cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A088218 Total number of leaves in all rooted ordered trees with n edges.

Original entry on oeis.org

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Views

Author

Michael Somos, Sep 24 2003

Keywords

Comments

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
() (11) (22) (33)
(121) (132)
(1111) (231)
(1122)
(1221)
(2112)
(2211)
(11121)
(12111)
(111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....
		

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Crossrefs

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218 (this sequence), 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 (this sequence), 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.

Programs

  • Magma
    [Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014
  • Mathematica
    a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
    c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
    Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
    a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
  • PARI
    {a(n) = sum( i=0, n, binomial(n+i-2,i))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
    
  • PARI
    {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
    
  • Sage
    def A088218(n):
        return rising_factorial(n,n)/falling_factorial(n,n)
    [A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012
    

Formula

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

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.

Original entry on oeis.org

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

Views

Author

Paul Barry, Aug 25 2004

Keywords

Comments

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 x
a(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)

Examples

			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)
		

References

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

Crossrefs

Case m=0 of the polynomials defined in A278073.
Cf. A000012 (diagonal), A011782 (row sums), A088218 (central terms).
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.

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {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 */
    
  • PARI
    T(n,k) = if (k==0, 0^n, binomial(n-1, k-1)); \\ Michel Marcus, May 06 2022
    
  • PARI
    row(n) = vector(n+1, k, k--; if (k==0, 0^n, binomial(n-1, k-1))); \\ Michel Marcus, May 06 2022
    
  • Python
    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
  • Sage
    # 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
    

Formula

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

Extensions

Corrected by Philippe Deléham, Oct 05 2005
New name using classical terminology by Peter Luschny, Feb 05 2019

A344616 Alternating sum of the integer partition with Heinz number n.

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 1, 0, 2, 5, 2, 6, 3, 1, 0, 7, 1, 8, 3, 2, 4, 9, 1, 0, 5, 2, 4, 10, 2, 11, 1, 3, 6, 1, 0, 12, 7, 4, 2, 13, 3, 14, 5, 3, 8, 15, 2, 0, 1, 5, 6, 16, 1, 2, 3, 6, 9, 17, 1, 18, 10, 4, 0, 3, 4, 19, 7, 7, 2, 20, 1, 21, 11, 2, 8, 1, 5, 22, 3, 0, 12
Offset: 1

Author

Gus Wiseman, Jun 03 2021

Keywords

Comments

The alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i, which is equal to the number of odd parts in the conjugate partition.
The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), giving a bijective correspondence between positive integers and integer partitions.
Also the reverse-alternating sum of the prime indices of n.

Examples

			The partition (6,4,3,2,2) has Heinz number 4095 and conjugate (5,5,3,2,1,1), so a(4095) = 5.
		

Crossrefs

Positions of nonzeros are A000037.
Positions of 0's are A000290.
The version for prime factors is A071321 (reverse: A071322).
A version for compositions is A124754.
The version for prime multiplicities is A316523.
The reverse version is A316524, with sign A344617.
A000041 counts partitions of 2n with alternating sum 0.
A056239 adds up prime indices, row sums of A112798.
A103919 counts partitions by sum and alternating sum.
A335433 ranks separable partitions.
A335448 ranks inseparable partitions.
A344606 counts wiggly permutations of prime indices with twins.
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.

Programs

  • Maple
    a:= n-> (l-> -add(l[i]*(-1)^i, i=1..nops(l)))(sort(map(
        i-> numtheory[pi](i[1])$i[2], ifactors(n)[2]), `>`)):
    seq(a(n), n=1..82);  # Alois P. Heinz, Jun 04 2021
  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    Table[ats[Reverse[primeMS[n]]],{n,100}]

Formula

a(n) = A257991(A122111(n)).
A057427(a(n)) = A049240(n).

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

Author

Gus Wiseman, Jun 01 2021

Keywords

Comments

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.

Examples

			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
		

Crossrefs

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.

Programs

  • Mathematica
    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}]
  • PARI
    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

Keywords

Comments

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

Examples

			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)
		

References

  • 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).

Crossrefs

First differences are in A024786.
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.
Shift of A093695.

Programs

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

Formula

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

Extensions

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

Author

Gus Wiseman, May 29 2021

Keywords

Comments

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)?

Examples

			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)
		

Crossrefs

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.

Programs

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

Formula

a(n) + A344608(n) = A000041(n).
a(2n+1) = A160786(n).

A116406 Expansion of ((1 + x - 2x^2) + (1+x)*sqrt(1-4x^2))/(2(1-4x^2)).

