A344651 -id:A344651 - cp's OEIS frontend

A097805 Number of compositions of n with k parts, T(n, k) = binomial(n-1, k-1) for n, k >= 1 and T(n, 0) = 0^n, triangle read by rows for n >= 0 and 0 <= k <= n.

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 5, 1, 0, 1, 6, 15, 20, 15, 6, 1, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Offset: 0

Paul Barry, Aug 25 2004

Previous name was: Riordan array (1, 1/(1-x)) read by rows.
Note this Riordan array would be denoted (1, x/(1-x)) by some authors.
Columns have g.f. (x/(1-x))^k. Reverse of A071919. Row sums are A011782. Antidiagonal sums are Fibonacci(n-1). Inverse as Riordan array is (1, 1/(1+x)). A097805=B*A059260*B^(-1), where B is the binomial matrix.
(0,1)-Pascal triangle. - Philippe Deléham, Nov 21 2006
(n+1) * each term of row n generates triangle A127952: (1; 0, 2; 0, 3, 3; 0, 4, 8, 4; ...). - Gary W. Adamson, Feb 09 2007
Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2008
From Paul Weisenhorn, Feb 09 2011: (Start)
Triangle read by rows: T(r,c) is the number of unordered partitions of n=r*(r+1)/2+c into (r+1) parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2.
Triangle read by rows: T(r,c) is the number of unordered partitions of the number n=r*(r+1)/2+(c-1) into r parts < (r+1) and at most pairs of equal parts and parts in neighboring pairs have difference 2. (End)
Triangle read by rows: T(r,c) is the number of ordered partitions (compositions) of r into c parts. - Juergen Will, Jan 04 2016
From Tom Copeland, Oct 25 2012: (Start)
Given a basis composed of a sequence of polynomials p_n(x) characterized by ladder (creation / annihilation, or raising / lowering) operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x) with p_0(x)=1, giving the number operator # p_n(x) = RL p_n(x) = n p_n(x), the lower triangular padded Pascal matrix Pd (A097805) serves as a matrix representation of the operator exp(R^2*L) = exp(R#) =
  1) exp(x^2D) for the set x^n  and
  2) D^(-1) exp(t*x)D for the set x^n/n! (see A218234).
  (End)
From James East, Apr 11 2014: (Start)
Square array a(m,n) with m,n=0,1,2,... read by off-diagonals.
a(m,n) gives the number of order-preserving functions f:{1,...,m}->{1,...,n}. Order-preserving means that xa(n,n)=A088218(n) is the size of the semigroup O_n of all order-preserving transformations of {1,...,n}.
Read as a triangle, this sequence may be obtained by augmenting Pascal's triangle by appending the column 1,0,0,0,... on the left.
(End)
A formula based on the partitions of n with largest part k is given as a Sage program below. The 'conjugate' formula leads to A048004. - Peter Luschny, Jul 13 2015
From Wolfdieter Lang, Feb 17 2017: (Start)
The transposed of this lower triangular Riordan matrix of the associated type T provides the transition matrix between the monomial basis {x^n}, n >= 0, and the basis {y^n}, n >= 0, with y = x/(1-x): x^0 = 1 = y^0, x^n = Sum_{m >= n} Ttrans(n,m) y^m, for n >= 1, with Ttrans(n,m) = binomial(m-1,n-1).
Therefore, if a transformation with this Riordan matrix from a sequence {a} to the sequence {b} is given by b(n) = Sum_{m=0..n} T(n, m)*a(m), with T(n, m) = binomial(n-1, m-1), for n >= 1, then Sum_{n >= 0} a(n)*x^n = Sum_{n >= 0} b(n)*y^n, with y = x/(1-x) and vice versa. This is a modified binomial transformation; the usual one belongs to the Pascal Riordan matrix A007318. (End)
From Gus Wiseman, Jan 23 2022: (Start)
Also the number of compositions of n with alternating sum k, with k ranging from -n to n in steps of 2. For example, row n = 6 counts the following compositions (empty column indicated by dot):
  .  (15)  (24)    (33)      (42)     (51)   (6)
           (141)   (132)     (123)    (114)
           (1113)  (231)     (222)    (213)
           (1212)  (1122)    (321)    (312)
           (1311)  (1221)    (1131)   (411)
                   (2112)    (2121)
                   (2211)    (3111)
                   (11121)   (11112)
                   (12111)   (11211)
                   (111111)  (21111)
The reverse-alternating version is the same. Counting compositions by all three parameters (sum, length, alternating sum) gives A345197. Compositions of 2n with alternating sum 2k with k ranging from -n + 1 to n are A034871. (End)
Also the convolution triangle of A000012. - Peter Luschny, Oct 07 2022
From Sergey Kitaev, Nov 18 2023: (Start)
Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with k right-to-left maxima. A right-to-left maximum in a permutation a(1)a(2)...a(n) is position i such that a(j) < a(i) for all i < j.
Number of permutations of length n avoiding simultaneously the patterns 231 and 312 with k right-to-left minima (resp., left-to-right maxima). A right-to-left minimum (resp., left-to-right maximum) in a permutation a(1)a(2)...a(n) is position i such that a(j) > a(i) for all j > i (resp., a(j) < a(i) for all j < i).
Number of permutations of length n avoiding simultaneously the patterns 213 and 312 with k right-to-left maxima (resp., left-to-right maxima).
Number of permutations of length n avoiding simultaneously the patterns 213 and 231 with k right-to-left maxima (resp., right-to-left minima). (End)

