A001400 -id:A001400 - cp's OEIS frontend

A117486 Expansion of 1/((1-x)(1-x^2)(1-x^3)*(1-x^4))^2.

1, 2, 5, 10, 20, 34, 59, 94, 149, 224, 334, 480, 685, 950, 1307, 1762, 2357, 3100, 4050, 5220, 6685, 8466, 10659, 13294, 16494, 20298, 24859, 30234, 36609, 44056, 52806, 62952, 74770, 88380, 104112, 122116, 142786, 166304, 193134, 223504, 257954, 296756, 340544

Offset: 0

Alford Arnold, Mar 22 2006

Molien series for S_4 X S_4, cf. A001400.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1,-4,4,4,2,0,-10,0,2,4,4,-4,-1,-2,1,2,-1).
Index entries for two-way infinite sequences

Column four of table A115994.
Cf. A001400, A001399, A117485, A117487.
Cf. A000027, A006918, A115994.

n:=4; G:=SymmetricGroup(n); H:=DirectProduct(G,G); MolienSeries(H);

CoefficientList[Series[1/((1-x)(1-x^2)(1-x^3)(1-x^4))^2,{x,0,50}],x] (* Harvey P. Dale, Jul 22 2012 *)

a(n):=floor(2*floor(-n/3)*cos(2*%pi*(n+1)/3)/81+(n+2)*cos(%pi*n/ 2)/128+(n+1)*(2835*(n^2+29*n+246)*(-1)^n+6*n^6+414*n^5+11556*n^4 +166944*n^3 +1320045*n^2+5489625*n+10008110)/ 17418240+1/2); /* Tani Akinari, Nov 14 2012 */

Vec(1 / ((1 - x)^8*(1 + x)^4*(1 + x^2)^2*(1 + x + x^2)^2) + O(x^40)) \\ Colin Barker, Apr 07 2019

G.f.: 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4))^2.

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) - 4*a(n-5) + 4*a(n-6) + 4*a(n-7) + 2*a(n-8) - 10*a(n-10) + 2*a(n-12) + 4*a(n-13) + 4*a(n-14) - 4*a(n-15) - a(n-16) - 2*a(n-17) + a(n-18) + 2*a(n-19) - a(n-20) for n>19. - Colin Barker, Apr 07 2019

Entry revised by N. J. A. Sloane, Mar 10 2007

A347586 Number of partitions of n into at most 4 distinct parts.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 26, 31, 36, 43, 49, 57, 65, 75, 84, 96, 107, 121, 134, 150, 165, 184, 201, 222, 242, 266, 288, 315, 340, 370, 398, 431, 462, 499, 533, 573, 611, 655, 696, 744, 789, 841, 890, 946, 999, 1060, 1117, 1182, 1244, 1314, 1380, 1455

Offset: 0

Ilya Gutkovskiy, Sep 08 2021

Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-2,0,0,1,1,-1).

Cf. A000578, A001400, A014591, A065033, A347587, A347588.

nmax = 60; CoefficientList[Series[Sum[x^(k (k + 1)/2)/Product[(1 - x^j), {j, 1, k}], {k, 0, 4}], {x, 0, nmax}], x]
Join[{1}, LinearRecurrence[{1, 1, 0, 0, -2, 0, 0, 1, 1, -1}, {1, 1, 2, 2, 3, 4, 5, 6, 8, 10}, 60]]

G.f.: Sum_{k=0..4} x^(k*(k + 1)/2) / Product_{j=1..k} (1 - x^j).

a(n) ~ A000578(n)/144. - Stefano Spezia, Sep 08 2021

A014126 Number of partitions of 2*n into at most 4 parts.

