + 2*(k + 1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k + 1)/(x*(6 - x)*(2*k + 1) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
a(n)^2 = Sum_{k=0..n} 2^k * C(2*k, k)^2 * C(n+k, n-k) = A243949(n). - Paul D. Hanna, Aug 17 2014
a(n) = hypergeom([-n, -n], [1], 2). - Peter Luschny, Nov 19 2014
a(n) = Sum_{k=0..n/2} C(n-k,k) * 3^(n-2*k) * 2^k * C(n,k). - Vladimir Kruchinin, Jun 29 2015
a(n) = A049600(n, n-1).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). Cf. A126086 and A274668. - Peter Bala, Jan 15 2020
a(n) ~ c * (3 + 2*sqrt(2))^n / sqrt(n), where c = 1/sqrt(4*Pi*(3*sqrt(2)-4)) = 0.572681... (Banderier and Schwer, 2005). - Amiram Eldar, Jun 07 2020
a(n+1) = 3*a(n) + 2*Sum_{l=1..n} A006318(l)*a(n-l). [Eq. (1.16) in Qi-Shi-Guo (2016)]
a(n) ~ (1 + sqrt(2))^(2*n+1) / (2^(5/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Jan 09 2023
a(n-1) + a(n) = A241023(n) for n >= 1. - Peter Bala, Sep 18 2024
a(n) = Sum_{k=0..n} C(n+k, 2*k) * C(2*k, k). - Greg Dresden and Leo Zhang, Jul 11 2025

Extensions

New name and reference Sep 15 1995
Formula and more references from Don Knuth, May 15 1996

A033999 a(n) = (-1)^n.

Original entry on oeis.org

1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1
Offset: 0

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Author

Vasiliy Danilov (danilovv(AT)usa.net), Jun 15 1998

Keywords

Comments

(-1)^(n+1) = signed area of parallelogram with vertices (0,0), U=(F(n),F(n+1)), V=(F(n+1),F(n+2)), where F = A000045 (Fibonacci numbers). The area of every such parallelogram is 1. The signed area is -1 if and only if F(n+1)^2 > F(n)*F(n+2), or, equivalently, n is even, or, equivalently, the vector U is "above" V, indicating that U and V "cross" as n -> n+1. - Clark Kimberling, Sep 09 2013
Periodic with period length 2. - Ray Chandler, Apr 03 2017
From Bernard Schott, May 11 2022: (Start)
Cesàro mean theorem: When a(n) has a limit (finite or infinite) in the usual sense, then c(n) = (a(1)+...+a(n))/n has the same Cesàro limit, but the converse is false. This sequence is a counterexample in the case of a finite Cesàro limit (see A237420 for counterexample with an infinite Cesàro limit).
This sequence is not convergent in the usual sense because a(2n) = 1 while a(2n+1) = -1; the successive arithmetic means c(n) of the first n terms of the sequence are 1/1, 0/2, 1/3, 0/4, 1/5, 0/6, ... so c(2n) = 1/(2n+1) and c(2n+1) = 0, hence the Cesàro limit is 0 because c(n) -> 0 when n -> oo.
In fact, when sequence a(n) is "Period k: [a1, a2, ..., ak]", then the Cesàro limit c of this sequence is (a1+a2+...+ak)/k.
Note that the converse of the theorem is true iff a(n) is monotonic (End).

Examples

			G.f. = 1 - x + x^2 - x^3 + x^4 - x^5 + x^6 - x^7 + x^8 - x^9 + x^10 - x^11 + x^12 + ...
		

References

  • J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Crossrefs

About Cesàro mean theorem: A114112, A237420.
Cf. A072691 (abs. val. Dgf at s=2), A197070 (abs. val. Dgf at s=3), A267315 (abs. val. Dgf at s=4).