			G.f. = 1 + x * (x + x^3 * (1 + x) + x^6 * (1 + x)^2 + x^10 * (1 + x)^3 + ...). - _Michael Somos_, Aug 20 2006
The triangle T(n, k) begins:
n\k  0 1 2  3  4   5   6  7  8 9 10 ...
0:   1
1:   0 1
2:   0 1 1
3:   0 1 2  1
4:   0 1 3  3  1
5:   0 1 4  6  4   1
6:   0 1 5 10 10   5   1
7:   0 1 6 15 20  15   6  1
8:   0 1 7 21 35  35  21  7  1
9:   0 1 8 28 56  70  56 28  8 1
10:  0 1 9 36 84 126 126 84 36 9  1
... reformatted _Wolfdieter Lang_, Jul 31 2017
From _Paul Weisenhorn_, Feb 09 2011: (Start)
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+c = 18 with (r+1)=6 summands: (5+5+4+2+1+1), (5+5+3+3+1+1), (5+4+4+3+1+1), (5+5+3+2+2+1), (5+4+4+2+2+1), (5+4+3+3+2+1).
T(r=5,c=3) = binomial(4,2) = 6 unordered partitions of the number n = r*(r+1)/2+(c-1) = 17 with r=5 summands: (5+5+4+2+1), (5+5+3+3+1), (5+5+3+2+2), (5+4+4+3+1), (5+4+4+2+2), (5+4+3+3+2).  (End)
From _James East_, Apr 11 2014: (Start)
a(0,0)=1 since there is a unique (order-preserving) function {}->{}.
a(m,0)=0 for m>0 since there is no function from a nonempty set to the empty set.
a(3,2)=4 because there are four order-preserving functions {1,2,3}->{1,2}: these are [1,1,1], [2,2,2], [1,1,2], [1,2,2]. Here f=[a,b,c] denotes the function defined by f(1)=a, f(2)=b, f(3)=c.
a(2,3)=6 because there are six order-preserving functions {1,2}->{1,2,3}: these are [1,1], [1,2], [1,3], [2,2], [2,3], [3,3].
(End)

D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Part 1, Section 7.2.1.3, 2011.

Alois P. Heinz, Rows n = 0..140, flattened
Tom Copeland, Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras
James East and Peter J. McNamara, On the work performed by a transformation semigroup, Australas. J. Combin. 49 (2011), 95-109.
Tian Han and Sergey Kitaev, Joint distributions of statistics over permutations avoiding two patterns of length 3, arXiv:2311.02974 [math.CO], 2023.
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 8.
P. A. MacMahon, Memoir on the Theory of the Compositions of Numbers, Phil. Trans. Royal Soc. London A, 184 (1893), 835-901. - _Juergen Will_, Jan 02 2016
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.

Case m=0 of the polynomials defined in A278073.
Cf. A000012 (diagonal), A011782 (row sums), A088218 (central terms).
Cf. A007318, A048004, A127952, A118800, A200139, A130595, A167374.
The terms just left of center in odd-indexed rows are A001791, even A002054.
The odd-indexed rows are A034871.
Row sums without the center are A058622.
The unordered version is A072233, without zeros A008284.
Right half without center has row sums A027306(n-1).
Right half with center has row sums A116406(n).
Left half without center has row sums A294175(n-1).
Left half with center has row sums A058622(n-1).
A025047 counts alternating compositions.
A098124 counts balanced compositions, unordered A047993.
A106356 counts compositions by number of maximal anti-runs.
A344651 counts partitions by sum and alternating sum.
A345197 counts compositions by sum, length, and alternating sum.
Cf. A000346, A000984, A001700, A008549, A008965, A114121, A124754, A238279, A345907, A345908, A346632.

b:= proc(n, i, p) option remember; `if`(n=0, p!, `if`(i<1, 0,
      expand(add(b(n-i*j, i-1, p+j)/j!*x^j, j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, 0)):
seq(T(n), n=0..20);  # Alois P. Heinz, May 25 2014
# Alternatively:
T := proc(k,n) option remember;
if k=n then 1 elif k=0 then 0 else
add(T(k-1,n-i), i=1..n-k+1) fi end:
A097805 := (n,k) -> T(k,n):
for n from 0 to 12 do seq(A097805(n,k), k=0..n) od; # Peter Luschny, Mar 12 2016
# Uses function PMatrix from A357368.
PMatrix(10, n -> 1);  # Peter Luschny, Oct 07 2022

T[0, 0] = 1; T[n_, k_] := Binomial[n-1, k-1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 03 2014, after Paul Weisenhorn *)
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&]],{n,0,10},{k,0,n}] (* Gus Wiseman, Jan 23 2022 *)

{a(n) = my(m); if( n<2, n==0, n--; m = (sqrtint(8*n + 1) - 1)\2; binomial(m-1, n - m*(m + 1)/2))}; /* Michael Somos, Aug 20 2006 */

T(n,k) = if (k==0, 0^n, binomial(n-1, k-1)); \\ Michel Marcus, May 06 2022

row(n) = vector(n+1, k, k--; if (k==0, 0^n, binomial(n-1, k-1))); \\ Michel Marcus, May 06 2022

from math import comb
def T(n, k): return comb(n-1, k-1) if k != 0 else k**n  # Peter Luschny, May 06 2022

# Illustrates a basic partition formula, is not efficient as a program for large n.
def A097805_row(n):
    r = []
    for k in (0..n):
        s = 0
        for q in Partitions(n, max_part=k, inner=[k]):
            s += mul(binomial(q[j],q[j+1]) for j in range(len(q)-1))
        r.append(s)
    return r
[A097805_row(n) for n in (0..9)] # Peter Luschny, Jul 13 2015

Number triangle T(n, k) defined by T(n,k) = Sum_{j=0..n} binomial(n, j)*if(k<=j, (-1)^(j-k), 0).
T(r,c) = binomial(r-1,c-1), 0 <= c <= r. - Paul Weisenhorn, Feb 09 2011
G.f.: (-1+x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015
a(0,0) = 1, a(n,k) = binomial(n-1,n-k) = binomial(n-1,k-1) Juergen Will, Jan 04 2016
G.f.: (x^1 + x^2 + x^3 + ...)^k = (x/(1-x))^k. - Juergen Will, Jan 04 2016
From Tom Copeland, Nov 15 2016: (Start)
E.g.f.: 1 + x*[e^((x+1)t)-1]/(x+1).
This padded Pascal matrix with the odd columns negated is NpdP = M*S = S^(-1)*M^(-1) = S^(-1)*M, where M(n,k) = (-1)^n A130595(n,k), the inverse Pascal matrix with the odd rows negated, S is the summation matrix A000012, the lower triangular matrix with all elements unity, and S^(-1) = A167374, a finite difference matrix. NpdP is self-inverse, i.e., (M*S)^2 = the identity matrix, and has the e.g.f. 1 - x*[e^((1-x)t)-1]/(1-x).
M = NpdP*S^(-1) follows from the well-known recursion property of the Pascal matrix, implying NpdP = M*S.
The self-inverse property of -NpdP is implied by the self-inverse relation of its embedded signed Pascal submatrix M (cf. A130595). Also see A118800 for another proof.
Let P^(-1) be A130595, the inverse Pascal matrix. Then T = A200139*P^(-1) and T^(-1) = padded P^(-1) = P*A097808*P^(-1). (End)
The center (n>0) is T(2n+1,n+1) = A000984(n) = 2*A001700(n-1) = 2*A088218(n) = A126869(2n) = 2*A138364(2n-1). - Gus Wiseman, Jan 25 2022