Original entry on oeis.org

1, 2, 5, 9, 15, 23, 34, 47, 64, 84, 108, 136, 169, 206, 249, 297, 351, 411, 478, 551, 632, 720, 816, 920, 1033, 1154, 1285, 1425, 1575, 1735, 1906, 2087, 2280, 2484, 2700, 2928, 3169, 3422, 3689, 3969, 4263, 4571, 4894, 5231, 5584, 5952, 6336, 6736, 7153, 7586

Offset: 0

N. J. A. Sloane

Bisection of A001400.
Molien series for 4-dimensional group of structure S_4 X C_2 and order 48, arising from complete weight enumerators of even trace-Hermitian self-dual additive codes over GF(4) containing the all-ones vector.
Partial sums of A156040. - Bob Selcoe, Feb 08 2014

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
L. Colmenarejo, Combinatorics on several families of Kronecker coefficients related to plane partitions, arXiv:1604.00803 [math.CO], 2016. See Table 1 p. 5.
H. R. Henze and C. M. Blair, The number of structurally isomeric Hydrocarbons of the Ethylene Series, J. Amer. Chem. Soc., 55 (2) (1933), 680-686.
H. R. Henze and C. M. Blair, The number of structurally isomeric Hydrocarbons of the Ethylene Series, J. Amer. Chem. Soc., 55 (2) (1933), 680-685. (Annotated scanned copy)
G. Nebe, E. M. Rains, and N. J. A. Sloane, Self-Dual Codes and Invariant Theory, Springer, Berlin, 2006.
Index entries for Molien series
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,-1,0,2,-1).

Cf. A000631, A001400, A014125, A049347, A092498, A156040.

with(combstruct): seq(count(Partition((2*n+4)), size=4), n=0..50); # Zerinvary Lajos, Mar 28 2008

CoefficientList[Series[(1 + x^2) / ((1 - x)^2 (1 - x^2) (1 - x^3)), {x, 0, 100}], x] (* Vincenzo Librandi, Aug 15 2013 *)
LinearRecurrence[{2,0,-1,-1,0,2,-1},{1,2,5,9,15,23,34},50] (* Harvey P. Dale, Aug 31 2015 *)

a(n)=(4*n^3+30*n^2+72*n+55+8*[1,-1,0][(n%3)+1]+9*(-1)^n)/72

G.f.: (1+x^2)/((1-x)^2*(1-x^2)*(1-x^3)). - James Sellers
a(n) = (1/72) * (4*n^3 + 30*n^2 + 72*n + 55 + 8*A049347(n) + 9*(-1)^n ). - Ralf Stephan, Aug 15 2013
E.g.f.: exp(-x)*(27 + 3*exp(2*x)*(55 + 106*x + 42*x^2 + 4*x^3) + 8*exp(x/2)*(3*cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2)))/216. - Stefano Spezia, Apr 05 2023

A047971 Triangle of coefficients of Gaussian polynomials [ n+3,3 ].

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 3, 4, 4, 5, 4, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 5, 6, 6, 6, 6, 5, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 5, 7, 7, 8, 8, 8, 7, 7, 5, 4, 3, 2, 1, 1

Offset: 0

N. J. A. Sloane

a(n) as illustrated is related to the following sequences: The row sum values are A001400. The column sums are A000292. The row lengths are the stuttering sequence A037915 (stutter values in A016777). The column lengths are the sequence A016777. The max values in each column are A001971. - Alford Arnold, Aug 16 2004
The Gaussian polynomial (or Gaussian binomial) [n,3]q is an example of a q-binomial coefficient (see the link) and may be defined for n >= 3 by [n,3]_q = ([n]_q * [n-1]_q * [n-2]_q)/([1]_q * [2]_q * [3]_q), where [n]_q := q^n - 1. The first few values are: [3,3]_q = 1; [4,3]_q = 1 + q + q^2 + q^3; [5,3]_q = 1 + q + 2q^2 + 2q^3 + 2q^4 + q^5 + q^6. - _Peter Bala, Sep 23 2007
The entry a(p,w), p >= 0, w = 0,1,...,3*p, of this irregular triangle is the number of nonnegative solutions of m_0 + m_1 + m_2 + m_3 = p and 1*m_1 + 2*m_2 + 3*m_3 = w. See the Hawkins reference given in A008967, p. 264, (4,7),(4.8), concerning Cayley's counting problem.  N(p,3,w) there equals a(p,w). The o.g.f. has been given in the formula section by Peter Bala.  See also the Cayley reference given in A008967, p. 110, 35. with m = 3, Theta = p and q = w. - Wolfdieter Lang, Dec 02 2012
The entry a(p,w) p >= 0, w = 0,1,...,3*p, of this array gives the number of partitions of w into at most p parts, each at most 3. This follows from the preceding comment with the two Diophantine equations. From Andrews, p. 33 and p. 35, a(p,w) (called there p(N,M,n) with N=p, M=3, n=w) gives also the number of partitions of w into at most 3 parts, each at most p. This is in accordance with the symmetry of the q-binomials [p+3,p] = [p+3,3]. - Wolfdieter Lang, Dec 04 2012