Programs

Formula

G.f.: 1/(1+x).
E.g.f.: exp(-x).
Linear recurrence: a(0)=1, a(n)=-a(n-1) for n>0. - Jaume Oliver Lafont, Mar 20 2009
Sum_{k=0..n} a(k) = A059841(n). - Jaume Oliver Lafont, Nov 21 2009
Sum_{k>=0} a(k)/(k+1) = log(2). - Jaume Oliver Lafont, Mar 30 2010
Euler transform of length 2 sequence [ -1, 1]. - Michael Somos, Mar 21 2011
Moebius transform is length 2 sequence [ -1, 2]. - Michael Somos, Mar 21 2011
a(n) = -b(n) where b(n) = multiplicative with b(2^e) = -1 if e>0, b(p^e) = 1 if p>2. - Michael Somos, Mar 21 2011
a(n) = a(-n) = a(n + 2) = cos(n * Pi). a(n) = c_2(n) if n>1 where c_k(n) is Ramanujan's sum. - Michael Somos, Mar 21 2011
a(n) = (1/2)*Product_{k=0..2*n-1} 2*cos((2*k+1)*Pi/(4*n)), n >= 1. See the product given in the Oct 21 2013 formula comment in A056594, and replace there n -> 2*n. - Wolfdieter Lang, Oct 23 2013
D.g.f.: (2^(1-s)-1)*zeta(s) = -eta(s) (the Dirichlet eta function). - Ralf Stephan, Mar 27 2015
From Ilya Gutkovskiy, Aug 17 2016: (Start)
a(n) = T_n(-1), where T_n(x) are the Chebyshev polynomials of the first kind.
Binomial transform of A122803. (End)
a(n) = exp(i*Pi*n) = exp(-i*Pi*n). - Carauleanu Marc, Sep 15 2016
a(n) = Sum_{k=0..n} (-1)^k*A063007(n, k), n >= 0. - Wolfdieter Lang, Sep 13 2016

A000172 The Franel number a(n) = Sum_{k = 0..n} binomial(n,k)^3.

Original entry on oeis.org

1, 2, 10, 56, 346, 2252, 15184, 104960, 739162, 5280932, 38165260, 278415920, 2046924400, 15148345760, 112738423360, 843126957056, 6332299624282, 47737325577620, 361077477684436, 2739270870994736, 20836827035351596, 158883473753259752, 1214171997616258240
Offset: 0

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Author

Keywords

Comments

Cusick gives a general method of deriving recurrences for the r-th order Franel numbers (this is the sequence of third-order Franel numbers), with floor((r+3)/2) terms.
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
An identity of V. Strehl states that a(n) = Sum_{k = 0..n} C(n,k)^2 * binomial(2*k,n). Zhi-Wei Sun conjectured that for every n = 2,3,... the polynomial f_n(x) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*k,n) * x^(n-k) is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Conjecture: a(n) == 2 (mod n^3) iff n is prime. - Gary Detlefs, Mar 22 2013
a(p) == 2 (mod p^3) for any prime p since p | C(p,k) for all k = 1,...,p-1. - Zhi-Wei Sun, Aug 14 2013
a(n) is the maximal number of totally mixed Nash equilibria in games of 3 players, each with n+1 pure options. - Raimundas Vidunas, Jan 22 2014
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Diagonal of rational functions 1/(1 - x*y - y*z - x*z - 2*x*y*z), 1/(1 - x - y - z + 4*x*y*z), 1/(1 + y + z + x*y + y*z + x*z + 2*x*y*z), 1/(1 + x + y + z + 2*(x*y + y*z + x*z) + 4*x*y*z). - Gheorghe Coserea, Jul 04 2018
a(n) is the constant term in the expansion of ((1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 27 2019
Diagonal of rational function 1 / ((1-x)*(1-y)*(1-z) - x*y*z). - Seiichi Manyama, Jul 11 2020
Named after the Swiss mathematician Jérôme Franel (1859-1939). - Amiram Eldar, Jun 15 2021
It appears that a(n) is equal to the coefficient of (x*y*z)^n in the expansion of (1 + x + y - z)^n * (1 + x - y + z)^n * (1 - x + y + z)^n. Cf. A036917. - Peter Bala, Sep 20 2021