Corrected by Philippe Deléham, Oct 05 2005

New name using classical terminology by Peter Luschny, Feb 05 2019

A103919 Triangle of numbers of partitions of n with total number of odd parts equal to k from {0,...,n}.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 2, 0, 1, 0, 4, 0, 2, 0, 1, 3, 0, 5, 0, 2, 0, 1, 0, 7, 0, 5, 0, 2, 0, 1, 5, 0, 9, 0, 5, 0, 2, 0, 1, 0, 12, 0, 10, 0, 5, 0, 2, 0, 1, 7, 0, 17, 0, 10, 0, 5, 0, 2, 0, 1, 0, 19, 0, 19, 0, 10, 0, 5, 0, 2, 0, 1, 11, 0, 28, 0, 20, 0, 10, 0, 5, 0, 2, 0, 1, 0, 30, 0, 33, 0, 20, 0, 10, 0, 5, 0, 2, 0, 1

Offset: 0

Wolfdieter Lang, Mar 24 2005

The partition (0) of n=0 is included. For n>0 no part 0 appears.
The first (k=0) column gives the number of partitions without odd parts, i.e., those with even parts only. See A035363.
Without the alternating zeros this becomes a triangle with columns given by the rows of the S_n(m) table shown in the Riordan reference.
From Gregory L. Simay, Oct 31 2015: (Start)
T(2n+k,k) = the number of partitions of n with parts 1..k of two kinds. If n<=k, then T(2n+k) = A000712(n), the number of partitions of n with parts of two kinds.
T(2n+k) = the convolution of A000041(n) and the number of partitions of n+k having exactly k parts.
T(2n+k) = d(n,k) where d(n,0) = p(n) and d(n,k) = d(n,k-1) + d(n-k,k-1) + d(n-2k,k-1) + ... (End)
From Emeric Deutsch, Oct 04 2016: (Start)
T(n,k) = number of partitions (p1 >= p2 >= p3 >= ...) of n having alternating sum p1 - p2 + p3 - ... = k. Example: T(5,3) = 2 because there are two partitions (3,1,1) and (4,1) of 5 with alternating sum 3.
The equidistribution of the partition statistics "alternating sum" and "total number of odd parts" follows by conjugation. (End)

			The triangle a(n,k) begins:
n\k 0  1  2  3  4  5  6  7  8  9 10
0:  1
1:  0  1
2:  1  0  1
3:  0  2  0  1
4:  2  0  2  0  1
5:  0  4  0  2  0  1
6:  3  0  5  0  2  0  1
7:  0  7  0  5  0  2  0  1
8:  5  0  9  0  5  0  2  0  1
9:  0 12  0 10  0  5  0  2  0  1
10: 7  0 17  0 10  0  5  0  2  0  1
... Reformatted - _Wolfdieter Lang_, Apr 28 2013
a(0,0) = 1 because n=0 has no odd part, only one even part, 0, by definition. a(5,3) = 2 because there are two partitions (1,1,3) and (1,1,1,2) of 5 with exactly 3 odd parts.
From _Gregory L. Simay_, Oct 31 2015: (Start)
T(10,4) = T(2*3+4,4) = d(3,4) = A000712(3) = 10.
T(10,2) = T(2*4+2,2) = d(4,2) = d(4,1)+d(2,1)+d(0,1) = d(4,0)+d(3,0)+d(2,0)+d(1,0)+d(0,0) + d(2,0)+d(1,0)+d(0,0) + d(0,0) = convolution sum p(4)+p(3)+2*p(2)+2*p(1)+3*p(0) = 5+3+2*2+2*1+3*1 = 17.
T(9,1) = T(8,0) + T(7,1) = 5 + 7 = 12.
(End)

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.

Alois P. Heinz, Rows n = 0..140, flattened
D. Kim, A. J. Yee, A note on partitions into distinct parts and odd parts, Ramanujan J. 3 (1999), 227-231. [_R. J. Mathar_, Nov 11 2008]
Wolfdieter Lang, First 11 rows.

Row sums gives A000041 (partition numbers). Columns: k=0: A035363 (with zero entries) A000041 (without zero entries), k=1: A000070, k=2: A000097, k=3: A000098, k=4: A000710, 3k>=n: A000712.
Cf. A066897.
The strict version (without zeros) is A152146 interleaved with A152157.
The rows (without zeros) are A239830 interleaved with A239829.
The reverse version (without zeros) is the right half of A344612.
Removing all zeros gives A344651.
The strict reverse version (without zeros) is the right half of A344739.
Cf. A006330, A027187, A116406, A344607, A344608, A344649, A344654.

g:=1/product((1-t*x^(2*j-1))*(1-x^(2*j)),j=1..20): gser:=simplify(series(g,x=0,22)): P[0]:=1: for n from 1 to 18 do P[n]:=coeff(gser,x^n) od: for n from 0 to 18 do seq(coeff(P[n],t,j),j=0..n) od; # yields sequence in triangular form # Emeric Deutsch, Feb 17 2006

T[n_, k_] := T[n, k] = Which[nJean-François Alcover, Mar 05 2014, after Paul D. Hanna *)
Table[Length[Select[IntegerPartitions[n],Count[#,?OddQ]==k&]],{n,0,15},{k,0,n}] (* _Gus Wiseman, Jun 20 2021 *)

{T(n, k)=if(n>=k, if(n==k, 1, if((n-k+1)%2==0, 0, if(k==0, sum(m=0, n, T(n\2, m)), T(n-1, k-1)+T(n-2*k, k)))))}
for(n=0, 20, for(k=0, n, print1(T(n, k), ", ")); print(""))
\\ Paul D. Hanna, Apr 27 2013

a(n, k) = number of partitions of n>=0, which have exactly k odd parts (and possibly even parts) for k from {0, ..., n}.
Sum_{k=0..n} k*T(n,k) = A066897(n). - Emeric Deutsch, Feb 17 2006
G.f.: G(t,x) = 1/Product_{j>=1} (1-t*x^(2*j-1))*(1-x^(2*j)). - Emeric Deutsch, Feb 17 2006
G.f. T(2n+k,k) = g.f. d(n,k) = (1/Product_{j=1..k} (1-x^j)) * g.f. p(n). - Gregory L. Simay, Oct 31 2015
T(n,k) = T(n-1,k-1) + T(n-2k,k). - Gregory L. Simay, Nov 01 2015

