A345907 -id:A345907 - 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

A345197 Concatenation of square matrices A(n), each read by rows, where A(n)(k,i) is the number of compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 2, 3, 0, 0, 2, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 3, 4, 0, 0, 3, 4, 3, 0, 0, 0, 0, 2, 3, 0, 0, 0, 0, 1, 0, 0, 0

Offset: 0

Gus Wiseman, Jul 03 2021

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

			The matrices for n = 1..7:
  1   0 1   0 0 1   0 0 0 1   0 0 0 0 1   0 0 0 0 0 1   0 0 0 0 0 0 1
      1 0   1 1 0   1 1 1 0   1 1 1 1 0   1 1 1 1 1 0   1 1 1 1 1 1 0
            0 1 0   0 1 2 0   0 1 2 3 0   0 1 2 3 4 0   0 1 2 3 4 5 0
                    0 1 0 0   0 2 2 0 0   0 3 4 3 0 0   0 4 6 6 4 0 0
                              0 0 1 0 0   0 0 2 3 0 0   0 0 3 6 6 0 0
                                          0 0 1 0 0 0   0 0 3 3 0 0 0
                                                        0 0 0 1 0 0 0
Matrix n = 5 counts the following compositions:
           i=-3:        i=-1:          i=1:            i=3:        i=5:
        -----------------------------------------------------------------
   k=1: |    0            0             0               0          (5)
   k=2: |   (14)         (23)          (32)            (41)         0
   k=3: |    0          (131)       (221)(122)   (311)(113)(212)    0
   k=4: |    0       (1211)(1112)  (2111)(1121)         0           0
   k=5: |    0            0          (11111)            0           0

Gus Wiseman, A raster plot of the zeros in A(16).

The number of nonzero terms in each matrix appears to be A000096.
The number of zeros in each matrix appears to be A000124.
Row sums and column sums both appear to be A007318 (Pascal's triangle).
The matrix sums are A131577.
Antidiagonal sums appear to be A163493.
The reverse-alternating version is also A345197 (this sequence).
Antidiagonals are A345907.
Traces are A345908.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Other tetrangles: A318393, A318816, A320808, A321912.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000097, A000346, A007318, A008549, A025047, A032443, A034871, A114121, A120452, A238279, A239830, A344604.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&&ats[#]==i&]],{n,0,6},{k,1,n},{i,-n+2,n,2}]

A163493 Number of binary strings of length n which have the same number of 00 and 01 substrings.

Original entry on oeis.org

1, 2, 2, 3, 6, 9, 15, 30, 54, 97, 189, 360, 675, 1304, 2522, 4835, 9358, 18193, 35269, 68568, 133737, 260802, 509132, 995801, 1948931, 3816904, 7483636, 14683721, 28827798, 56637969, 111347879, 219019294, 431043814, 848764585, 1672056525, 3295390800, 6497536449

Offset: 0

Christopher Carl Heckman, Jul 29 2009

nonn

A variation of problem 11424 in the American Mathematical Monthly. Terms were brute-force calculated using Maple 10.
Proposed Problem 11610 in the Dec 2011 A.M.M.
From Gus Wiseman, Jul 27 2021: (Start)
Also the antidiagonal sums of the matrices counting integer compositions by length and alternating sum (A345197). So a(n) is the number of integer compositions of n + 1 of length (n - s + 3)/2, where s is the alternating sum of the composition. For example, the a(0) = 1 through a(6) = 7 compositions are:
  (1)  (2)   (3)   (4)    (5)     (6)     (7)
       (11)  (21)  (31)   (41)    (51)    (61)
                   (121)  (122)   (123)   (124)
                          (221)   (222)   (223)
                          (1112)  (321)   (322)
                          (1211)  (1122)  (421)
                                  (1221)  (1132)
                                  (2112)  (1231)
                                  (2211)  (2122)
                                          (2221)
                                          (3112)
                                          (3211)
                                          (11131)
                                          (12121)
                                          (13111)
For a bijection with the main (binary string) interpretation, take the run-lengths of each binary string of length n + 1 that satisfies the condition and starts with 1.
(End)

