A024493 -id:A024493 - cp's OEIS frontend

A094717 a(n) = n! * Sum_{i+2j+3k=n} 1/(i!(2j)!(3k)!).

1, 1, 2, 5, 12, 36, 113, 351, 1080, 3281, 9882, 29646, 88817, 266085, 797526, 2391485, 7173360, 21520080, 64563521, 193700403, 581120892, 1743392201, 5230206126, 15690618378, 47071766561, 141215033961, 423644570442, 1270932914165, 3812797945332, 11438393835996

Offset: 0

Benoit Cloitre, May 23 2004

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-12,10,-6,12,-9).

Cf. A024493, A049347, A057083, A057682, A094715, A099837.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-5*x+8*x^2-5*x^3+2*x^4-2*x^5)/((1-x)*(1-3*x)*(1+x+x^2)*(1-3*x+3*x^2)) )); // G. C. Greubel, Jul 14 2023

A094717_list := proc(n) local i; exp(z)*cosh(z)*(exp(z)+2*exp(-z/2)* cos(z*sqrt(3/4)))/3; series(%,z,n+2); seq(simplify(i!*coeff(%,z,i)),i=0..n) end: A094717_list(27); # Peter Luschny, Jul 11 2012

a[n_]:= n! Sum[Boole[i +2j +3k ==n]/(i! (2j)! (3k)!), {i,0,n}, {j,0,n}, {k,0,n}]; Table[a[n], {n,0,27}] (* Jean-François Alcover, Jul 06 2019 *)
LinearRecurrence[{6,-12,10,-6,12,-9}, {1,1,2,5,12,36}, 40] (* G. C. Greubel, Jul 14 2023 *)

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,n,if(n-i-2*j-3*k,0,n!/(i)!/(2*j)!/(3*k)!))))

def A094717_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-5*x+8*x^2-5*x^3+2*x^4-2*x^5)/((1-x)*(1-3*x)*(1+x+x^2)*(1-3*x+3*x^2)) ).list()
A094717_list(40) # G. C. Greubel, Jul 14 2023

Limit_{n->oo} a(n)/3^n = 1/6.
E.g.f.: exp(z)*cosh(z)*(exp(z) + 2*exp(-z/2)*cos(z*sqrt(3/4)))/3. - Peter Luschny, Jul 11 2012
G.f.: (1-5*x+8*x^2-5*x^3+2*x^4-2*x^5)/((1-x)*(1-3*x)*(1+x+x^2)*(1-3*x+3*x^2)). - Colin Barker, Dec 24 2012
From G. C. Greubel, Jul 14 2023: (Start)
a(n) = (1/6)*(1 + 3^n + 2*A049347(n) + A049347(n-1) + 2*A057083(n) - 3*A057083(n-1)).
a(n) = (1/6)*(1 + 3^n + A099837(n+3) + A057682(n+3)). (End)

A098172 Triangle T(n,k) with diagonals T(n,n-k) = binomial(n,3k).

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 4, 1, 0, 0, 0, 0, 10, 1, 0, 0, 0, 0, 1, 20, 1, 0, 0, 0, 0, 0, 7, 35, 1, 0, 0, 0, 0, 0, 0, 28, 56, 1, 0, 0, 0, 0, 0, 0, 1, 84, 84, 1, 0, 0, 0, 0, 0, 0, 0, 10, 210, 120, 1, 0, 0, 0, 0, 0, 0, 0, 0, 55, 462, 165, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 220, 924, 220, 1

Offset: 0

Paul Barry, Aug 30 2004

Row sums are A024493.
From R. J. Mathar, Mar 22 2013: (Start)
The matrix inverse starts
  1;
  0, 1;
  0, 0,      1;
  0, 0,     -1,     1;
  0, 0,      4,    -4,     1;
  0, 0,    -40,    40,   -10,   1;
  0, 0,    796,  -796,   199, -20,   1;
  0, 0, -27580, 27580, -6895, 693, -35, 1;
  ... (End)

			Rows begin
  {1},
  {0,1},
  {0,0,1},
  {0,0,1,1},
  {0,0,0,4,1},
  {0,0,0,0,10,1},
  ...