A344612 Triangle read by rows where T(n,k) is the number of integer partitions of n with reverse-alternating sum k ranging from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 2, 3, 3, 1, 1, 0, 1, 2, 4, 3, 3, 1, 1, 0, 1, 2, 4, 5, 5, 3, 1, 1, 0, 1, 2, 4, 7, 5, 6, 3, 1, 1, 0, 1, 2, 4, 8, 7, 9, 6, 3, 1, 1, 0, 1, 2, 4, 8, 12, 7, 11, 6, 3, 1, 1, 0, 1, 2, 4, 8, 14, 11, 14, 12, 6, 3, 1, 1

Offset: 0

Gus Wiseman, Jun 01 2021

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i. This is also (-1)^(k-1) times the sum of the even-indexed parts minus the sum of the odd-indexed parts.
Also the number of reversed integer partitions of n with alternating sum k ranging from -n to n in steps of 2.
Also the number of integer partitions of n with (-1)^(m-1) * b = k where m is the greatest part and b is the number of odd parts, with k ranging from -n to n in steps of 2.

			Triangle begins:
                                1
                              0   1
                            0   1   1
                          0   1   1   1
                        0   1   2   1   1
                      0   1   2   2   1   1
                    0   1   2   3   3   1   1
                  0   1   2   4   3   3   1   1
                0   1   2   4   5   5   3   1   1
              0   1   2   4   7   5   6   3   1   1
            0   1   2   4   8   7   9   6   3   1   1
          0   1   2   4   8  12   7  11   6   3   1   1
        0   1   2   4   8  14  11  14  12   6   3   1   1
      0   1   2   4   8  15  19  11  18  12   6   3   1   1
    0   1   2   4   8  15  24  15  23  20  12   6   3   1   1
  0   1   2   4   8  15  26  30  15  31  21  12   6   3   1   1
For example, row n = 7 counts the following partitions:
  (61)  (52)    (43)      (331)      (322)    (511)  (7)
        (4111)  (2221)    (22111)    (421)
                (3211)    (1111111)  (31111)
                (211111)
Row n = 9 counts the following partitions:
  81  72    63      54        441        333      522    711  9
      6111  4221    3222      22221      432      621
            5211    3321      33111      531      51111
            411111  4311      2211111    32211
                    222111    111111111  42111
                    321111               3111111
                    21111111

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Row sums are A000041.
The midline k = n/2 is also A000041.
The right half (i.e., k >= 0) for even n is A344610.
The rows appear to converge to A344611 (from left) and A006330 (from right).
The non-reversed version is A344651 (A239830 interleaved with A239829).
The strict version is A344739.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A120452 counts partitions of 2n with rev-alt sum 2 (negative: A344741).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A325534/A325535 count separable/inseparable partitions.
A344618 gives reverse-alternating sums of standard compositions.
Cf. A000070, A000097, A003242, A027187, A124754, A152146, A344607, A344608, A344649, A344650, A344654.

sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Table[Length[Select[IntegerPartitions[n],sats[#]==k&]],{n,0,15},{k,-n,n,2}]

row(n)={my(v=vector(n+1)); forpart(p=n, my(s=-sum(i=1, #p, p[i]*(-1)^i)); v[(s+n)/2+1]++); v} \\ Andrew Howroyd, Jan 06 2024

A000097 Number of partitions of n if there are two kinds of 1's and two kinds of 2's.

Original entry on oeis.org

1, 2, 5, 9, 17, 28, 47, 73, 114, 170, 253, 365, 525, 738, 1033, 1422, 1948, 2634, 3545, 4721, 6259, 8227, 10767, 13990, 18105, 23286, 29837, 38028, 48297, 61053, 76926, 96524, 120746, 150487, 187019, 231643, 286152, 352413, 432937, 530383, 648245

Offset: 0

N. J. A. Sloane

Also number of partitions of 2*n with exactly 2 odd parts (offset 1). - Vladeta Jovovic, Jan 12 2005
Also number of transitions from one partition of n+2 to another, where a transition consists of replacing any two parts with their sum. Remove all 1' and 2' from the partition, replacing them with ((number of 2') + 1) and ((number of 1') + (number of 2') + 1); these are the two parts being summed. Number of partitions of n into parts of 2 kinds with at most 2 parts of the second kind, or of n+2 into parts of 2 kinds with exactly 2 parts of the second kind. - Franklin T. Adams-Watters, Mar 20 2006
From Christian Gutschwager (gutschwager(AT)math.uni-hannover.de), Feb 10 2010: (Start)
a(n) is also the number of pairs of partitions of n+2 which differ by only one box (for bijection see the first Gutschwager link).
a(n) is also the number of partitions of n with two parts marked.
a(n) is also the number of partitions of n+1 with two different parts marked. (End)
Convolution of A000041 and A008619. - Vaclav Kotesovec, Aug 18 2015
a(n) = P(/2,n), a particular case of P(/k,n) defined as follows: P(/0,n) = A000041(n) and P(/k,n) = P(/k-1, n) + P(/k-1,n-k) + P(/k-1, n-2k) + ... Also, P(/k,n) = the convolution of A000041 and the partitions of n with exactly k parts, and g.f. P(/k,n) = (g.f. for P(n)) * 1/(1-x)...(1-x^k). - Gregory L. Simay, Mar 22 2018
a(n) is also the sum of binomial(D(p),2) in partitions p of (n+3), where D(p)= number of different sizes of parts in p. - Emily Anible, Apr 03 2018
Also partitions of 2*(n+1) with alternating sum 2. Also partitions of 2*(n+1) with reverse-alternating sum -2 or 2. - Gus Wiseman, Jun 21 2021
Define the distance graph of the partitions of n using the distance function in A366156 as follows: two vertices (partitions) share an edge if and only if the distance between the vertices is 2. Then a(n) is the number of edges in the distance graph of the partitions of n. - Clark Kimberling, Oct 12 2023

			a(3) = 9 because we have 3, 2+1, 2+1', 2'+1, 2'+1', 1+1+1, 1+1+1', 1+1'+1' and 1'+1'+1'.
From _Gus Wiseman_, Jun 22 2021: (Start)
The a(0) = 1 through a(4) = 9 partitions of 2*(n+1) with exactly 2 odd parts:
  (1,1)  (3,1)    (3,3)      (5,3)
         (2,1,1)  (5,1)      (7,1)
                  (3,2,1)    (3,3,2)
                  (4,1,1)    (4,3,1)
                  (2,2,1,1)  (5,2,1)
                             (6,1,1)
                             (3,2,2,1)
                             (4,2,1,1)
                             (2,2,2,1,1)
The a(0) = 1 through a(4) = 9 partitions of 2*(n+1) with alternating sum 2:
  (2)  (3,1)    (4,2)        (5,3)
       (2,1,1)  (2,2,2)      (3,3,2)
                (3,2,1)      (4,3,1)
                (3,1,1,1)    (3,2,2,1)
                (2,1,1,1,1)  (4,2,1,1)
                             (2,2,2,1,1)
                             (3,2,1,1,1)
                             (3,1,1,1,1,1)
                             (2,1,1,1,1,1,1)
(End)

H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
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).