Examples

			O.g.f.: A(x) = 1 + 2*x + 10*x^2 + 56*x^3 + 346*x^4 + 2252*x^5 + ...
O.g.f.: A(x) = 1/(1-2*x) + 3!*x^2/(1-2*x)^4 + (6!/2!^3)*x^4/(1-2*x)^7 + (9!/3!^3)*x^6/(1-2*x)^10 + (12!/4!^3)*x^8/(1-2*x)^13 + ... - _Paul D. Hanna_, Oct 30 2010
Let g.f. A(x) = Sum_{n >= 0} a(n)*x^n/n!^3, then
A(x) = 1 + 2*x + 10*x^2/2!^3 + 56*x^3/3!^3 + 346*x^4/4!^3 + ... where
A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 + ...]^2. - _Paul D. Hanna_
		

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • Jérôme Franel, On a question of Laisant, Intermédiaire des Mathématiciens, vol 1 1894 pp 45-47
  • H. W. Gould, Combinatorial Identities, Morgantown, 1972, (X.14), p. 56.
  • Murray Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 148-149.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 193.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A002893, A052144, A005260, A096191, A033581, A189791. Second row of array A094424.
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.
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Column k=3 of A372307.

Programs

  • Haskell
    a000172 = sum . map a000578 . a007318_row
    -- Reinhard Zumkeller, Jan 06 2013
    
  • Maple
    A000172 := proc(n)
        add(binomial(n,k)^3,k=0..n) ;
    end proc:
    seq(A000172(n),n=0..10) ; # R. J. Mathar, Jul 26 2014
    A000172_list := proc(len) series(hypergeom([], [1, 1], x)^2, x, len);
    seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
    A000172_list(21); # Peter Luschny, May 31 2017
  • Mathematica
    Table[Sum[Binomial[n,k]^3,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Aug 24 2011 *)
    Table[ HypergeometricPFQ[{-n, -n, -n}, {1, 1}, -1], {n, 0, 20}]  (* Jean-François Alcover, Jul 16 2012, after symbolic sum *)
    a[n_] := Sum[ Binomial[2k, n]*Binomial[2k, k]*Binomial[2(n-k), n-k], {k, 0, n}]/2^n; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 20 2013, after Zhi-Wei Sun *)
    a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/3, 2/3, 1, 27 x^2 / (1 - 2 x)^3] / (1 - 2 x), {x, 0, n}]; (* Michael Somos, Jul 16 2014 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)/(1-2*x+x*O(x^n))^(3*m+1)),n)} \\ Paul D. Hanna, Oct 30 2010
    
  • PARI
    {a(n)=n!^3*polcoeff(sum(m=0,n,x^m/m!^3+x*O(x^n))^2,n)} \\ Paul D. Hanna, Jan 19 2011
    
  • PARI
    A000172(n)={sum(k=0,(n-1)\2,binomial(n,k)^3)*2+if(!bittest(n,0),binomial(n,n\2)^3)} \\ M. F. Hasler, Sep 21 2015
    
  • Sage
    def A000172():
        x, y, n = 1, 2, 1
        while True:
            yield x
            n += 1
            x, y = y, (8*(n-1)^2*x + (7*n^2-7*n + 2)*y) // n^2
    a = A000172()
    [next(a) for i in range(21)]   # Peter Luschny, Oct 12 2013