Seiichi Manyama, Rows n = 0..139, flattened

Cf. A098158.

Flat(List([0..12], n-> List([0..n], k-> Binomial(n, 3*(n-k)) ))); # G. C. Greubel, Mar 15 2019

[[Binomial(n, 3*(n-k)): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Mar 15 2019

Table[Binomial[n, 3(n-k)], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 15 2019 *)

{T(n, k) = binomial(n, 3*(n-k))}; \\ G. C. Greubel, Mar 15 2019

[[binomial(n, 3*(n-k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Mar 15 2019

Triangle T(n, k) = binomial(n, 3(n-k)).

A138635 a(n) =3a(n-3)-3a(n-6)+2*a(n-9).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 2, 1, 3, 3, 2, 6, 5, 5, 11, 10, 11, 21, 21, 22, 42, 43, 43, 85, 86, 85, 171, 171, 170, 342, 341, 341, 683, 682, 683, 1365, 1365, 1366, 2730, 2731, 2731, 5461, 5462, 5461, 10923, 10923, 10922, 21846, 21845, 21845, 43691, 43690, 43691, 87381

Offset: 0

Paul Curtz, May 14 2008

As the recurrence shows, these are three interleaved sequences which obey recurrences b(n)=3*b(n-1)-3*b(n-2)+2*b(n-3), indicating that the b(n) equal their third differences.
These three sequences are A024495, A024494 (or A131708) and A024493 (or A130781).
Their starting "vectors" b(0,1,2) are 0,0,1 and 0,1,2 and 1,1,1, respectively, therefore linearly independent, such that other sequences with the same recursion as b(n) can be written as linear combinations of these.

Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,2).

Cf. A024493, A024494, A024495, A131708, A130781.

a(18*n) = 21*A133853(n).

G.f.: -x^2*(1+x^2-2*x^3+x^4-x^5+x^6)/((2*x^3-1)*(x^6-x^3+1)). - R. J. Mathar, May 17 2009

Edited by R. J. Mathar, May 17 2009

A290285 Determinant of circulant matrix of order 3 with entries in the first row (-1)^j * Sum_{k>=0} binomial(n,3*k+j), j=0,1,2.

Original entry on oeis.org

1, 0, 0, 62, 666, 5292, 39754, 307062, 2456244, 19825910, 159305994, 1274445900, 10184391946, 81430393590, 651443132340, 5212260963062, 41700950994186, 333607607822412, 2668815050206474, 21350337149539062, 170802697195263924, 1366424509598012150

Offset: 0

Vladimir Shevelev, Jul 26 2017

nonn

In the Shevelev link the author proved that, for even N>=2 and every n>=1, the determinant of circulant matrix of order N with entries in the first row being (-1)^j*Sum_{k>=0} binomial(n,N*k+j), j=0..N-1, is 0. This sequence shows what happens for the first odd N>2.

Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Wikipedia, Circulant matrix

Cf. A024493, A024495, A131708, A290286.

a:= n-> LinearAlgebra[Determinant](Matrix(3, shape=Circulant[seq(
        (-1)^j*add(binomial(n, 3*k+j), k=0..(n-j)/3), j=0..2)])):
seq(a(n), n=0..25);  # Alois P. Heinz, Jul 27 2017

ro[n_] := Table[(-1)^j Sum[Binomial[n, 3k+j], {k, 0, n/3}], {j, 0, 2}];
M[n_] := Table[RotateRight[ro[n], m], {m, 0, 2}];
a[n_] := Det[M[n]];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Aug 09 2018 *)