			The table a(p,w) = [q^w][p+3,3]_q starts:
p\w 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 ...
0:  1
1:  1  1  1  1
2:  1  1  2  2  2  1  1
3:  1  1  2  3  3  3  3  2  1  1
4:  1  1  2  3  4  4  5  4  4  3  2  1  1
5:  1  1  2  3  4  5  6  6  6  6  5  4  3  2  1  1
6:  1  1  2  3  4  5  7  7  8  8  8  7  7  5  4  3  2  1  1
... Reformatted and extended by _Wolfdieter Lang_, Dec 04 2012
Partition example: Row p=2 is 1 1 2 2 2 1 1 because there are ten solution for (m_0, m_1, m_2, m_3) of the first equation given in a comment above, namely (2,0,0,0), (0,2,0,0), (0,0,2,0), (0,0,0,2), (1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1) and (0,0,1,1) which have the w = 1*m_1 + 2*m_2 + 3*m_3 values 0, 2, 4, 6, 1, 2, 3, 3, 4 and 5, respectively. Therefore there are 1, 1, 2, 2, 2, 1, 1 solutions for w = 0, 1, 2, 3, 4, 5, 6, respectively. - _Wolfdieter Lang_, Dec 03 2012
a(4,5) = 4 because there are 4 partitions of 5 with 1, 2, 3  or 4 parts, each being <= 3, namely all partitions of 5 excluding 5, 14 and 11111. There are also 4 partitions of 5 with 1, 2, or 3 parts, each being <= 4, namely all partitions of 5 excluding 5, 1112 and 11111. - _Wolfdieter Lang_, Dec 04 2012
The table may also be arranged as follows (see the Alford Arnold comment above):
1
..1
..1..1
..1..1..1
..1..2..1..1
.....2..2..1..1
.....2..3..2..1..1
.....1..3..3..2..1..1

G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 242.

Eric Weisstein's World of Mathematics, q-Binomial Coefficient.

Cf. A008967.

Cf. A001400.

nmax = 6;
se = Series[ 1/Product[1 - q^k*x, {k, 0, 3}], {x, 0, nmax}];
row[n_] := CoefficientList[ SeriesCoefficient[se, n], q];
Flatten[ Table[ row[n], {n, 0, nmax}]] (* Jean-François Alcover, Dec 19 2011 *)

O.g.f.: 1/((1-x)(1-qx)(1-q^2x)(1-q^3x)) = 1 + x(1 + q + q^2 + q^3) + x^2(1 + q + 2q^2 + 2q^3 + 2q^4 + q^5 + q^6) + .... - Peter Bala, Sep 23 2007

A117487 G.f.: 1/((1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5))^2.