Formula

A002893(n) = Sum_{m = 0..n} binomial(n, m)*a(m) [Barrucand].
Sum_{k = 0..n} C(n, k)^3 = (-1)^n*Integral_{x = 0..infinity} L_k(x)^3 exp(-x) dx. - from Askey's book, p. 43.
D-finite with recurrence (n + 1)^2*a(n+1) = (7*n^2 + 7*n + 2)*a(n) + 8*n^2*a(n-1) [Franel]. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jan 31 2001
a(n) ~ 2*3^(-1/2)*Pi^-1*n^-1*2^(3*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
O.g.f.: A(x) = Sum_{n >= 0} (3*n)!/n!^3 * x^(2*n)/(1 - 2*x)^(3*n+1). - Paul D. Hanna, Oct 30 2010
G.f.: hypergeom([1/3, 2/3], [1], 27 x^2 / (1 - 2x)^3) / (1 - 2x). - Michael Somos, Dec 17 2010
G.f.: Sum_{n >= 0} a(n)*x^n/n!^3 = [ Sum_{n >= 0} x^n/n!^3 ]^2. - Paul D. Hanna, Jan 19 2011
G.f.: A(x) = 1/(1-2*x)*(1+6*(x^2)/(G(0)-6*x^2)),
with G(k) = 3*(x^2)*(3*k+1)*(3*k+2) + ((1-2*x)^3)*((k+1)^2) - 3*(x^2)*((1-2*x)^3)*((k+1)^2)*(3*k+4)*(3*k+5)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
In 2011 Zhi-Wei Sun found the formula Sum_{k = 0..n} C(2*k,n)*C(2*k,k)*C(2*(n-k),n-k) = (2^n)*a(n) and proved it via the Zeilberger algorithm. - Zhi-Wei Sun, Mar 20 2013
0 = a(n)*(a(n+1)*(-2048*a(n+2) - 3392*a(n+3) + 768*a(n+4)) + a(n+2)*(-1280*a(n+2) - 2912*a(n+3) + 744*a(n+4)) + a(n+3)*(+288*a(n+3) - 96*a(n+4))) + a(n+1)*(a(n+1)*(-704*a(n+2) - 1232*a(n+3) + 288*a(n+4)) + a(n+2)*(-560*a(n+2) - 1372*a(n+3) + 364*a(n+4)) + a(n+3)*(+154*a(n+3) - 53*a(n+4))) + a(n+2)*(a(n+2)*(+24*a(n+2) + 70*a(n+3) - 20*a(n+4)) + a(n+3)*(-11*a(n+3) + 4*a(n+4))) for all n in Z. - Michael Somos, Jul 16 2014
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^3*C(n,k)^3 = C(n,r)^3*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^2. - Peter Luschny, May 31 2017
From Gheorghe Coserea, Jul 04 2018: (Start)
a(n) = Sum_{k=0..floor(n/2)} (n+k)!/(k!^3*(n-2*k)!) * 2^(n-2*k).
G.f. y=A(x) satisfies: 0 = x*(x + 1)*(8*x - 1)*y'' + (24*x^2 + 14*x - 1)*y' + 2*(4*x + 1)*y. (End)
a(n) = [x^n] (1 - x^2)^n*P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. See Gould, p. 56. - Peter Bala, Mar 24 2022
a(n) = (2^n/(4*Pi^2)) * Integral_{x,y=0..2*Pi} (1+cos(x)+cos(y)+cos(x+y))^n dx dy = (8^n/(Pi^2)) * Integral_{x,y=0..Pi} (cos(x)*cos(y)*cos(x+y))^n dx dy (Pla, 1995). - Amiram Eldar, Jul 16 2022
a(n) = Sum_{k = 0..n} m^(n-k)*binomial(n,k)*binomial(n+2*k,n)*binomial(2*k,k) at m = -4. Cf. A081798 (m = 1), A006480 (m = 0), A124435 (m = -1), A318109 (m = -2) and A318108 (m = -3). - Peter Bala, Mar 16 2023
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=2*(1 + 4*x)*T(x) + (-1 + 14*x + 24*x^2)*T'(x) + x*(1 + x)*(-1 + 8*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (4/243)*(1 - 8*x + 240*x^2 - 464*x^3 + 16*x^4);
g3 = -(8/19683)*(1 - 12*x - 480*x^2 + 3080*x^3 - 12072*x^4 + 4128*x^5 +
64*x^6);
which determine an elliptic surface with four singular fibers. (End)
From Peter Bala, Oct 31 2024: (Start)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^2 * binomial(n-1, k). Cf. A361716.
For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n + 1], [1, 1], -1). (End)