mj(j,n) = (-1)^j*sum(k=0, n\3, binomial(n, 3*k+j));
a(n) = {m = matrix(3, 3); for (j=1, 3, m[1, j] = mj(j-1,n)); for (j=2, 3, m[2, j] = m[1, j-1]); m[2, 1] = m[1, 3]; for (j=2, 3, m[3, j] = m[2, j-1]); m[3, 1] = m[2, 3]; matdet(m);} \\ Michel Marcus, Jul 26 2017

from sympy.matrices import Matrix
from sympy import binomial
def mj(j, n):
    return (-1)**j*sum(binomial(n, 3*k + j) for k in range(n//3 + 1))
def a(n):
    m=Matrix(3, 3, [0]*9)
    for j in range(3):m[0, j]=mj(j, n)
    for j in range(1, 3):m[1, j]=m[0, j - 1]
    m[1, 0]=m[0, 2]
    for j in range(1, 3):m[2, j] = m[1, j - 1]
    m[2, 0]=m[1, 2]
    return m.det()
print([a(n) for n in range(22)]) # Indranil Ghosh, Jul 31 2017

G.f.: (1-12*x+48*x^2-73*x^3+6*x^4-60*x^5+736*x^6-576*x^7)/((1+x)*(-1+2*x)*(-1+8*x)* (1-x+x^2)*(1+2*x+4*x^2)*(1-4*x+16*x^2)). - Peter J. C. Moses, Jul 26 2017

More terms from Peter J. C. Moses, Jul 26 2017

A373905 a(n) = Sum_{k=0..floor(n/3)} binomial(n+3k,n-3k).

Original entry on oeis.org

1, 1, 1, 2, 8, 29, 86, 224, 554, 1381, 3556, 9382, 24901, 65737, 172321, 450017, 1174985, 3072365, 8044478, 21074012, 55199573, 144535714, 378366976, 990441502, 2592800365, 6787973872, 17771619370, 46527959417, 121813193825, 318910531073, 834913179137

Offset: 0

Seiichi Manyama, Jun 22 2024

Index entries for linear recurrences with constant coefficients, signature (6,-15,21,-15,6,-1).

Cf. A000930, A005251, A024493, A099099, A369845.

Cf. A107025, A373904, A373906.

a(n) = sum(k=0, n\3, binomial(n+3*k,n-3*k));

a(n) = 6*a(n-1) - 15*a(n-2) + 21*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).

G.f.: 1/(1 - x - x^3/(1 - x)^5).

A086953 Binomial transform of (-1)^mod(n,3) (A257075).

Original entry on oeis.org

1, 0, 0, 2, 6, 12, 22, 42, 84, 170, 342, 684, 1366, 2730, 5460, 10922, 21846, 43692, 87382, 174762, 349524, 699050, 1398102, 2796204, 5592406, 11184810, 22369620, 44739242, 89478486, 178956972, 357913942, 715827882, 1431655764, 2863311530, 5726623062

Offset: 0

Paul Barry, Jul 25 2003

Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

Cf. A024493, A024495, A131370, A131708, A257075.

Join[{1, a = 0, b = 0}, Table[c = 2^n - a + b; a = b; b = c, {n, 1, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
LinearRecurrence[{3,-3,2},{1,0,0},40] (* Harvey P. Dale, Aug 02 2017 *)

a(n+3)/2 = A024495(n+2). - corrected by Vladimir Shevelev, Aug 08 2017
a(n) = 0^n + Sum{k=0..floor((n-1)/3)} C(n-1, 3*k+2).
a(n) = Sum{k=0..n} C(n, k)(-1)^mod(k, 3).
G.f.: (1 - 3*x + 3*x^2)/((1 - 2*x)*(1 - x + x^2)). - Paul Barry, Dec 14 2004
From Vladimir Shevelev, Aug 02 2017: (Start)
a(n) = A024493(n) - A131708(n) + A024495(n);
a(n) = A024495(n) if and only if n == 1 (mod 3);
a(n) = A024495(n) - 1 if and only if n == 2 or 3 (mod 6);
a(n) = A024495(n) + 1 if and only if n == 0 or 5 (mod 6);
a(3*k+1) = 2*A024495(3*k). (End)
a(n) = A131370(n+1)/2. - Rick L. Shepherd, Aug 02 2017
3*a(n) = 2^n + 2*A057079(n+2). - R. J. Mathar, Aug 04 2017