Original entry on oeis.org

1, 1, 2, 3, 7, 11, 26, 42, 99, 163, 382, 638, 1486, 2510, 5812, 9908, 22819, 39203, 89846, 155382, 354522, 616666, 1401292, 2449868, 5546382, 9740686, 21977516, 38754732, 87167164, 154276028, 345994216, 614429672, 1374282019, 2448023843
Offset: 0

Author

Paul Barry, Feb 13 2006

Keywords

Comments

Interleaving of A114121 and A032443. Row sums of A116405. Binomial transform is A116409.
Appears to be the number of n-digit binary numbers not having more zeros than ones; equivalently, the number of unrestricted Dyck paths of length n not going below the axis. - Ralf Stephan, Mar 25 2008
From Gus Wiseman, Jun 20 2021: (Start)
Also the number compositions of n with alternating sum >= 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. The a(0) = 1 through a(5) = 11 compositions are:
() (1) (2) (3) (4) (5)
(11) (21) (22) (32)
(111) (31) (41)
(112) (113)
(121) (122)
(211) (212)
(1111) (221)
(311)
(1121)
(2111)
(11111)
(End)
From J. Stauduhar, Jan 14 2022: (Start)
Also, for n >= 2, first differences of partial row sums of Pascal's triangle. The first ceiling(n/2)+1 elements of rows n=0 to n=4 in Pascal's triangle are:
1
1 1
1 2
1 3 3
1 4 6
...
The cumulative sums of these partial rows form the sequence 1,3,6,13,24,..., and its first differences are a(2),a(3),a(4),... in this sequence.
(End)

Crossrefs

The alternating sum = 0 case is A001700 or A088218.
The alternating sum > 0 case appears to be A027306.
The bisections are A032443 (odd) and A114121 (even).
The alternating sum <= 0 version is A058622.
The alternating sum < 0 version is A294175.
The restriction to reversed partitions is A344607.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A124754 gives the alternating sum of standard compositions.
A344610 counts partitions by sum and positive reverse-alternating sum.
A344616 lists the alternating sums of partitions by Heinz number.

Programs

  • Mathematica
    CoefficientList[Series[((1+x-2x^2)+(1+x)Sqrt[1-4x^2])/(2(1-4x^2)),{x,0,40}],x] (* Harvey P. Dale, Aug 16 2012 *)
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],ats[#]>=0&]],{n,0,15}] (* Gus Wiseman, Jun 20 2021 *)

Formula

a(n) = A114121(n/2)*(1+(-1)^n)/2 + A032443((n-1)/2)*(1-(-1)^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-1,k). - Paul Barry, Oct 06 2007
Conjecture: n*(n-3)*a(n) +2*(-n^2+4*n-2)*a(n-1) -4*(n-2)^2*a(n-2) +8*(n-2)*(n-3)*a(n-3)=0. - R. J. Mathar, Nov 28 2014
a(n) ~ 2^(n-2) * (1 + (3+(-1)^n)/sqrt(2*Pi*n)). - Vaclav Kotesovec, May 30 2016
a(n) = 2^(n-1) - A294175(n). - Gus Wiseman, Jun 27 2021

A344651 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with alternating sum k, with k ranging from n mod 2 to n in steps of 2.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 2, 1, 4, 2, 1, 3, 5, 2, 1, 7, 5, 2, 1, 5, 9, 5, 2, 1, 12, 10, 5, 2, 1, 7, 17, 10, 5, 2, 1, 19, 19, 10, 5, 2, 1, 11, 28, 20, 10, 5, 2, 1, 30, 33, 20, 10, 5, 2, 1, 15, 47, 35, 20, 10, 5, 2, 1, 45, 57, 36, 20, 10, 5, 2, 1, 22, 73, 62, 36, 20, 10, 5, 2, 1
Offset: 0

Author

Gus Wiseman, Jun 05 2021

Keywords

Comments

The alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. This is equal to the number of odd parts in the conjugate partition, so T(n,k) is the number of integer partitions of n with k odd parts in the conjugate partition, which is also the number of partitions of n with k odd parts.
Also the number of integer partitions of n with odd-indexed parts (odd bisection) summing to k, ceiling(n/2) <= k <= n. The even-indexed version is A346633. - Gus Wiseman, Nov 29 2021

Examples

			Triangle begins:
   1
   1
   1   1
   2   1
   2   2   1
   4   2   1
   3   5   2   1
   7   5   2   1
   5   9   5   2   1
  12  10   5   2   1
   7  17  10   5   2   1
  19  19  10   5   2   1
  11  28  20  10   5   2   1
  30  33  20  10   5   2   1
  15  47  35  20  10   5   2   1
  45  57  36  20  10   5   2   1
  22  73  62  36  20  10   5   2   1
  67  92  64  36  20  10   5   2   1
  30 114 102  65  36  20  10   5   2   1
  97 147 107  65  36  20  10   5   2   1
Row n = 10 counts the following partitions (A = 10):
  (55)          (64)         (73)       (82)     (91)   (A)
  (3322)        (442)        (433)      (622)    (811)
  (4411)        (541)        (532)      (721)
  (222211)      (3331)       (631)      (7111)
  (331111)      (4222)       (5221)     (61111)
  (22111111)    (4321)       (6211)
  (1111111111)  (5311)       (42211)
                (22222)      (52111)
                (32221)      (511111)
                (33211)      (4111111)
                (43111)
                (322111)
                (421111)
                (2221111)
                (3211111)
                (31111111)
                (211111111)
The conjugate version is:
  (A)      (55)      (3331)     (331111)    (31111111)   (1111111111)
  (64)     (73)      (5311)     (511111)    (211111111)
  (82)     (91)      (7111)     (3211111)
  (442)    (433)     (33211)    (4111111)
  (622)    (532)     (43111)    (22111111)
  (4222)   (541)     (52111)
  (22222)  (631)     (61111)
           (721)     (322111)
           (811)     (421111)
           (3322)    (2221111)
           (4321)
           (4411)
           (5221)
           (6211)
           (32221)
           (42211)
           (222211)
		