A005259 Apery (Apéry) numbers: Sum_{k=0..n} (binomial(n,k)*binomial(n+k,k))^2.

Original entry on oeis.org

1, 5, 73, 1445, 33001, 819005, 21460825, 584307365, 16367912425, 468690849005, 13657436403073, 403676083788125, 12073365010564729, 364713572395983725, 11111571997143198073, 341034504521827105445, 10534522198396293262825, 327259338516161442321485
Offset: 0

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Author

Keywords

Comments

Conjecture: For each n = 1,2,3,... the Apéry polynomial A_n(x) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
The expansions of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + 7038*x^4 + 144550*x^5 + ... both appear to have integer coefficients. See A267220. - Peter Bala, Jan 12 2016
Diagonal of the rational function R(x, y, z, w) = 1 / (1 - (w*x*y*z + w*x*y + w*z + x*y + x*z + y + z)); also diagonal of rational function H(x, y, z, w) = 1/(1 - w*(1+x)*(1+y)*(1+z)*(x*y*z + y*z + y + z + 1)). - Gheorghe Coserea, Jun 26 2018
Named after the French mathematician Roger Apéry (1916-1994). - Amiram Eldar, Jun 10 2021

Examples

			G.f. = 1 + 5*x + 73*x^2 + 1445*x^3 + 33001*x^4 + 819005*x^5 + 21460825*x^6 + ...
a(2) = (binomial(2,0) * binomial(2+0,0))^2 + (binomial(2,1) * binomial(2+1,1))^2 + (binomial(2,2) * binomial(2+2,2))^2 = (1*1)^2 + (2*3)^2 + (1*6)^2 = 1 + 36 + 36 = 73. - _Michael B. Porter_, Jul 14 2016
		

References

  • Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 137-153.
  • Wolfram Koepf, Hypergeometric Identities. Ch. 2 in Hypergeometric Summation: An Algorithmic Approach to Summation and Special Function Identities. Braunschweig, Germany: Vieweg, pp. 55, 119 and 146, 1998.
  • Maxim Kontsevich and Don Zagier, Periods, pp. 771-808 of B. Engquist and W. Schmid, editors, Mathematics Unlimited - 2001 and Beyond, 2 vols., Springer-Verlag, 2001.
  • Leonard Lipshitz and Alfred van der Poorten, "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), pp. 339-358.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Apéry's number or Apéry's constant zeta(3) is A002117. - N. J. A. Sloane, Jul 11 2023
Related to diagonal of rational functions: A268545-A268555.
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.
Cf. A092826 (prime terms).

Programs

  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k)^2)); # Muniru A Asiru, Sep 28 2018
    
  • Haskell
    a005259 n = a005259_list !! n
    a005259_list = 1 : 5 : zipWith div (zipWith (-)
       (tail $ zipWith (*) a006221_list a005259_list)
       (zipWith (*) (tail a000578_list) a005259_list)) (drop 2 a000578_list)
    -- Reinhard Zumkeller, Mar 13 2014
    
  • Magma
    [&+[Binomial(n, k) ^2 *Binomial(n+k, k)^2: k in [0..n]]:n in  [0..17]]; // Marius A. Burtea, Jan 20 2020
    