Original entry on oeis.org

1, 2, 5, 10, 20, 36, 63, 104, 169, 264, 405, 604, 888, 1278, 1815, 2536, 3502, 4772, 6437, 8586, 11352, 14866, 19315, 24890, 31851, 40466, 51089, 64092, 79952, 99172, 122386, 150264, 183639, 223394, 270605, 326422, 392225, 469490, 559970, 665542, 788412

Offset: 1

Alford Arnold, Mar 22 2006

nonn

Molien series for S_5 X S_5, cf. A001401.
Molien series for S_k X S_k approaches A000712 as k increases.
Column 5 of table A115994.
Note that a(5) is 20, the scalar product of (1 1 2 3 5) and (5 3 2 1 1 ). a(6) is 36, the scalar product of (1 1 2 3 5 7) and (7 5 3 2 1 1 ).

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2, 1, -2, -1, -2, 0, 2, 6, 2, -3, -6, -5, -2, 3, 12, 3, -2, -5, -6, -3, 2, 6, 2, 0, -2, -1, -2, 1, 2, -1).

Cf. A000027, A000712, A001399, A001400, A006918.

Cf. A115994, A117485, A117486, A117488, A117489.

n:=5; G:=SymmetricGroup(n); H:=DirectProduct(G,G); MolienSeries(H); // N. J. A. Sloane

# adapted from A115994 kmax := 120 : qmax := kmax/2 : g:=sum(t^k*q^(k^2)/product((1-q^j)^2, j=1..k), k=1..kmax): gser:=series(g, q=0, qmax): for n from 25 to qmax-1 do P :=coeff(gser, q^n) : printf("%a,",coeff(P, t^5)); od: # R. J. Mathar, Apr 07 2006

CoefficientList[Series[1/(Product[(1-x^j), {j,5}])^2, {x,0,45}], x] (* G. C. Greubel, Jan 01 2020 *)

my(x='x+O('x^45)); Vec( 1/(prod(j=1,5, 1-x^j))^2 ) \\ G. C. Greubel, Jan 01 2020

def A117487_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/(product(1-x^j for j in (1..5)))^2 ).list()
A117487_list(45) # G. C. Greubel, Jan 01 2020

More terms from R. J. Mathar, Apr 07 2006

Entry revised by N. J. A. Sloane, Mar 10 2007

A145574 Array a(n,m) for number of partitions of n>=2 with m parts having no part 1. Hence m=1..floor(n/2).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 2, 1, 1, 3, 3, 1, 1, 4, 4, 2, 1, 1, 4, 5, 3, 1, 1, 5, 7, 5, 2, 1, 1, 5, 8, 6, 3, 1, 1, 6, 10, 9, 5, 2, 1, 1, 6, 12, 11, 7, 3, 1, 1, 7, 14, 15, 10, 5, 2, 1, 1, 7, 16, 18, 13, 7, 3, 1, 1, 8, 19, 23, 18, 11, 5, 2, 1, 1, 8, 21, 27, 23, 14, 7, 3, 1, 1, 9, 24, 34, 30

Offset: 2

Wolfdieter Lang and Malin Sjodahl, Mar 06 2009

The row lengths sequence is floor(n/2) = [1,1,2,2,3,3,4,4,...], see A008619(n-1), n>=2.
Obtained from the characteristic partition array A145573 by summing in row n>=2 over entries belonging to like parts number m.
The column sequences give A000012, A004526, A001399, A001400, A001401, A001402, A026813 for m=1..7.

			1;
1;
1, 1;
1, 1;
1, 2, 1;
1, 2, 1;
1, 3, 2, 1;
1, 3, 3, 1;
1, 4, 4, 2, 1;

Alois P. Heinz, Rows n = 2..200, flattened
W. Lang and M. Sjodahl, First 20 rows of the array and row sums.

