A082601 -id:A082601 - cp's OEIS frontend

A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852

Offset: 0

N. J. A. Sloane

The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Also (for n > 1) number of ordered trees with n+1 edges and having all leaves at level three. Example: a(4)=2 because we have two ordered trees with 5 edges and having all leaves at level three: (i) one edge emanating from the root, at the end of which two paths of length two are hanging and (ii) one path of length two emanating from the root, at the end of which three edges are hanging. - Emeric Deutsch, Jan 03 2004
a(n) is the number of compositions of n-2 with no part greater than 3. Example: a(5)=4 because we have 1+1+1 = 1+2 = 2+1 = 3. - Emeric Deutsch, Mar 10 2004
Let A denote the 3 X 3 matrix [0,0,1;1,1,1;0,1,0]. a(n) corresponds to both the (1,2) and (3,1) entries in A^n. - Paul Barry, Oct 15 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-2, with k=1, r=2. - Vladimir Baltic, Jan 17 2005
Number of binary sequences of length n-3 that have no three consecutive 0's. Example: a(7)=13 because among the 16 binary sequences of length 4 only 0000, 0001 and 1000 have 3 consecutive 0's. - Emeric Deutsch, Apr 27 2006
Therefore, the complementary sequence to A050231 (n coin tosses with a run of three heads). a(n) = 2^(n-3) - A050231(n-3) - Toby Gottfried, Nov 21 2010
Convolved with the Padovan sequence = row sums of triangle A153462. - Gary W. Adamson, Dec 27 2008
For n > 1: row sums of the triangle in A157897. - Reinhard Zumkeller, Jun 25 2009
a(n+2) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 0, 1; 1, 0, 0] or [1, 1, 0; 1, 0, 1; 1, 0, 0] or [1, 1, 1; 1, 0, 0; 0, 1, 0] or [1, 0, 1; 1, 0, 0; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 0, 1; 1, 1, 1; 0, 1, 0], [0, 1, 0; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 0, 1; 0, 1, 1] or [0, 1, 0; 0, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Also row sums of A082601 and of A082870. - Reinhard Zumkeller, Apr 13 2014
Least significant bits are given in A021913 (a(n) mod 2 = A021913(n)). - Andres Cicuttin, Apr 04 2016
The nonnegative powers of the tribonacci constant t = A058265 are t^n = a(n)*t^2 + (a(n-1) + a(n-2))*t + a(n-1)*1, for  n >= 0, with  a(-1) = 1 and a(-2) = -1. This follows from the recurrences derived from t^3 = t^2 + t + 1. See the example in A058265 for the first nonnegative powers. For the negative powers see A319200. - Wolfdieter Lang, Oct 23 2018
The term "tribonacci number" was coined by Mark Feinberg (1963), a 14-year-old student in the 9th grade of the Susquehanna Township Junior High School in Pennsylvania. He died in 1967 in a motorcycle accident. - Amiram Eldar, Apr 16 2021
Andrews, Just, and Simay (2021, 2022) remark that it has been suggested that this sequence is mentioned in Charles Darwin's Origin of Species as bearing the same relation to elephant populations as the Fibonacci numbers do to rabbit populations. - N. J. A. Sloane, Jul 12 2022

			G.f. = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 24*x^8 + 44*x^9 + 81*x^10 + ...

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Index entries for linear recurrences with constant coefficients, signature (1,1,1).

Cf. A000045, A000078, A000213, A000931, A001590 (first differences, also a(n)+a(n+1)), A001644, A008288 (tribonacci triangle), A008937 (partial sums), A021913, A027024, A027083, A027084, A046738 (Pisano periods), A050231, A054668, A062544, A063401, A077902, A081172, A089068, A118390, A145027, A153462, A230216.
A057597 is this sequence run backwards: A057597(n) = a(1-n).
Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Partitions: A240844 and A117546.
Cf. also A092836 (subsequence of primes), A299399 = A092835 + 1 (indices of primes).

a:=[0,0,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 24 2018

a000073 n = a000073_list !! n
a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail
                          (zipWith (+) a000073_list $ tail a000073_list))
-- Reinhard Zumkeller, Dec 12 2011

[n le 3 select Floor(n/3) else Self(n-1)+Self(n-2)+Self(n-3): n in [1..70]]; // Vincenzo Librandi, Jan 29 2016

a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1,3]:
seq(a(n), n=0..40);  # Alois P. Heinz, Dec 19 2016
# second Maple program:
A000073:=proc(n) option remember; if n <= 1 then 0 elif n=2 then 1 else procname(n-1)+procname(n-2)+procname(n-3); fi; end; # N. J. A. Sloane, Aug 06 2018

CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Array[a, 36, 0] (* Robert G. Wilson v, Nov 07 2010 *)
LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, May 24 2011 *)
a[n_] := SeriesCoefficient[If[ n < 0, x/(1 + x + x^2 - x^3), x^2/(1 - x - x^2 - x^3)], {x, 0, Abs @ n}] (* Michael Somos, Jun 01 2013 *)
Table[-RootSum[-1 - # - #^2 + #^3 &, -#^n - 9 #^(n + 1) + 4 #^(n + 2) &]/22, {n, 0, 20}] (* Eric W. Weisstein, Nov 09 2017 *)

A000073[0]:0$
A000073[1]:0$
A000073[2]:1$
A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$
  makelist(A000073[n], n, 0, 40);  /* Emanuele Munarini, Mar 01 2011 */

{a(n) = polcoeff( if( n<0, x / ( 1 + x + x^2 - x^3), x^2 / ( 1 - x - x^2 - x^3) ) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Sep 03 2007 */

my(x='x+O('x^99)); concat([0, 0], Vec(x^2/(1-x-x^2-x^3))) \\ Altug Alkan, Apr 04 2016

a(n)=([0,1,0;0,0,1;1,1,1]^n)[1,3] \\ Charles R Greathouse IV, Apr 18 2016, simplified by M. F. Hasler, Apr 18 2018

def a(n, adict={0:0, 1:0, 2:1}):
    if n in adict:
        return adict[n]
    adict[n]=a(n-1)+a(n-2)+a(n-3)
    return adict[n] # David Nacin, Mar 07 2012
from functools import cache
@cache
def A000073(n: int) -> int:
    if n <= 1: return 0
    if n == 2: return 1
    return A000073(n-1) + A000073(n-2) + A000073(n-3) # Peter Luschny, Nov 21 2022

G.f.: x^2/(1 - x - x^2 - x^3).
G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+2) *[ Product_{k = 1..n} (k + k*x + x^2)/(1 + k*x + k*x^2) ] = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + ... may be proved by the method of telescoping sums. - Peter Bala, Jan 04 2015
a(n+1)/a(n) -> A058265.  a(n-1)/a(n) -> A192918.
a(n) = central term in M^n * [1 0 0] where M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 1]. (M^n * [1 0 0] = [a(n-1) a(n) a(n+1)].) a(n)/a(n-1) tends to the tribonacci constant, 1.839286755... = A058265, an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n+2) = Sum_{k=0..n} T(n-k, k), where T(n, k) = trinomial coefficients (A027907). - Paul Barry, Feb 15 2005
A001590(n) = a(n+1) - a(n); A001590(n) = a(n-1) + a(n-2) for n > 1; a(n) = (A000213(n+1) - A000213(n))/2; A000213(n-1) = a(n+2) - a(n) for n > 0. - Reinhard Zumkeller, May 22 2006
Let C = the tribonacci constant, 1.83928675...; then C^n = a(n)*(1/C) + a(n+1)*(1/C + 1/C^2) + a(n+2)*(1/C + 1/C^2 + 1/C^3). Example: C^4 = 11.444...= 2*(1/C) + 4*(1/C + 1/C^2) + 7*(1/C + 1/C^2 + 1/C^3). - Gary W. Adamson, Nov 05 2006
a(n) = j*C^n + k*r1^n + L*r2^n where C is the tribonacci constant (C = 1.8392867552...), real root of x^3-x^2-x-1=0, and r1 and r2 are the two other roots (which are complex), r1 = m+p*i and r2 = m-p*i, where i = sqrt(-1), m = (1-C)/2 (m = -0.4196433776...) and p = ((3*C-5)*(C+1)/4)^(1/2) = 0.6062907292..., and where j = 1/((C-m)^2 + p^2) = 0.1828035330..., k = a+b*i, and L = a-b*i, where a = -j/2 = -0.0914017665... and b = (C-m)/(2*p*((C-m)^2 + p^2)) = 0.3405465308... . - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n+1) = 3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3), c=(586+102*sqrt(33))^(1/3). Round to the nearest integer. - Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009
a(n) = round(3*((a+b+1)/3)^n/(a^2+b^2+4)) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3).. - Anton Nikonov
Another form of the g.f.: f(z) = (z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as a sum: a(n) = Sum_{i=0..floor((n-2)/4)} ((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i)) - Sum_{i=0..floor((n-3)/4)} ((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i)) with natural convention: Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+2) = Sum_{k=0..n} Sum_{i=k..n, mod(4*k-i,3)=0} binomial(k,(4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1,k-1). - Vladimir Kruchinin, Aug 18 2010
a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4). - Gary Detlefs, Sep 13 2010
Sum_{k=0..2*n} a(k+b)*A027907(n,k) = a(3*n+b), b >= 0 (see A099464, A074581).
a(n) = 2*a(n-1) - a(n-4), with a(0)=a(1)=0, a(2)=a(3)=1. - Vincenzo Librandi, Dec 20 2010
Starting (1, 2, 4, 7, ...) is the INVERT transform of (1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x + x^2)/( x*(4*k+3 + x + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013
a(n+2) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2*j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
Sum_{k=0..n} (n-k)*a(k) = (a(n+2) + a(n+1) - n - 1)/2. See A062544. - Yichen Wang, Aug 20 2020
a(n) = A008937(n-1) - A008937(n-2) for n >= 2. - Peter Luschny, Aug 20 2020
From Yichen Wang, Aug 27 2020: (Start)
Sum_{k=0..n} a(k) = (a(n+2) + a(n) - 1)/2. See A008937.
Sum_{k=0..n} k*a(k) = ((n-1)*a(n+2) - a(n+1) + n*a(n) + 1)/2. See A337282. (End)
For n > 1, a(n) = b(n) where b(1) = 1 and then b(n) = Sum_{k=1..n-1} b(n-k)*A000931(k+2). - J. Conrad, Nov 24 2022
Conjecture: the congruence a(n*p^(k+1)) + a(n*p^k) + a(n*p^(k-1)) == 0 (mod p^k) holds for positive integers k and n and for all the primes p listed in A106282. - Peter Bala, Dec 28 2022
Sum_{k=0..n} k^2*a(k) = ((n^2-4*n+6)*a(n+1) - (2*n^2-2*n+5)*a(n) + (n^2-2*n+3)*a(n-1) - 3)/2. - Prabha Sivaramannair, Feb 10 2024
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r^2-2*r-1). - Fabian Pereyra, Nov 23 2024