T. D. Noe and Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Christian Gutschwager, The skew diagram poset and components of skew characters, arXiv:1104.0008 [math.CO], 2011.
Christian Gutschwager, Reduced Kronecker products which are multiplicity free or contain only a few components, Eur. J. Combinat. 31 (2010) 1996-2005. doi:10.1016/j.ejc.2010.05.008.
J. P. Robinson, Edges in the poset of partitions of an integer, J. Combin. Theory Ser. A, 48 (1988), 236-238.
N. J. A. Sloane, Transforms

First differences are in A024786.
Cf. A000098, A000710.
Third column of Riordan triangle A008951 and of triangle A103923.
The case of reverse-alternating sum 1 or alternating sum 0 is A000041.
The case of reverse-alternating sum -1 or alternating sum 1 is A000070.
The normal case appears to be A004526 or A065033.
The strict case is A096914.
The case of reverse-alternating sum 2 is A120452.
The case of reverse-alternating sum -2 is A344741.
A001700 counts compositions with alternating sum 2.
A035363 counts partitions into even parts.
A058696 counts partitions of 2n.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A124754 gives alternating sums of standard compositions (reverse: A344618).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Cf. A006330, A027187, A239830, A306145, A343941, A344607, A344608, A344619, A344650, A344651, A344740.
Shift of A093695.

with(numtheory): etr:= proc(p) local b; b:=proc(n) option remember; local d,j; if n=0 then 1 else add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n fi end end: a:= etr(n->`if`(n<3,2,1)): seq(a(n), n=0..40); # Alois P. Heinz, Sep 08 2008

CoefficientList[Series[1/((1 - x) (1 - x^2) Product[1 - x^k, {k, 1, 100}]), {x, 0, 100}], x] (* Ben Branman, Mar 07 2012 *)
etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[j]}]*b[n - j], {j, 1, n}]/n]; b]; a = etr[If[# < 3, 2, 1]&]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Apr 09 2014, after Alois P. Heinz *)
(1/((1 - x) (1 - x^2) QPochhammer[x]) + O[x]^50)[[3]] (* Vladimir Reshetnikov, Nov 22 2016 *)
Table[Length@IntegerPartitions[n,All,Join[{1,2},Range[n]]],{n,0,15}] (* Robert Price, Jul 28 2020 and Jun 21 2021 *)
T[n_, 0] := PartitionsP[n];
T[n_, m_] /; (n >= m (m + 1)/2) := T[n, m] = T[n - m, m - 1] + T[n - m, m];
T[, ] = 0;
a[n_] := T[n + 3, 2];
Table[a[n], {n, 0, 60}] (* Jean-François Alcover, May 30 2021 *)
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[IntegerPartitions[n],ats[#]==2&]],{n,0,30,2}] (* Gus Wiseman, Jun 21 2021 *)

my(x = 'x + O('x^66)); Vec( 1/((1-x)*(1-x^2)*eta(x)) ) \\ Joerg Arndt, Apr 29 2013

Euler transform of 2 2 1 1 1 1 1...
G.f.: 1/( (1-x) * (1-x^2) * Product_{k>=1} (1-x^k) ).
a(n) = Sum_{j=0..floor(n/2)} A000070(n-2*j), n>=0.
a(n) = A014153(n)/2 + A087787(n)/4 + A000070(n)/4. - Vaclav Kotesovec, Nov 05 2016
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (4*Pi^2) * (1 + 35*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Aug 18 2015, extended Nov 05 2016
a(n) = A120452(n) + A344741(n). - Gus Wiseman, Jun 21 2021

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 04 2004
Edited by Emeric Deutsch, Mar 23 2005
More terms from Franklin T. Adams-Watters, Mar 20 2006
Edited by Charles R Greathouse IV, Apr 20 2010

A344607 Number of integer partitions of n with reverse-alternating sum >= 0.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 8, 8, 15, 16, 27, 29, 48, 52, 81, 90, 135, 151, 220, 248, 352, 400, 553, 632, 859, 985, 1313, 1512, 1986, 2291, 2969, 3431, 4394, 5084, 6439, 7456, 9357, 10836, 13479, 15613, 19273, 22316, 27353, 31659, 38558, 44601, 53998, 62416, 75168

Offset: 0

Gus Wiseman, May 29 2021

nonn

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
Also the number of reversed integer partitions of n with alternating sum >= 0.
A formula for the reverse-alternating sum of a partition is: (-1)^(k-1) times the number of odd parts in the conjugate partition, where k is the number of parts. So a(n) is the number of integer partitions of n whose conjugate parts are all even or whose length is odd. By conjugation, this is also the number of integer partitions of n whose parts are all even or whose greatest part is odd.
All integer partitions have alternating sum >= 0, so the non-reversed version is A000041.
Is this sequence weakly increasing? In particular, is A344611(n) <= A160786(n)?