			1 + 2*x + 2*x^2 + 3*x^3 + 6*x^4 + 9*x^5 + 15*x^6 + 30*x^7 + 54*x^8 + 97*x^9 + ...
From _Gus Wiseman_, Jul 27 2021: (Start)
The a(0) = 1 though a(6) = 15 binary strings:
  ()  (0)  (1,0)  (0,0,1)  (0,0,1,0)  (0,0,1,1,0)  (0,0,0,1,0,1)
      (1)  (1,1)  (1,1,0)  (0,0,1,1)  (0,0,1,1,1)  (0,0,1,0,0,1)
                  (1,1,1)  (0,1,0,0)  (0,1,1,0,0)  (0,0,1,1,1,0)
                           (1,0,0,1)  (1,0,0,1,0)  (0,0,1,1,1,1)
                           (1,1,1,0)  (1,0,0,1,1)  (0,1,0,0,0,1)
                           (1,1,1,1)  (1,0,1,0,0)  (0,1,1,1,0,0)
                                      (1,1,0,0,1)  (1,0,0,1,1,0)
                                      (1,1,1,1,0)  (1,0,0,1,1,1)
                                      (1,1,1,1,1)  (1,0,1,1,0,0)
                                                   (1,1,0,0,1,0)
                                                   (1,1,0,0,1,1)
                                                   (1,1,0,1,0,0)
                                                   (1,1,1,0,0,1)
                                                   (1,1,1,1,1,0)
                                                   (1,1,1,1,1,1)
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..3328 (first 501 terms from R. H. Hardin)
Shalosh B. Ekhad and Doron Zeilberger, Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type, arXiv preprint arXiv:1112.6207 [math.CO], 2011. See subpages for rigorous derivations of g.f., recurrence, asymptotics for this sequence. [From _N. J. A. Sloane_, Apr 07 2012]
R. Stanley, Problem 11610, Amer. Math. Monthly, 118 (2011), 937; 120 (2013), 943-944.

Antidiagonal sums of the matrices A345197.
Row sums of A345907.
Taking diagonal instead of antidiagonal sums gives A345908.
A011782 counts compositions (or binary strings).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000041, A000070, A000096, A000097, A000124, A000346, A007318, A008549, A025047, A131577, A238279.

with(combinat): count := proc(n) local S, matches, A, k, i; S := subsets(\{seq(i, i=1..n)\}): matches := 0: while not S[finished] do A := S[nextvalue](): k := 0: for i from 1 to n-1 do: if not (i in A) and not (i+1 in A) then k := k + 1: fi: if not (i in A) and (i+1 in A) then k := k - 1: fi: od: if (k = 0) then matches := matches + 1: fi: end do; return(matches); end proc:
# second Maple program:
b:= proc(n, l, t) option remember; `if`(n-abs(t)<0, 0, `if`(n=0, 1,
      add(b(n-1, i, t+`if`(l=0, (-1)^i, 0)), i=0..1)))
    end:
a:= n-> b(n, 1, 0):
seq(a(n), n=0..36);  # Alois P. Heinz, Mar 20 2024

a[0] = 1; a[n_] := Sum[Binomial[2*k - 1, k]*Binomial[n - 2*k, k] + Binomial[2*k, k]*Binomial[n - 2*k - 1, k], {k, 0, n/3}];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 28 2017, after Joel B. Lewis *)
Table[Length[Select[Tuples[{0,1},n],Count[Partition[#,2,1],{0,0}]==Count[Partition[#,2,1],{0,1}]&]],{n,0,10}] (* Gus Wiseman, Jul 27 2021 *)
a[0]:=1; a[n_]:=(1 + 3*HypergeometricPFQ[{1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3},{1, (1-n)/2, 1-n/2, -3*n/8}, -27])/2; Array[a,37,0] (* Stefano Spezia, Apr 26 2024 *)

from math import comb
def A163493(n): return 2+sum((x:=comb((k:=m<<1)-1,m)*comb(n-k,m))+(x*(n-3*m)<<1)//(n-k) for m in range(1,n//3+1)) if n else 1 # Chai Wah Wu, May 01 2024

G.f.: 1/2/(1-x) + (1+2*x)/2/sqrt((1-x)*(1-2*x)*(1+x+2*x^2)). - Richard Stanley, corrected Apr 29 2011
G.f.: (1 + sqrt( 1 + 4*x / ((1 - x) * (1 - 2*x) * (1 + x + 2*x^2)))) / (2*(1 - x)). - Michael Somos, Jan 30 2012
a(n) = sum( binomial(2*k-1, k)*binomial(n-2*k,k) + binomial(2*k, k)*binomial(n-2*k-1, k), k=0..floor(n/3)). - Joel B. Lewis, May 21 2011
Conjecture: -n*a(n) +(2+n)*a(n-1) +(3n-12)*a(n-2) +(12-n)*a(n-3) +(2n-18)*a(n-4)+(56-12n)*a(n-5) +(8n-40)*a(n-6)=0. - R. J. Mathar, Nov 28 2011
G.f. y = A(x) satisfies x = (1 - x) * (1 - 2*x) * (1 + x + 2*x^2) * y * (y * (1 - x) - 1). - Michael Somos, Jan 30 2012
Sequence a(n) satisfies 0 = a(n) * (n^2-2*n) + a(n-1) * (-3*n^2+8*n-2) + a(n-2) * (3*n^2-10*n+2) + a(n-3) * (-5*n^2+18*n-6) + a(n-4) * (8*n^2-34*n+22) + a(n-5) * (-4*n^2+20*n-16) except if n=1 or n=2. - Michael Somos, Jan 30 2012
a(n) = (1 + 3*hypergeom([1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3],[1, (1-n)/2, 1-n/2, -3*n/8],-27))/2 for n > 0. - Stefano Spezia, Apr 26 2024
a(n) ~ 2^n / sqrt(Pi*n). - Vaclav Kotesovec, Apr 26 2024