Minor edits by M. F. Hasler, Apr 18 2018

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

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

N. J. A. Sloane, Simon Plouffe

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

			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

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Index entries for sequences of k-nomial coefficients
Index entries for "core" sequences

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.

a002426 n = a027907 n n  -- Reinhard Zumkeller, Jan 22 2013

P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2)^n)): n in [0..26]]; // Bruno Berselli, Jul 05 2011

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

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 *)

trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
makelist(trinomial(n,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */

makelist(ultraspherical(n,-n,-1/2),n,0,12); /* Emanuele Munarini, Dec 20 2016 */

{a(n) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, n))};

/* as lattice paths: same as in A092566 but use */
steps=[[2, 0], [0, 2], [1, 1]];
/* Joerg Arndt, Jul 01 2011 */

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

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

A002426 = lambda n: hypergeometric([-n/2, (1-n)/2], [1], 4)
[simplify(A002426(n)) for n in (0..29)]
# Peter Luschny, Sep 17 2014

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

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

Row sums are tribonacci numbers.
From Gary W. Adamson, Nov 15 2016: (Start)
With an alternative format:
1, 0, 0, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, 0, ...
1, 2, 3, 2, 1, 0, 0, ...
1, 3, 6, 7, 6, 3, 1, ...
... (where the k-th row is (1 + x + x^2)^k), let q(x) = (r(x) * r(x^3) * r(x^9) * r(x^27) * ...). Then q(x) is the binomial sequence beginning (1, k, ...). Example: (1, 3, 6, 10, ...) = q(x) with r(x) = (1, 3, 6, 7, 3, 1, 0, 0, 0). (End)

The rows of T are essentially the antidiagonals of A027907 (trinomial coefficients). Reversing the rows produces A078803. Row sums: A000073.

Also, the diagonals of T are essentially the rows of A027907, so diagonal sums = 3^n. Antidiagonal sums are essentially A060961 (number of ordered partitions of n into 1's, 3's and 5's). - Gerald McGarvey, May 13 2005

cp's OEIS Frontend

A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

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

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A082870 Tribonacci array.

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A078802 Triangular array T given by T(n,k) = number of 01-words of length n containing k 1's, no three of which are consecutive.

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