			The a(1) = 1 through a(8) = 15 partitions:
  (1)  (2)   (3)    (4)     (5)      (6)       (7)        (8)
       (11)  (111)  (22)    (221)    (33)      (322)      (44)
                    (211)   (311)    (222)     (331)      (332)
                    (1111)  (11111)  (321)     (421)      (422)
                                     (411)     (511)      (431)
                                     (2211)    (22111)    (521)
                                     (21111)   (31111)    (611)
                                     (111111)  (1111111)  (2222)
                                                          (3311)
                                                          (22211)
                                                          (32111)
                                                          (41111)
                                                          (221111)
                                                          (2111111)
                                                          (11111111)

The non-reversed version is A000041.
The opposite version (rev-alt sum <= 0) is A027187, ranked by A028260.
The strict case for n > 0 is A067659 (even bisection: A344650).
The ordered version appears to be A116406 (even bisection: A114121).
The odd bisection is A160786.
The complement is counted by A344608.
The Heinz numbers of these partitions are A344609 (complement: A119899).
The even bisection is A344611.
A000070 counts partitions with alternating sum 1 (reversed: A000004).
A000097 counts partitions with alternating sum 2 (reversed: A120452).
A035363 counts partitions with alternating sum 0, ranked by A000290.
A103919 counts partitions by sum and alternating sum.
A316524 is the alternating sum of prime indices of n (reversed: A344616).
A325534/A325535 count separable/inseparable partitions.
A344610 counts partitions by sum and positive reverse-alternating sum.
A344612 counts partitions by sum and reverse-alternating sum.
A344618 gives reverse-alternating sums of standard compositions.
Cf. A006330, A071321, A071322, A124754, A239829, A239830, A344604, A344651, A344654, A344739, A344742.

sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Table[Length[Select[IntegerPartitions[n],sats[#]>=0&]],{n,0,30}]

a(n) + A344608(n) = A000041(n).

a(2n+1) = A160786(n).

A344610 Triangle read by rows where T(n,k) is the number of integer partitions of 2n with reverse-alternating sum 2k.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 5, 3, 1, 1, 7, 9, 6, 3, 1, 1, 11, 14, 12, 6, 3, 1, 1, 15, 23, 20, 12, 6, 3, 1, 1, 22, 34, 35, 21, 12, 6, 3, 1, 1, 30, 52, 56, 38, 21, 12, 6, 3, 1, 1, 42, 75, 91, 62, 38, 21, 12, 6, 3, 1, 1, 56, 109, 140, 103, 63, 38, 21, 12, 6, 3, 1, 1

Offset: 0

Gus Wiseman, May 31 2021

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i. This is equal to (-1)^(k-1) times the number of odd parts in the conjugate partition, where k is the number of parts.

Also the number of reversed integer partitions of 2n with alternating sum 2k.

			Triangle begins:
   1
   1   1
   2   1   1
   3   3   1   1
   5   5   3   1   1
   7   9   6   3   1   1
  11  14  12   6   3   1   1
  15  23  20  12   6   3   1   1
  22  34  35  21  12   6   3   1   1
  30  52  56  38  21  12   6   3   1   1
  42  75  91  62  38  21  12   6   3   1   1
  56 109 140 103  63  38  21  12   6   3   1   1
  77 153 215 163 106  63  38  21  12   6   3   1   1
Row n = 5 counts the following partitions:
  (55)          (442)        (433)      (622)    (811)  (10)
  (3322)        (541)        (532)      (721)
  (4411)        (22222)      (631)      (61111)
  (222211)      (32221)      (42211)
  (331111)      (33211)      (52111)
  (22111111)    (43111)      (4111111)
  (1111111111)  (2221111)
                (3211111)
                (211111111)

The columns with initial 0's removed appear to converge to A006330.
The odd version is A239829.
The non-reversed version is A239830.
Row sums are A344611, odd bisection of A344607.
Including odd n and negative k gives A344612 (strict: A344739).
The strict case is A344649 (row sums: A344650).
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A103919 counts partitions by sum and alternating sum.
A120452 counts partitions of 2n with rev-alt sum 2 (negative: A344741).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A325534/A325535 count separable/inseparable partitions.
A344604 counts wiggly compositions with twins.
A344618 gives reverse-alternating sums of standard compositions.
Cf. A000070, A000097, A001250, A003242, A027187, A028260, A124754, A152146, A344608, A344651, A344654.

sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Table[Length[Select[IntegerPartitions[n],k==sats[#]&]],{n,0,15,2},{k,0,n,2}]

A344611 Number of integer partitions of 2n with reverse-alternating sum >= 0.

Original entry on oeis.org

1, 2, 4, 8, 15, 27, 48, 81, 135, 220, 352, 553, 859, 1313, 1986, 2969, 4394, 6439, 9357, 13479, 19273, 27353, 38558, 53998, 75168, 104022, 143172, 196021, 267051, 362086, 488733, 656802, 879026, 1171747, 1555997, 2058663, 2714133, 3566122, 4670256, 6096924, 7935184

Offset: 0

Gus Wiseman, May 30 2021

nonn

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
Also the number of reversed integer partitions of 2n with alternating sum >= 0.
The reverse-alternating sum of a partition is equal to (-1)^(k-1) times the number of odd parts in the conjugate partition, where k is the number of parts. So a(n) is the number of partitions of 2n whose conjugate parts are all even or whose length is odd. By conjugation, this is also the number of partitions of 2n whose parts are all even or whose greatest part is odd.

			The a(0) = 1 through a(4) = 15 partitions:
  ()  (2)   (4)     (6)       (8)
      (11)  (22)    (33)      (44)
            (211)   (222)     (332)
            (1111)  (321)     (422)
                    (411)     (431)
                    (2211)    (521)
                    (21111)   (611)
                    (111111)  (2222)
                              (3311)
                              (22211)
                              (32111)
                              (41111)
                              (221111)
                              (2111111)
                              (11111111)

The non-reversed version is A058696 (partitions of 2n).
The ordered version appears to be A114121.
Odd bisection of A344607.
Row sums of A344610.
The strict case is A344650.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A000070 counts partitions with alternating sum 1.
A000097 counts partitions with alternating sum 2.
A103919 counts partitions by sum and alternating sum.
A120452 counts partitions of 2n with reverse-alternating sum 2.
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A325534/A325535 count separable/inseparable partitions.
A344612 counts partitions by sum and rev-alt sum (strict: A344739).
A344618 gives reverse-alternating sums of standard compositions.
A344741 counts partitions of 2n with reverse-alternating sum -2.
Cf. A001250, A027187, A028260, A116406, A119899, A124754, A152146, A239829, A344608, A344609, A344649, A344651, A344654.

sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Table[Length[Select[IntegerPartitions[n],sats[#]>=0&]],{n,0,30,2}]

Conjecture: a(n) <= A160786(n). The difference is 0, 0, 0, 0, 1, 2, 4, 9, 16, 28, 48, 79, ...

More terms from Bert Dobbelaere, Jun 12 2021

A120452 Number of partitions of n-1 boys and one girl with no couple.