A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.

Original entry on oeis.org

1, 1, 0, 1, 3, 3, 6, 15, 24, 43, 92, 171, 315, 629, 1218, 2313, 4523, 8835, 17076, 33299, 65169

Offset: 0

Gus Wiseman, Jul 26 2021

The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. Here, the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. So a(n) is the number of compositions of n of length (n + s)/2, where s is the alternating sum of the composition.

			The a(0) = 1 through a(7) = 15 compositions of n = 0..7 of length (n + s)/2 where s = alternating sum (empty column indicated by dot):
  ()  (1)  .  (2,1)  (2,2)    (2,3)    (2,4)      (2,5)
                     (1,1,2)  (1,2,2)  (1,3,2)    (1,4,2)
                     (2,1,1)  (2,2,1)  (2,3,1)    (2,4,1)
                                       (1,1,3,1)  (1,1,3,2)
                                       (2,1,2,1)  (1,2,3,1)
                                       (3,1,1,1)  (2,1,2,2)
                                                  (2,2,2,1)
                                                  (3,1,1,2)
                                                  (3,2,1,1)
                                                  (1,1,1,1,3)
                                                  (1,1,2,1,2)
                                                  (1,1,3,1,1)
                                                  (2,1,1,1,2)
                                                  (2,1,2,1,1)
                                                  (3,1,1,1,1)

Traces of the matrices given by A345197.
Diagonals and antidiagonals of the same matrices are A346632 and A345907.
Row sums of A346632.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Other diagonals are A008277 of A318393 and A055884 of A320808.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A001405, A007318, A008549, A025047, A114121, A163493.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==(n+ats[#])/2&]],{n,0,15}]

A346632 Triangle read by rows giving the main diagonals of the matrices counting integer compositions by length and alternating sum (A345197).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 1, 2, 0, 0, 0, 1, 2, 3, 0, 0, 0, 1, 2, 6, 6, 0, 0, 0, 1, 2, 9, 12, 0, 0, 0, 0, 1, 2, 12, 18, 10, 0, 0, 0, 0, 1, 2, 15, 24, 30, 20, 0, 0, 0, 0, 1, 2, 18, 30, 60, 60, 0, 0, 0, 0, 0, 1, 2, 21, 36, 100, 120, 35, 0, 0, 0, 0

Offset: 0

Gus Wiseman, Jul 26 2021

The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			Triangle begins:
   1
   0   0
   0   1   0
   0   1   2   0
   0   1   2   0   0
   0   1   2   3   0   0
   0   1   2   6   6   0   0
   0   1   2   9  12   0   0   0
   0   1   2  12  18  10   0   0   0
   0   1   2  15  24  30  20   0   0   0
   0   1   2  18  30  60  60   0   0   0   0
   0   1   2  21  36 100 120  35   0   0   0   0
   0   1   2  24  42 150 200 140  70   0   0   0   0
   0   1   2  27  48 210 300 350 280   0   0   0   0   0
   0   1   2  30  54 280 420 700 700 126   0   0   0   0   0

The first nonzero element in each column appears to be A001405.
These are the diagonals of the matrices given by A345197.
Antidiagonals of the same matrices are A345907.
Row sums are A345908.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Other diagonals are A008277 of A318393 and A055884 of A320808.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A007318, A008549, A025047, A163493, A344610.

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

Showing 1-5 of 5 results.

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.

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A345197 Concatenation of square matrices A(n), each read by rows, where A(n)(k,i) is the number of compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2.

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A163493 Number of binary strings of length n which have the same number of 00 and 01 substrings.

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A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.

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A346632 Triangle read by rows giving the main diagonals of the matrices counting integer compositions by length and alternating sum (A345197).

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