A161869 Convergent of an infinite product of Pascal's triangles aerated by rows.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 33, 71, 160, 376, 912, 2256, 5633, 14093, 35170, 87344, 215680, 529568, 1293633, 3146515, 7627208, 18441476, 44510160, 107310480, 258566402, 622900466, 1500717220, 3616471960, 8717948688, 21023129472, 50713990918, 122374025914, 295366777856

Offset: 0

Gary W. Adamson, Jun 20 2009

The sequence may be the binomial transform of A024493 interleaved with zeros. A024493 = (1, 1, 1, 2, 5, 11, 22,...); so the conjecture succeeds through a(12) = A007318 * [1, 0, 1, 0, 1, 0, 2, 0, 5, 0, 11,...].

Calculating more terms of the sequence shows the above conjecture is incorrect. - Peter Bala, Jul 07 2015

			Pascal(1)     Pascal(2)    Pascal(3)
1              1            1
1 1            0 0          0 0
1 2 1          1 1 0        0 0 0
1 3 3 1        0 0 0 0      1 1 0 0
1 4 6 4 1      1 2 1 0 0    0 0 0 0 0
...
First columns of
Pascal(1):
1, 1, 1, 1, 1, 1, 1, 1,...
Pascal(1)*Pascal(2):
1, 1, 2, 4, 8, 16, 32, 64,...
Pascal(1)*Pascal(2)*Pascal(3):
1, 1, 2, 4, 8, 16, 33, 71,...
Pascal(1)*Pascal(2)*Pascal(3)*Pascal(4):
1, 1, 2, 4, 8, 16, 33, 71,...
...
converging to A161869.

Cf. A024493, A027826 (from first column of Pascal(2)^n as n -> inf).

#A161869
#define aerated Pascal matrices (note indexing starts at 1)
Pascal := proc (n) local i, j, r;
Matrix(33, 33, (i, j) -> (product(r-(mod(i-1, n)), r = 1 .. n-1))*binomial(floor((i-1)/n), j-1) )/factorial(n-1) end proc:
#it suffices to take the product of the first four aerated
#Pascal arrays to get 33 correct terms of the sequence
seq((Pascal(1).Pascal(2).Pascal(3).Pascal(4))(n, 1), n = 1 .. 33);
# Peter Bala, Jul 07 2015

From Peter Bala, Jul 07 2015: (Start)
Construct an infinite set of Pascal's triangles aerated by rows, denoted Pascal(1), Pascal(2), ..., where Pascal(1) = A007318, Pascal(2) is an aerated version of Pascal(1) with alternate rows 1, 3, 5, ... set equal to (0, 0, 0, ...), Pascal(3) is a further aeration of Pascal(1) with now two adjacent rows set equal to (0, 0, 0, ...), and so on.
The infinite product Pascal(1)*Pascal(2)*Pascal(3)*... is well-defined. This sequence is the first column of the infinite product - all the other entries in the product are zero. (End)

Terms a(12) through a(32) added by Peter Bala, Jul 07 2015

A329479 Number of degree n polynomials f with all nonzero coefficients equal to 1 such that f(k) is divisible by 3 for all integers k.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 6, 15, 30, 66, 121, 242, 462, 903, 1806, 3570, 7225, 14450, 29070, 58311, 116622, 233586, 466489, 932978, 1864590, 3727815, 7455630, 14908530, 29822521, 59645042, 119301006, 238612935, 477225870, 954473586, 1908903481, 3817806962, 7635526542

Offset: 1

Peter Kagey, Nov 13 2019

nonn

Equivalently, this counts strings of numbers of length n that start with a 1 and which yield a multiple of 3 when read in any base.