Original entry on oeis.org

1, 1, 3, 5, 9, 14, 23, 34, 52, 75, 109, 153, 216, 296, 407, 549, 739, 981, 1300, 1702, 2224, 2879, 3716, 4761, 6083, 7721, 9774, 12306, 15450, 19307, 24064, 29867, 36978, 45614, 56130, 68846, 84250, 102793, 125148, 151955, 184123, 222553, 268482

Offset: 1

Yasutoshi Kohmoto, Jul 20 2006

From Gus Wiseman, Jun 08 2021: (Start)
Also the number of:
- integer partitions of 2n with reverse-alternating sum 2;
- reversed integer partitions of 2n with alternating sum 2;
- integer partitions of 2n with exactly two odd parts, one of which is the greatest;
- odd-length integer partitions of 2n whose conjugate partition has exactly two odd parts.
Note that integer partitions of 2n with alternating or reverse-alternating sum 0 are counted by A000041, ranked by A000290.
(End)

			n=5:
If partitions have no pair "o*", then a(5)=9 ("o" means a boy, "*" means a girl): {o, o, o, o, *}, {o, o, *, oo}, {*, oo, oo}, {o, *, ooo}, {o, o, oo*}, {oo, oo*}, {*, oooo}, {o, ooo*}, {oooo*}.
From _Gus Wiseman_, Jun 08 2021: (Start)
The a(1) = 1 through a(6) = 14 partitions of 2n with reverse-alternating sum 2:
  (2)  (211)  (222)    (332)      (442)        (552)
              (321)    (431)      (541)        (651)
              (21111)  (22211)    (22222)      (33222)
                       (32111)    (32221)      (33321)
                       (2111111)  (33211)      (43221)
                                  (43111)      (44211)
                                  (2221111)    (54111)
                                  (3211111)    (2222211)
                                  (211111111)  (3222111)
                                               (3321111)
                                               (4311111)
                                               (222111111)
                                               (321111111)
                                               (21111111111)
For example, the partition (43221) has reverse-alternating sum 1 - 2 + 2 - 3 + 4 = 2, so is counted under a(6).
The a(1) = 1 through a(6) = 14 partitions of 2n with exactly two odd parts, one of which is the greatest:
  (11)  (31)  (33)   (53)    (55)     (75)
              (51)   (71)    (73)     (93)
              (321)  (332)   (91)     (111)
                     (521)   (532)    (543)
                     (3221)  (541)    (552)
                             (721)    (732)
                             (3322)   (741)
                             (5221)   (921)
                             (32221)  (5322)
                                      (5421)
                                      (7221)
                                      (33222)
                                      (52221)
                                      (322221)
(End)

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000

Cf. A000041, A000070, A002865.
A diagonal of A103919.
A diagonal of A344612.
A000097 counts partitions of 2n with alternating sum 2.
A001700/A088218 appear to count compositions with reverse-alternating sum 2.
A058696 counts partitions of 2n, ranked by A300061.
A344610 counts partitions of 2n by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A344741 counts partitions of 2n with reverse-alternating sum -2.
Cf. A001250, A027187, A119899, A124754, A239830, A316524, A325535, A344618, A344651, A344739.

a[n_] := Total[PartitionsP[Range[0, n-3]]] + PartitionsP[n-1];
Array[a, 50] (* Jean-François Alcover, Jun 05 2021 *)

a(n) = A000070(n-2) + A002865(n-1). - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 15 2006
a(n) = A000070(n-1) - A000041(n-2) = A000070(n-3) + A000041(n-1). - Max Alekseyev, Aug 23 2006
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2^(3/2)*Pi*sqrt(n)) * (1 - 37*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Oct 25 2016

More terms from Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 15 2006

More terms from Max Alekseyev, Aug 23 2006

A026810 Number of partitions of n in which the greatest part is 4.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 23, 27, 34, 39, 47, 54, 64, 72, 84, 94, 108, 120, 136, 150, 169, 185, 206, 225, 249, 270, 297, 321, 351, 378, 411, 441, 478, 511, 551, 588, 632, 672, 720, 764, 816, 864, 920, 972, 1033, 1089, 1154, 1215, 1285, 1350

Offset: 0

Clark Kimberling

Also number of partitions of n into exactly 4 parts.
Also the number of weighted cubic graphs on 4 nodes (=the tetrahedron) with weight n. - R. J. Mathar, Nov 03 2018
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of strict integer partitions of 2n with alternating sum 4, or (by conjugation) partitions of 2n covering an initial interval of positive integers with exactly 4 odd parts. The strict partitions with alternating sum 4 are:
  (4)  (5,1)  (6,2)    (7,3)    (8,4)      (9,5)      (10,6)
              (5,2,1)  (5,3,2)  (5,4,3)    (6,5,3)    (7,6,3)
                       (6,3,1)  (6,4,2)    (7,5,2)    (8,6,2)
                                (7,4,1)    (8,5,1)    (9,6,1)
                                (6,3,2,1)  (6,4,3,1)  (6,5,4,1)
                                           (7,4,2,1)  (7,4,3,2)
                                                      (7,5,3,1)
                                                      (8,5,2,1)
                                                      (6,4,3,2,1)
(End)

			From _Gus Wiseman_, Jun 27 2021: (Start)
The a(4) = 1 through a(10) = 9 partitions of length 4:
  (1111)  (2111)  (2211)  (2221)  (2222)  (3222)  (3322)
                  (3111)  (3211)  (3221)  (3321)  (3331)
                          (4111)  (3311)  (4221)  (4222)
                                  (4211)  (4311)  (4321)
                                  (5111)  (5211)  (4411)
                                          (6111)  (5221)
                                                  (5311)
                                                  (6211)
                                                  (7111)
(End)

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 275.
D. E. Knuth, The Art of Computer Programming, vol. 4,fascicle 3, Generating All Combinations and Partitions, Section 7.2.1.4., p. 56, exercise 31.

Washington Bomfim, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-2,0,0,1,1,-1).

Cf. A001400, A026811, A026812, A026813, A026814, A026815, A026816, A069905 (3 positive parts), A002621 (partial sums), A005044 (first differences).
A non-strict version is A000710 or A088218.
This is column k = 2 of A152146.
A reverse version is A343941.
Cf. A000041, A000070, A000097, A067659, A103919, A120452, A236559, A239830, A306145, A343942, A344616, A344649, A344651.