  • Maple
    a := proc(n) option remember; if n=0 then 1 elif n=1 then 5 else (n^(-3))* ( (34*(n-1)^3 + 51*(n-1)^2 + 27*(n-1) +5)*a((n-1)) - (n-1)^3*a((n-1)-1)); fi; end;
    # Alternative:
    a := n -> hypergeom([-n, -n, 1+n, 1+n], [1, 1, 1], 1):
    seq(simplify(a(n)), n=0..17); # Peter Luschny, Jan 19 2020
  • Mathematica
    Table[HypergeometricPFQ[{-n, -n, n+1, n+1}, {1,1,1}, 1],{n,0,13}] (* Jean-François Alcover, Apr 01 2011 *)
    Table[Sum[(Binomial[n,k]Binomial[n+k,k])^2,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 15 2011 *)
    a[ n_] := SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ 1 / (1 - t (1 + x ) (1 + y ) (1 + z ) (x y z + (y + 1) (z + 1))), {t, 0, n}], {x, 0, n}], {y, 0, n}], {z, 0, n}]; (* Michael Somos, May 14 2016 *)
  • PARI
    a(n)=sum(k=0,n,(binomial(n,k)*binomial(n+k,k))^2) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Python
    def A005259(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= ((n+k+1)*(n-k))**2
            m //=(k+1)**4
        return g # Chai Wah Wu, Oct 02 2022

Formula

D-finite with recurrence (n+1)^3*a(n+1) = (34*n^3 + 51*n^2 + 27*n + 5)*a(n) - n^3*a(n-1), n >= 1.
Representation as a special value of the hypergeometric function 4F3, in Maple notation: a(n)=hypergeom([n+1, n+1, -n, -n], [1, 1, 1], 1), n=0, 1, ... - Karol A. Penson Jul 24 2002
a(n) = Sum_{k >= 0} A063007(n, k)*A000172(k). A000172 = Franel numbers. - Philippe Deléham, Aug 14 2003
G.f.: (-1/2)*(3*x - 3 + (x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2 - 7*x + 1)*(x+1)^(-3)*(x^2 - 34*x + 1)^(1/2)+(1/2)*(x^3 + 30*x^2 - 24*x + 1)*(x+1)^(-3))^2. - Mark van Hoeij, Oct 29 2011
Let g(x, y) = 4*cos(2*x) + 8*sin(y)*cos(x) + 5 and let P(n,z) denote the Legendre polynomial of degree n. Then G. A. Edgar posted a conjecture of Alexandru Lupas that a(n) equals the double integral 1/(4*Pi^2)*int {y = -Pi..Pi} int {x = -Pi..Pi} P(n,g(x,y)) dx dy. (Added Jan 07 2015: Answered affirmatively in Math Overflow question 178790) - Peter Bala, Mar 04 2012; edited by G. A. Edgar, Dec 10 2016
a(n) ~ (1+sqrt(2))^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 01 2012
a(n) = Sum_{k=0..n} C(n,k)^2 * C(n+k,k)^2. - Joerg Arndt, May 11 2013
0 = (-x^2+34*x^3-x^4)*y''' + (-3*x+153*x^2-6*x^3)*y'' + (-1+112*x-7*x^2)*y' + (5-x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Peter Bala, Jan 18 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j) * C(n,k)^2 * C(n+k,k)^2 * C(n,j) * C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} C(n,k) * C(n+k,k) * C(k,j)^3 (see Koepf, p. 55).
a(n) = Sum_{0 <= j, k <= n} C(n,k)^2 * C(n,j)^2 * C(3*n-j-k,2*n) (see Koepf, p. 119).
Diagonal coefficients of the rational function 1/((1 - x - y)*(1 - z - t) - x*y*z*t) (Straub, 2014). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 22 2020
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k). - Peter Bala, Jul 18 2024
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,k)^2 * C(n,j)^2 * C(k+j,k), see Labelle et al. link. - Max Alekseyev, Mar 12 2025

A005258 Apéry numbers: a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,k).