Crossrefs

This is A103919 with all zeros removed.
The strict version is A152146 interleaved with A152157.
The rows are those of A239830 interleaved with those of A239829.
The reverse version is the right half of A344612.
The strict reverse version is the right half of A344739.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A027187 counts partitions with rev-alternating sum <= 0, ranked by A028260.
A124754 lists alternating sums of standard compositions (reverse: A344618).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A325534/A325535 count separable/inseparable partitions.
A344607 counts partitions with rev-alternating sum >= 0, ranked by A344609.
A344608 counts partitions with rev-alternating sum < 0, ranked by A119899.
A344610 counts partitions of n by positive rev-alternating sum.
A344611 counts partitions of 2n with rev-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
A346697 gives the sum of odd-indexed prime indices (reverse: A346699).
A346702 represents the odd bisection of compositions, sums A209281.

Programs

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

A344618 Reverse-alternating sums of standard compositions (A066099). Alternating sums of the compositions ranked by A228351.

Original entry on oeis.org

0, 1, 2, 0, 3, -1, 1, 1, 4, -2, 0, 2, 2, 0, 2, 0, 5, -3, -1, 3, 1, 1, 3, -1, 3, -1, 1, 1, 3, -1, 1, 1, 6, -4, -2, 4, 0, 2, 4, -2, 2, 0, 2, 0, 4, -2, 0, 2, 4, -2, 0, 2, 2, 0, 2, 0, 4, -2, 0, 2, 2, 0, 2, 0, 7, -5, -3, 5, -1, 3, 5, -3, 1, 1, 3, -1, 5, -3, -1, 3
Offset: 0

Author

Gus Wiseman, Jun 03 2021

Keywords

Comments

Up to sign, same as A124754.
The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The sequence of nonnegative integers together with the corresponding standard compositions and their reverse-alternating sums begins:
  0:     () ->  0    15: (1111) ->  0    30:  (1112) ->  1
  1:    (1) ->  1    16:    (5) ->  5    31: (11111) ->  1
  2:    (2) ->  2    17:   (41) -> -3    32:     (6) ->  6
  3:   (11) ->  0    18:   (32) -> -1    33:    (51) -> -4
  4:    (3) ->  3    19:  (311) ->  3    34:    (42) -> -2
  5:   (21) -> -1    20:   (23) ->  1    35:   (411) ->  4
  6:   (12) ->  1    21:  (221) ->  1    36:    (33) ->  0
  7:  (111) ->  1    22:  (212) ->  3    37:   (321) ->  2
  8:    (4) ->  4    23: (2111) -> -1    38:   (312) ->  4
  9:   (31) -> -2    24:   (14) ->  3    39:  (3111) -> -2
  10:  (22) ->  0    25:  (131) -> -1    40:    (24) ->  2
  11: (211) ->  2    26:  (122) ->  1    41:   (231) ->  0
  12:  (13) ->  2    27: (1211) ->  1    42:   (222) ->  2
  13: (121) ->  0    28:  (113) ->  3    43:  (2211) ->  0
  14: (112) ->  2    29: (1121) -> -1    44:   (213) ->  4
Triangle begins (row lengths A011782):
  0
  1
  2  0
  3 -1  1  1
  4 -2  0  2  2  0  2  0
  5 -3 -1  3  1  1  3 -1  3 -1  1  1  3 -1  1  1
		

Crossrefs

Up to sign, same as the reverse version A124754.
The version for Heinz numbers of partitions is A344616.
Positions of zeros are A344619.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A116406 counts compositions with alternating sum >= 0.
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
All of the following pertain to compositions in standard order:
- The length is A000120.
- Converting to reversed ranking gives A059893.
- The rows are A066099.
- The sum is A070939.
- The runs are counted by A124767.
- The reversed version is A228351.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- The Heinz number is A333219.
- Anti-run compositions are ranked by A333489.

Programs

  • Mathematica
    sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
    stc[n_]:=Reverse[Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]]
    Table[sats[stc[n]],{n,0,100}]

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

Author

Gus Wiseman, May 31 2021

Keywords

Comments

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.

Examples

			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)
		

Crossrefs

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.

Programs

  • Mathematica
    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}]
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