Cf. A145573, A002865 (row sums).

b:= proc(n, i, t) option remember; `if`(2*t>n or t*i b(n, n, m):
seq(seq(a(n, m), m=1..iquo(n, 2)), n=2..30); # Alois P. Heinz, Oct 18 2012

nn=15; f[list_]:=Select[list,#>0&]; p=Product[1/(1-y x^i), {i, 2, nn}]; Drop[Map[f, CoefficientList[Series[p, {x, 0, nn}], {x, y}]], 1]//Grid  (* Geoffrey Critzer, Sep 23 2012 *)

# Prints the table; cf. A011973.
for n in (2..20): [Partitions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012

a(n,m) = sum over entries of A145573(n,k) array which belong to partitions with part number m, for m=1..floor(n/2)). Note that partitions with parts number m>floor(n/2) have always at least one part 1.

G.f.: Product_{i>=2} 1/(1- y*x^i). - Geoffrey Critzer, Sep 23 2012

A274100 Number of partitions of 2^n into at most four parts.

Original entry on oeis.org

1, 2, 5, 15, 64, 351, 2280, 16335, 123464, 959631, 7566280, 60090255, 478968264, 3824743311, 30569959880, 244447781775, 1955134763464, 15639288341391, 125107148059080, 1000828550570895, 8006513870533064, 64051652831273871, 512411390124519880

Offset: 0

N. J. A. Sloane, Jun 11 2016

nonn

Colin Barker, Table of n, a(n) for n = 0..1000

A subsequence of A001400. Cf. A274099.

\\ b(n) is the coefficient of x^n in the g.f. 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)).
b(n) = round(real((68+36*(-1)^n+18*((-I)^n+I^n)+(16*exp(-2/3*I*n*Pi)*(1+I*sqrt(3)+2*exp((4*I*n*Pi)/3)))/(1+(-1)^(1/3))+59*(1+n)+9*(-1)^n*(1+n)+18*(1+n)*(2+n)+2*(1+n)*(2+n)*(3+n))/288))
vector(50, n, n--; b(2^n)) \\ Colin Barker, Jun 12 2016

Coefficient of x^(2^n) in 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)).
Conjectures from Colin Barker, Jun 12 2016: (Start)
a(n) = 14*a(n-1)-55*a(n-2)+50*a(n-3)+56*a(n-4)-64*a(n-5) for n>6.
G.f.: (1-12*x+32*x^2+5*x^3-27*x^4-18*x^5-16*x^6) / ((1-x)*(1+x)*(1-2*x)*(1-4*x)*(1-8*x)).
(End)

More terms from Colin Barker, Jun 12 2016

A324602 Irregular triangle read by rows: T(n, k) gives the coefficients of the Girard-Waring formula for the sum of n-th power of four indeterminates in terms of their elementary symmetric functions (reverse Abramowitz-Stegun order of partitions).

Original entry on oeis.org

1, 1, -2, 1, -3, 3, 1, -4, 2, 4, -4, 1, -5, 5, 5, -5, -5, 1, -6, 9, -2, 6, -12, 3, -6, 6, 1, -7, 14, -7, 7, -21, 7, 7, -7, 14, -7, 1, -8, 20, -16, 2, 8, -32, 24, 12, -8, -8, 24, -8, -16, 4, 1, -9, 27, -30, 9, 9, -45, 54, -9, 18, -27, 3, -9, 36, -27, -27, 18, 9, 1, -10, 35, -50, 25, -2, 10, -60, 100, -40, 25, -60, 15, 10, -10, 50, -60, 10, -40, 60, -10, 15, -10

Offset: 1

Wolfdieter Lang, May 03 2019

The length of row n is A001400(n), n >= 1.
The Girard-Waring formula for the power sum p(4,n) := Sum_{j=1..4} (x_j)^n in terms of the elementary symmetric functions e_j(x_1, x_2, x_3, x_4), for j = 1, 2, 3, 4, is given by Sum_{i1=0..floor(n/4)} Sum_{i2=0...floor((n-4*i1)/3)} Sum_{i3=0...floor((n-4*i1-3*i2)/2)} ((-1)^(i1 + i3))*n*(n-1-i3-2*i2-3*i1)!/(i1!*i2!*i3!*(n-2*i3-3*i2-4*i1)!)*e_1^(n-2*i3-3*i2-4*i1)*(e_2)^i3*(e_3)^i2*(e_4)^i1, n >= 1 (the arguments of e_j have been omitted). See the W. Lang reference, Theorem 1, case N = 4, with r -> n.
This is an array using the partitions of n, in the reverse Abramowitz-Stegun order, with all partitions with a part >= 5 eliminated. See row n of the array of Waring numbers A115131, read backwards, with these partitions omitted.