			For n = 7, the a(7) = 6 (0,1)-polynomials of degree seven such that f(0) = f(1) = f(2) = 0 (mod 3) are
x^7 + x^5 + x^3,
x^7 + x^6 + x^5 + x^4 + x^3 + x^2,
x^7 + x^5 + x,
x^7 + x^3 + x,
x^7 + x^6 + x^5 + x^4 + x^2 + x, and
x^7 + x^6 + x^4 + x^3 + x^2 + x.

Peter Kagey, Table of n, a(n) for n = 1..1000

Cf. A024493, A024495, A329126.

A008776(n) gives the number of polynomials of degree n+3 without the coefficient restriction.

a(2n) = A024495(n-1) * A024493(n).
a(2n+1) = A024495(n) * A024493(n).
Conjectured recurrence: a(n) = 2a(n-1) + 2a(n-2) - 5a(n-3) - 2a(n-4) + 10a(n-5) - 4a(n-6) - 4a(n-7) + 8a(n-8).

A102517 Expansion of (1+x^2)/((1-x+x^2)(1+2x^2)).

Original entry on oeis.org

1, 1, -1, -2, 1, 3, -2, -5, 5, 10, -11, -21, 22, 43, -43, -86, 85, 171, -170, -341, 341, 682, -683, -1365, 1366, 2731, -2731, -5462, 5461, 10923, -10922, -21845, 21845, 43690, -43691, -87381, 87382, 174763, -174763, -349526, 349525, 699051, -699050, -1398101, 1398101, 2796202, -2796203, -5592405

Offset: 0

Paul Barry, Jan 13 2005

Index entries for linear recurrences with constant coefficients, signature (1,-3,2,-2).

Cf. A001045, A078008.

CoefficientList[Series[(1+x^2)/((1-x+x^2)(1+2x^2)),{x,0,50}],x] (* or *) LinearRecurrence[{1,-3,2,-2},{1,1,-1,-2},50] (* Harvey P. Dale, Oct 28 2011 *)

G.f.: (1+x^2)^2/((1+x^2)^3+x^6)+x(1+x^2)/((1+x^2)^3+x^6).
a(n) = Sum_{k=0..floor(n/2)} T(n-k, k)*(-1)^k, T(n, k) = Sum_{i=0..k} C(n, i) (A008949).
a(n) = (-1)^(n/2)*(Sum_{k=0..floor(n/6)} C(n/2, 3*k))*(1+(-1)^n)/2 + (-1)^((n-1)/2)*(Sum_{k=0..floor((n+1)/6)} C((n+1)/2, 3*k+1))*(1-(-1)^n)/2.
a(n) = 2^(n/2)*(cos(Pi*n/2)/3+sqrt(2)*sin(Pi*n/2)/3)+cos(Pi*n/3+Pi/3)/3+sqrt(3)*sin(Pi*n/3+Pi/3)/3.
a(2*n) = (-1)^n*A024493(n); a(2*n+1) = (-1)^n*A024494(n).
a(0)=1, a(1)=1, a(2)=-1, a(3)=-2, a(n) = a(n-1)-3*a(n-2)+2*a(n-3)-2*a(n-4). - Harvey P. Dale, Oct 28 2011

A202349 Lexicographically earliest sequence such that the sequence and its first and second differences share no terms, and the 3rd differences are equal to the original sequence.

Original entry on oeis.org

1, 3, 9, 20, 39, 75, 148, 297, 597, 1196, 2391, 4779, 9556, 19113, 38229, 76460, 152919, 305835, 611668, 1223337, 2446677, 4893356, 9786711, 19573419, 39146836, 78293673, 156587349, 313174700, 626349399, 1252698795, 2505397588, 5010795177, 10021590357