Original entry on oeis.org

1, 3, 19, 147, 1251, 11253, 104959, 1004307, 9793891, 96918753, 970336269, 9807518757, 99912156111, 1024622952993, 10567623342519, 109527728400147, 1140076177397091, 11911997404064793, 124879633548031009, 1313106114867738897, 13844511065506477501
Offset: 0

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Author

Keywords

Comments

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Equals the main diagonal of square array A108625. - Paul D. Hanna, Jun 14 2005
This sequence is t_5 in Cooper's paper. - Jason Kimberley, Nov 25 2012
Conjecture: For each n=1,2,3,... the polynomial a_n(x) = Sum_{k=0..n} C(n,k)^2*C(n+k,k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Diagonal of rational functions 1/(1 - x - x*y - y*z - x*z - x*y*z), 1/(1 + y + z + x*y + y*z + x*z + x*y*z), 1/(1 - x - y - z + x*y + x*y*z), 1/(1 - x - y - z + y*z + x*z - x*y*z). - Gheorghe Coserea, Jul 07 2018

Examples

			G.f. = 1 + 3*x + 19*x^2 + 147*x^3 + 1251*x^4 + 11253*x^5 + 104959*x^6 + ...
		

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • S. Melczer, An Invitation to Analytic Combinatorics, 2021; p. 129.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A007318.
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.

Programs

  • GAP
    a:=n->Sum([0..n],k->(-1)^(n-k)*Binomial(n,k)*Binomial(n+k,k)^2);;
    A005258:=List([0..20],n->a(n));; # Muniru A Asiru, Feb 11 2018
    
  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k))); # Muniru A Asiru, Jul 29 2018
    
  • Haskell
    a005258 n = sum [a007318 n k ^ 2 * a007318 (n + k) k | k <- [0..n]]
    -- Reinhard Zumkeller, Jan 04 2013
    
  • Magma
    [&+[Binomial(n,k)^2 * Binomial(n+k,k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Nov 28 2018
    
  • Maple
    with(combinat): seq(add((multinomial(n+k,n-k,k,k))*binomial(n,k), k=0..n), n=0..18); # Zerinvary Lajos, Oct 18 2006
    a := n -> binomial(2*n, n)*hypergeom([-n, -n, -n], [1, -2*n], 1):
    seq(simplify(a(n)), n=0..20); # Peter Luschny, Feb 10 2018
  • Mathematica
    a[n_] := HypergeometricPFQ[ {n+1, -n, -n}, {1, 1}, 1]; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Jan 20 2012, after Vladeta Jovovic *)
    Table[Sum[Binomial[n,k]^2 Binomial[n+k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Aug 25 2019 *)
  • PARI
    {a(n) = if( n<0, -(-1)^n * a(-1-n), sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)))} /* Michael Somos, Sep 18 2013 */
    
  • Python
    def A005258(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= (n+k+1)*(n-k)**2
            m //= (k+1)**3
        return g # Chai Wah Wu, Oct 02 2022