			The irregular triangle T(n, k) begins:
n\k 1   2  3  4   5   6  7  8   9  10  11  12  13   14   15  16  17 18 ...
-----------------------------------------------------------------------------
1:  1
2:  1  -2
3:  1  -3  3
4:  1  -4  2  4  -4
5:  1  -5  5  5  -5  -5
6:  1  -6  9  6  -2 -12 -6  3   6
7:  1  -7 14  7  -7 -21 -7  7   7  14  -7
8:  1  -8 20  8 -16 -32 -8  2  24  12  24  -8  -8  -16    4
9:  1  -9 27  9 -30 -45 -9  9  54  18  36  -9 -27  -27  -27   3  18  9
...
n = 10: 1 -10 35 10 -50 -60 -10 25 100 25 50 -2 -40 -60 -60 -40 15 10 10 60 15 -10 -10.
...
-----------------------------------------------------------------------------
Row n = 5: p(4,5) = x_1^5 + x_2^5 + x_3^5 + x_4^5 =  1*e_1^5  - 5* e_1^3*e_2  + 5*e_1*e_2^2 + 5*e_1^2*e_3 - 5*e_2*e_3 - 5*e_1*e_4,
  with e_1 = Sum_{j=1..4} x_j, e_2 = x1*(x_2 + x_3 + x_4) + x_2*(x_3 + x_4) + x_3*x_4, e_3 = x_1*x_2*x_3 + x_1*x_2*x_4 + x_2*x_3*x_4, e_4 = Product_{i=1..4} x_j.

Wolfdieter Lang, On sums of powers of zeros of polynomials, J. Comp. Appl. Math. 89 (1998) 237-356.

Cf. A001400, A115131, A132460 (N=2), A325477 (N=3).

T(n, k) is the k-th coefficient of the Waring number partition array A115131(n, m) (k there is replaced here by m), read backwards, omitting all partitions which have a part >= 5.

A165188 Interleaving of A014125 and zero followed by A014125.

Original entry on oeis.org

1, 0, 3, 1, 6, 3, 11, 6, 18, 11, 27, 18, 39, 27, 54, 39, 72, 54, 94, 72, 120, 94, 150, 120, 185, 150, 225, 185, 270, 225, 321, 270, 378, 321, 441, 378, 511, 441, 588, 511, 672, 588, 764, 672, 864, 764, 972, 864, 1089, 972, 1215, 1089, 1350, 1215, 1495, 1350

Offset: 1

Alford Arnold, Sep 13 2009

nonn

This sequence convolved with A000217 (without initial term 0) yields A164680.
See A164680 for similar convolutions.
A165188 convolved with A000217 yields sequence A164680. This is to be expected since A000217 can be associated with partition 1+1+1, A164680 with partition 1+1+1+2+2+2+3 and A165188 with partition 2+2+2+3 by observing their unreduced generating functions and verified by generating the sequences by converting the partitions into finite sequences and using Euler's Transform. Thus partition 1+1+1 yields the finite sequence (3); partition 2+2+2+3 yields the finite sequence (0,3,1); and, when combined, partition 1+1+1+2+2+2+3 yields (3,3,1). - Alford Arnold, Sep 24 2009

			A014125 begins 1,3,6,11,18,27,..., thus this sequence begins 1,0,3,1,6,3,11,6,18,11,27,18,... .
G.f. = x + 3*x^3 + x^4 + 6*x^5 + 3*x^6 + 11*x^7 + 6*x^8 + 18*x^9 + 11*x^10 + ...