[Round((n^3+3*n^2-9*n*(n mod 2))/144): n in [0..60]]; // Vincenzo Librandi, Oct 14 2015

A049347 := proc(n)
        op(1+(n mod 3),[1,-1,0]) ;
end proc:
A056594 := proc(n)
        op(1+(n mod 4),[1,0,-1,0]) ;
end proc:
A026810 := proc(n)
        1/288*(n+1)*(2*n^2+4*n-13+9*(-1)^n) ;
        %-A049347(n)/9 ;
        %+A056594(n)/8 ;
end proc: # R. J. Mathar, Jul 03 2012

Table[Count[IntegerPartitions[n], {4, _}], {n, 0, 60}]
LinearRecurrence[{1, 1, 0, 0, -2, 0, 0, 1, 1, -1}, {0, 0, 0, 0, 1, 1, 2, 3, 5, 6}, 60] (* Vincenzo Librandi, Oct 14 2015 *)
Table[Length[IntegerPartitions[n, {4}]], {n, 0, 60}] (* Eric Rowland, Mar 02 2017 *)
CoefficientList[Series[x^4/Product[1 - x^k, {k, 1, 4}], {x, 0, 60}], x] (* Robert A. Russell, May 13 2018 *)

for(n=0, 60, print(n, " ", round((n^3 + 3*n^2 -9*n*(n % 2))/144))); \\ Washington Bomfim, Jul 03 2012

x='x+O('x^60); concat([0, 0, 0, 0], Vec(x^4/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)))) \\ Altug Alkan, Oct 14 2015

vector(60, n, n--; (n+1)*(2*n^2+4*n-13+9*(-1)^n)/288 + real(I^n)/8 - ((n+2)%3-1)/9) \\ Altug Alkan, Oct 26 2015

print1(0,", "); for(n=1,60,j=0;forpart(v=n,j++,,[4,4]); print1(j,", ")) \\ Hugo Pfoertner, Oct 01 2018

G.f.: x^4/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) = x^4/((1-x)^4*(1+x)^2*(1+x+x^2)*(1+x^2)).
a(n+4) = A001400(n). - Michael Somos, Apr 07 2012
a(n) = round( (n^3 + 3*n^2 -9*n*(n mod 2))/144 ). - Washington Bomfim, Jan 06 2021 and Jul 03 2012
a(n) = (n+1)*(2*n^2+4*n-13+9*(-1)^n)/288 -A049347(n)/9 +A056594(n)/8. - R. J. Mathar, Jul 03 2012
From Gregory L. Simay, Oct 13 2015: (Start)
a(n) = (n^3 + 3*n^2 - 9*n)/144 + a(m) - (m^3 + 3*m^2 - 9*m)/144 if n = 12k + m and m is odd. For example, a(23) = a(12*1 + 11) = (23^3 + 3*23^2 - 9*23)/144 + a(11) - (11^3 + 3*11^2 - 9*11)/144 = 94.
a(n) = (n^3 + 3*n^2)/144 + a(m) - (m^3 + 3*m^2)/144 if n = 12k + m and m is even. For example, a(22) = a(12*1 + 10) = (22^3 + 3*22^2)/144 + a(10) - (10^3 + 3*10^2)/144 = 84. (End)
a(n) = A008284(n,4). - Robert A. Russell, May 13 2018
From Gregory L. Simay, Jul 28 2019: (Start)
a(2n+1) = a(2n) + a(n+1) - a(n-3) and
a(2n) = a(2n-1) + a(n+2) - a(n-2). (End)

A357136 Triangle read by rows where T(n,k) is the number of integer compositions of n with alternating sum k = 0..n. Part of the full triangle A097805.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 3, 0, 3, 0, 1, 0, 6, 0, 4, 0, 1, 10, 0, 10, 0, 5, 0, 1, 0, 20, 0, 15, 0, 6, 0, 1, 35, 0, 35, 0, 21, 0, 7, 0, 1, 0, 70, 0, 56, 0, 28, 0, 8, 0, 1, 126, 0, 126, 0, 84, 0, 36, 0, 9, 0, 1, 0, 252, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1

Offset: 0

Gus Wiseman, Sep 30 2022

A composition of n is a finite sequence of positive integers summing to n.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			Triangle begins:
    1
    0   1
    1   0   1
    0   2   0   1
    3   0   3   0   1
    0   6   0   4   0   1
   10   0  10   0   5   0   1
    0  20   0  15   0   6   0   1
   35   0  35   0  21   0   7   0   1
    0  70   0  56   0  28   0   8   0   1
  126   0 126   0  84   0  36   0   9   0   1
    0 252   0 210   0 120   0  45   0  10   0   1
  462   0 462   0 330   0 165   0  55   0  11   0   1
    0 924   0 792   0 495   0 220   0  66   0  12   0   1
For example, row n = 5 counts the following compositions:
  .  (32)     .  (41)   .  (5)
     (122)       (113)
     (221)       (212)
     (1121)      (311)
     (2111)
     (11111)

The full triangle counting compositions by alternating sum is A097805.
The version for partitions is A103919, full triangle A344651.
This is the right-half of even-indexed rows of A260492.
The triangle without top row and left column is A108044.
Ranking and counting compositions:
- product = sum: A335404, counted by A335405.
- sum = twice alternating sum: A348614, counted by A262977.
- length = alternating sum: A357184, counted by A357182.
- length = absolute value of alternating sum: A357185, counted by A357183.
A003242 counts anti-run compositions, ranked by A333489.
A011782 counts compositions.
A025047 counts alternating compositions, ranked by A345167.
A032020 counts strict compositions, ranked by A233564.
A124754 gives alternating sums of standard compositions.
A238279 counts compositions by sum and number of maximal runs.
Cf. A000120, A051159, A070939, A114220, A114901, A242882, A262046.

Prepend[Table[If[EvenQ[nn],Prepend[#,0],#]&[Riffle[Table[Binomial[nn,k],{k,Floor[nn/2],nn}],0]],{nn,0,10}],{1}]

Showing 1-10 of 54 results. Next

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A097805 Number of compositions of n with k parts, T(n, k) = binomial(n-1, k-1) for n, k >= 1 and T(n, 0) = 0^n, triangle read by rows for n >= 0 and 0 <= k <= n.

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A103919 Triangle of numbers of partitions of n with total number of odd parts equal to k from {0,...,n}.

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A344612 Triangle read by rows where T(n,k) is the number of integer partitions of n with reverse-alternating sum k ranging from -n to n in steps of 2.

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A000097 Number of partitions of n if there are two kinds of 1's and two kinds of 2's.

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A344607 Number of integer partitions of n with reverse-alternating sum >= 0.

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A344610 Triangle read by rows where T(n,k) is the number of integer partitions of 2n with reverse-alternating sum 2k.

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A344611 Number of integer partitions of 2n with reverse-alternating sum >= 0.