Formula

a(n) = hypergeom([n+1, -n, -n], [1, 1], 1). - Vladeta Jovovic, Apr 24 2003
D-finite with recurrence: (n+1)^2 * a(n+1) = (11*n^2+11*n+3) * a(n) + n^2 * a(n-1). - Matthijs Coster, Apr 28 2004
Let b(n) be the solution to the above recurrence with b(0) = 0, b(1) = 5. Then the b(n) are rational numbers with b(n)/a(n) -> zeta(2) very rapidly. The identity b(n)*a(n-1) - b(n-1)*a(n) = (-1)^(n-1)*5/n^2 leads to a series acceleration formula: zeta(2) = 5 * Sum_{n >= 1} 1/(n^2*a(n)*a(n-1)) = 5*(1/(1*3) + 1/(2^2*3*19) + 1/(3^2*19*147) + ...). Similar results hold for the constant e: see A143413. - Peter Bala, Aug 14 2008
G.f.: hypergeom([1/12, 5/12],[1], 1728*x^5*(1-11*x-x^2)/(1-12*x+14*x^2+12*x^3+x^4)^3) / (1-12*x+14*x^2+12*x^3+x^4)^(1/4). - Mark van Hoeij, Oct 25 2011
a(n) ~ ((11+5*sqrt(5))/2)^(n+1/2)/(2*Pi*5^(1/4)*n). - Vaclav Kotesovec, Oct 05 2012
1/Pi = 5*(sqrt(47)/7614)*Sum_{n>=0} (-1)^n a(n)*binomial(2n,n)*(682n+71)/15228^n. [Cooper, equation (4)] - Jason Kimberley, Nov 26 2012
a(-1 - n) = (-1)^n * a(n) if n>=0. a(-1 - n) = -(-1)^n * a(n) if n<0. - Michael Somos, Sep 18 2013
0 = a(n)*(a(n+1)*(+4*a(n+2) + 83*a(n+3) - 12*a(n+4)) + a(n+2)*(+32*a(n+2) + 902*a(n+3) - 147*a(n+4)) + a(n+3)*(-56*a(n+3) + 12*a(n+4))) + a(n+1)*(a(n+1)*(+17*a(n+2) + 374*a(n+3) - 56*a(n+4)) + a(n+2)*(+176*a(n+2) + 5324*a(n+3) - 902*a(n+4)) + a(n+3)*(-374*a(n+3) + 83*a(n+4))) + a(n+2)*(a(n+2)*(-5*a(n+2) - 176*a(n+3) + 32*a(n+4)) + a(n+3)*(+17*a(n+3) - 4*a(n+4))) for all n in Z. - Michael Somos, Aug 06 2016
a(n) = binomial(2*n, n)*hypergeom([-n, -n, -n],[1, -2*n], 1). - Peter Luschny, Feb 10 2018
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*binomial(n+k,k)^2. - Peter Bala, Feb 10 2018
G.f. y=A(x) satisfies: 0 = x*(x^2 + 11*x - 1)*y'' + (3*x^2 + 22*x - 1)*y' + (x + 3)*y. - Gheorghe Coserea, Jul 01 2018
From Peter Bala, Jan 15 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(j+k)*C(n,k)*C(n+k,k)^2*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)^2*C(n+k,k)*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^j*C(n,k)^2*C(n,j)*C(3*n-j-k,2*n). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 1. At m = 2 we get the Apéry numbers A005259. - Peter Bala, Dec 22 2020
a(n) = (-1)^n*Sum_{j=0..n} (1 - 5*j*H(j) + 5*j*H(n - j))*binomial(n, j)^5, where H(n) denotes the n-th harmonic number, A001008/A002805. (Paule/Schneider). - Peter Luschny, Jul 23 2021
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=(3 + x)*T(x) + (-1 + 22*x + 3*x^2)*T'(x) + x*(-1 + 11*x + x^2)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = 3*(1 - 12*x + 14*x^2 + 12*x^3 + x^4);
g3 = 1 - 18*x + 75*x^2 + 75*x^4 + 18*x^5 + x^6;
which determine an elliptic surface with four singular fibers. (End)
Conjecture: a(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n, k). - Peter Bala, Jul 08 2024

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

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Author

Keywords

Comments

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

Examples

			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
		

References

  • 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).

Crossrefs

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.

Programs

  • Magma
    [&+[Binomial(n, k)^2 * Binomial(2*k, k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 26 2018
    
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {a(n) = if( n<0, 0, n!^2 * polcoeff( besseli(0, 2*x + O(x^(2*n+1)))^3, 2*n))};
    
  • PARI
    {a(n) = sum(k=0, n, binomial(n, k)^2 * binomial(2*k, k))}; /* Michael Somos, Jul 25 2007 */
    
  • PARI
    {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
    
  • PARI
    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
    
  • SageMath
    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

Formula

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
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