Index entries for linear recurrences with constant coefficients, signature (0,3,1,-3,-3,1,3,0,-1).

Cf. A014125, A000217, A164680, A001400.

I:=[1,0,3,1,6,3,11,6]; [n le 8 select I[n] else -Self(n-1)+2*Self(n-2)+3*Self(n-3)-3*Self(n-5)-2*Self(n-6)+Self(n-7)+Self(n-8)+1: n in [1..60]]; // Vincenzo Librandi, Jun 24 2015

a[ n_] := Module[{s = 1, m = n}, If[ n < 0, s = -1; m = -7 - n]; s SeriesCoefficient[ x / ((1 - x^2)^3 (1 - x^3)), {x, 0, m}]]; (* Michael Somos, Feb 01 2015 *)
LinearRecurrence[{0,3,1,-3,-3,1,3,0,-1},{1,0,3,1,6,3,11,6,18},60] (* Harvey P. Dale, Apr 14 2018 *)

/* first computes u = A014125 as second bisection of A001400, then interleaves */ {m=28; u=vector(m, n, polcoeff(1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4))+O(x^(2*n)), 2*n-1)); vector(2*m, k, if(k%2==1,u[(k+1)/2], if(k==2, 0, u[k/2-1])))} /* Klaus Brockhaus, Sep 15 2009 */

{a(n) = my(s=1); if( n<0, s=-1; n=-7-n); s * polcoeff( x / ((1 - x^2)^3 * (1 - x^3)) + x * O(x^n), n)}; /* Michael Somos, Feb 01 2015 */

a(n) = -a(n-1)+2*a(n-2)+3*a(n-3)-3*a(n-5)-2*a(n-6)+a(n-7)+a(n-8)+1 for n > 8; a(1)=1, a(2)=0, a(3)=3, a(4)=1, a(5)=6, a(6)=3, a(7)=11, a(8)=6. - Klaus Brockhaus, Sep 15 2009
G.f.: x/((1-x)^4*(1+x)^3*(1+x+x^2)). - Klaus Brockhaus, Sep 15 2009
a(n) = (2*n^3+21*n^2+63*n+49)/288-(-1)^n*(9+7*n+n^2)/32+A057078(n)/9. - R. J. Mathar, Sep 17 2009
Euler transform of length 3 sequence [ 0, 3, 1]. - Michael Somos, Feb 01 2015
G.f.: x / ((1 - x^2)^3 * (1 - x^3)). - Michael Somos, Feb 01 2015
a(n) = -a(-7 - n) for all n in Z.
a(n+3) - a(n) = 0 if n odd else (n+6) * (n+4) / 8. - Michael Somos, Feb 01 2015
a(2*n + 1) = a(2*n + 4) = A014125(n) for all n in Z. - Michael Somos, Feb 01 2015

Edited and extended by Klaus Brockhaus, Sep 15 2009

Linear recurrence link and signature correct by Michel Marcus, Jun 25 2015

A254594 Expansion of 1 / ((1 - x^2)^2 * (1 - x^3) * (1 - x^4)) in powers of x.

Original entry on oeis.org

1, 0, 2, 1, 4, 2, 7, 4, 11, 7, 16, 11, 23, 16, 31, 23, 41, 31, 53, 41, 67, 53, 83, 67, 102, 83, 123, 102, 147, 123, 174, 147, 204, 174, 237, 204, 274, 237, 314, 274, 358, 314, 406, 358, 458, 406, 514, 458, 575, 514, 640, 575, 710, 640, 785, 710, 865, 785, 950

Offset: 0

Michael Somos, Feb 02 2015

Partitions of n into parts of size 3 and size 4 and two kinds of parts of size 2.
The number of quadruples of integers [x, u, v, w] which satisfy x > u > v > w >=0, n+5 = x+u, u+v >= x+w, and x+u+v+w is even.
Euler transform of length 4 sequence [ 0, 2, 1, 1].