Offset: 1

Eric Angelini, Jun 21 2016

The sequence is completely determined by its first 3 terms. If the first terms are x, y, z, then the following terms are 2*x-3*y+3*z, 6*x-7*y+6*z, 12*x-12*y+11*z, 22*x-21*y+21*z, 42*x-41*y+42*z, 84*x-84*y+85*z, 170*x-171*y+171*z, 342*x-343*y+342*z. - Giovanni Resta, Jun 21 2016
Is it a theorem that, if x,y,z = 1,3,9, the sequence has the desired properties, or is it just a conjecture? - N. J. A. Sloane, Jun 21 2016
From Charlie Neder, Jan 10 2019: (Start)
No two terms among this sequence and its first and second differences are equal.
Proof: Representing the first and second differences by b(n) and c(n), we have that a-b is [-1, -3, -2, 1, 3, 2] with period 6, a-c is [-3, -2, -1, 3, 2, 1] with period 6, and b-c is [-2, 1, 3, 2, -1, -3] with period 6. Therefore, no two terms at the same index are equal. Since the sequence is forced to grow exponentially, only the first few terms need to be checked to confirm that no two terms at different indices are equal, proving the criterion always holds. (End)

			  1 3 9  20  39  75  148   297   597   1196
   2 6 11  19  36  73   149   300   599
    4 5   8  17  37  76   151    299
     1  3   9  20  39   75    148   <-- the starting sequence

David A. Corneth, Table of n, a(n) for n = 1..3318
Mathematics Stack Exchange, Conjectured recurrence/generating function for a binomial sum.
Sequence and first differences include all numbers, etc.
Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

Cf. A024493, A130781, A069705 (inverse binomial transform assuming offset 0).

For many similar sequences, see the Index link.

d = Differences; i = Intersection; sol = Solve[d@ d@ d@ Array[x, 50] == Array[x, 47], Array[x, 47, 4]][[1]]; a = (Array[x, 50] /. sol) /. {x[1] -> 1, x[2] -> 3, x[3] -> 9}; Print["Check = ", {i[a, d@ a], i[a, d@ d@ a], i[d@ a, d@ d@ a]}]; a (* Giovanni Resta, Jun 21 2016 *)

first(n) = {n = max(n, 4); my(res = vector(n)); for(i = 1, 3, res[i] = 3^(i - 1)); for(i = 4, n, res[i] = 3 * res[i - 1] - 3 * res[i - 2] + 2 * res[i - 3]); res } \\ David A. Corneth, Jan 11 2019

Vec(x*(1 + 3*x^2) / ((1 - 2*x)*(1 - x + x^2)) + O(x^40)) \\ Colin Barker, Jan 12 2019

From Colin Barker, Jan 11 2019: (Start)
G.f.: x*(1 + 3*x^2) / ((1 - 2*x)*(1 - x + x^2)).
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) for n>2. (End)
a(n) = Sum_{k=0..n-1} binomial(n-1,k)*(2^k mod 7). - Fabio Visonà, Sep 05 2023

a(18)-a(33) from Giovanni Resta, Jun 21 2016

cp's OEIS Frontend

A094717 a(n) = n! * Sum_{i+2j+3k=n} 1/(i!*(2j)!*(3k)!).

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A098172 Triangle T(n,k) with diagonals T(n,n-k) = binomial(n,3k).

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A138635 a(n) =3*a(n-3)-3*a(n-6)+2*a(n-9).

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A290285 Determinant of circulant matrix of order 3 with entries in the first row (-1)^j * Sum_{k>=0} binomial(n,3*k+j), j=0,1,2.

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A373905 a(n) = Sum_{k=0..floor(n/3)} binomial(n+3*k,n-3*k).

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A086953 Binomial transform of (-1)^mod(n,3) (A257075).

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Mathematica

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A161869 Convergent of an infinite product of Pascal's triangles aerated by rows.

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Programs

Maple

Formula

A094717 a(n) = n! * Sum_{i+2j+3k=n} 1/(i!(2j)!(3k)!).

A138635 a(n) =3a(n-3)-3a(n-6)+2*a(n-9).

A373905 a(n) = Sum_{k=0..floor(n/3)} binomial(n+3k,n-3k).

A102517 Expansion of (1+x^2)/((1-x+x^2)(1+2x^2)).