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

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,1,0,-2,-2,0,1,2,0,-1).

Cf. A001400, A002620, A005044, A115264, A165188.

Cf. A241526, A000601, A181120.

I:=[1,0,2,1,4,2,7,4,11,7,16]; [n le 11 select I[n] else 2*Self(n-2)+Self(n-3)-2*Self(n-5)-2*Self(n-6)+Self(n-8)+2*Self(n-9)-Self(n-11): n in [1..60]]; // Vincenzo Librandi, Feb 03 2015

a[ n_] := Quotient[ n^3 + If[ OddQ[n], 12 n^2 + 33 n + 54, 21 n^2 + 132 n + 288], 288];
a[ n_] := Module[{s = 1, m = n}, If[ n < 0, s = -1; m = -11 - n]; s SeriesCoefficient[ 1 / ((1 - x^2)^2 (1 - x^3) (1 - x^4)), {x, 0, m}]];
a[ n_] := Length @ FindInstance[ {x > u, u > v, v > w, w >= 0, x + u == n + 5, u + v >= x + w, x + u + v + w == 2 k}, {x, u, v, w, k}, Integers, 10^9];
CoefficientList[Series[1 / (1 - 2 x^2 - x^3 + 2 x^5 + 2 x^6 - x^8 - 2 x^9 + x^11), {x, 0, 60}], x] (* Vincenzo Librandi, Feb 03 2015 *)

{a(n) = (n^3 + if(n%2, 12*n^2 + 33*n + 54, 21*n^2 + 132*n + 288)) \ 288};

{a(n) = my(s=1); if( n<0, s=-1; n=-11-n); s * polcoeff( 1 / ((1 - x^2)^2 * (1 - x^3) * (1 - x^4)) + x * O(x^n), n)};

G.f.: 1 / (1 - 2*x^2 - x^3 + 2*x^5 + 2*x^6 - x^8 - 2*x^9 + x^11).
a(n) = -a(-11-n) for all n in Z.
a(n+3) - a(n) = 0 if n even else floor((n+7)^2 / 16).
0 = a(n) - 2*a(n+2) - a(n+3) + 2*a(n+5) + 2*a(n+6) - a(n+8) - 2*a(n+9) + a(n+11) for all n in Z.
a(n) - a(n-2) = A005044(n+3) for all n in Z.
a(n) + a(n-1) = A001400(n) for all n in Z.
a(n) + a(n-2) = A165188(n+1) for all n in Z.
a(n) = A115264(n) - A115264(n-1) for all n in Z.
a(2*n) - a(2*n-6) = a(2*n+3) - a(2*n-3) = A002620(n+2) for all n in Z. - Michael Somos, Feb 11 2015
a(n) = (2*n^3+33*n^2+181*n+234+3*(3*n^2+33*n+86)*(-1)^n+84*(-1)^((2*n+1-(-1)^n)/4)-96*((1+(-1)^n)*floor(((2*n+9+(-1)^n-6*(-1)^((2*n+3+(-1)^n)/4))/24))+(1-(-1)^n)*floor(((2*n+5+(-1)^n-6*(-1)^((2*n-1+(-1)^n)/4))/24))))/576. - Luce ETIENNE, May 22 2015

cp's OEIS Frontend

A117486 Expansion of 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4))^2.

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A347586 Number of partitions of n into at most 4 distinct parts.

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A014126 Number of partitions of 2*n into at most 4 parts.

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A047971 Triangle of coefficients of Gaussian polynomials [ n+3,3 ].

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A117487 G.f.: 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5))^2.

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A145574 Array a(n,m) for number of partitions of n>=2 with m parts having no part 1. Hence m=1..floor(n/2).

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A274100 Number of partitions of 2^n into at most four parts.

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A117486 Expansion of 1/((1-x)(1-x^2)(1-x^3)*(1-x^4))^2.

A117487 G.f.: 1/((1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5))^2.