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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A001003 Schroeder's second problem (generalized parentheses); also called super-Catalan numbers or little Schroeder numbers.

Original entry on oeis.org

1, 1, 3, 11, 45, 197, 903, 4279, 20793, 103049, 518859, 2646723, 13648869, 71039373, 372693519, 1968801519, 10463578353, 55909013009, 300159426963, 1618362158587, 8759309660445, 47574827600981, 259215937709463, 1416461675464871
Offset: 0

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Comments

If you are looking for the Schroeder numbers (a.k.a. large Schroder numbers, or big Schroeder numbers), see A006318.
Yang & Jiang (2021) call these the small 2-Schroeder numbers. - N. J. A. Sloane, Mar 28 2021
There are two schools of thought about the index for the first term. I prefer the indexing a(0) = a(1) = 1, a(2) = 3, a(3) = 11, etc.
a(n) is the number of ways to insert parentheses in a string of n+1 symbols. The parentheses must be balanced but there is no restriction on the number of pairs of parentheses. The number of letters inside a pair of parentheses must be at least 2. Parentheses enclosing the whole string are ignored.
Also length of list produced by a variant of the Catalan producing iteration: replace each integer k with the list 0,1,..,k,k+1,k,...,1,0 and get the length a(n) of the resulting (flattened) list after n iterations. - Wouter Meeussen, Nov 11 2001
Stanley gives several other interpretations for these numbers.
Number of Schroeder paths of semilength n (i.e., lattice paths from (0,0) to (2n,0), with steps H=(2,0), U=(1,1) and D(1,-1) and not going below the x-axis) with no peaks at level 1. Example: a(2)=3 because among the six Schroeder paths of semilength two HH, UHD, UUDD, HUD, UDH and UDUD, only the first three have no peaks at level 1. - Emeric Deutsch, Dec 27 2003
a(n+1) is the number of Dyck n-paths in which the interior vertices of the ascents are colored white or black. - David Callan, Mar 14 2004
Number of possible schedules for n time slots in the first-come first-served (FCFS) printer policy.
Also row sums of A086810, A033282, A126216. - Philippe Deléham, May 09 2004
a(n+1) is the number of pairs (u,v) of same-length compositions of n, 0's allowed in u but not in v and u dominates v (meaning u_1 >= v_1, u_1 + u_2 >= v_1 + v_2 and so on). For example, with n=2, a(3) counts (2,2), (1+1,1+1), (2+0,1+1). - David Callan, Jul 20 2005
The big Schroeder number (A006318) is the number of Schroeder paths from (0,0) to (n,n) (subdiagonal paths with steps (1,0) (0,1) and (1,1)). These paths fall in two classes: those with steps on the main diagonal and those without. These two classes are equinumerous and the number of paths in either class is the little Schroeder number a(n) (half the big Schroeder number). - Marcelo Aguiar (maguiar(AT)math.tamu.edu), Oct 14 2005
With offset 0, a(n) = number of (colored) Motzkin (n-1)-paths with each upstep U getting one of 2 colors and each flatstep F getting one of 3 colors. Example. With their colors immediately following upsteps/flatsteps, a(2) = 3 counts F1, F2, F3 and a(3)=11 counts U1D, U2D, F1F1, F1F2, F1F3, F2F1, F2F2, F2F3, F3F1, F3F2, F3F3. - David Callan, Aug 16 2006
Shifts left when INVERT transform applied twice. - Alois P. Heinz, Apr 01 2009
Number of increasing tableaux of shape (n,n). An increasing tableau is a semistandard tableaux with strictly increasing rows and columns, and set of entries an initial segment of the positive integers. Example: a(2) = 3 because of the three tableaux (12)(34), (13)(24), (12)(23). - Oliver Pechenik, Apr 22 2014
Number of ordered trees with no vertex of outdegree 1 and having n+1 leaves (called sometimes Schröder trees). - Emeric Deutsch, Dec 13 2014
Number of dissections of a convex (n+2)-gon by nonintersecting diagonals. Example: a(2)=3 because for a square ABCD we have (i) no diagonal, (ii) dissection with diagonal AC, and (iii) dissection with diagonal BD. - Emeric Deutsch, Dec 13 2014
The little Schroeder numbers are the moments of the Marchenko-Pastur law for the case c=2 (although the moment m0 is 1/2 instead of 1): 1/2, 1, 3, 11, 45, 197, 903, ... - Jose-Javier Martinez, Apr 07 2015
Number of generalized Motzkin paths with no level steps at height 0, from (0,0) to (2n,0), and consisting of steps U=(1,1), D=(1,-1) and H2=(2,0). For example, for n=3, we have the 11 paths: UDUDUD, UUDDUD, UDUUDD, UH2DUD, UDUH2D, UH2H2D, UUDUDD, UUUDDD, UUH2DD, UUDH2D, UH2UDD. - José Luis Ramírez Ramírez, Apr 20 2015
REVERT transform of A225883. - Vladimir Reshetnikov, Oct 25 2015
Total number of (nonempty) faces of all dimensions in the associahedron K_{n+1} of dimension n-1. For example, K_4 (a pentagon) includes 5 vertices and 5 edges together with K_4 itself (5 + 5 + 1 = 11), while K_5 includes 14 vertices, 21 edges and 9 faces together with K_5 itself (14 + 21 + 9 + 1 = 45). - Noam Zeilberger, Sep 17 2018
a(n) is the number of interval posets of permutations with n minimal elements that have exactly two realizers, up to a shift by 1 in a(4). See M. Bouvel, L. Cioni, B. Izart, Theorem 17 page 13. - Mathilde Bouvel, Oct 21 2021
a(n) is the number of sequences of nonnegative integers (u_1, u_2, ..., u_n) such that (i) u_1 = 1, (ii) u_i <= i for all i, (iii) the nonzero u_i are weakly increasing. For example, a(2) = 3 counts 10, 11, 12, and a(3) = 11 counts 100, 101, 102, 103, 110, 111, 112, 113, 120, 122, 123. See link below. - David Callan, Dec 19 2021
a(n) is the number of parking functions of size n avoiding the patterns 132 and 213. - Lara Pudwell, Apr 10 2023
a(n+1) is the number of Schroeder paths from (0,0) to (2n,0) in which level steps at height 0 come in 2 colors. - Alexander Burstein, Jul 23 2023

Examples

			G.f. = 1 + x + 3*x^2 + 11*x^3 + 45*x^4 + 197*x^5 + 903*x^6 + 4279*x^7 + ...
a(2) = 3: abc, a(bc), (ab)c; a(3) = 11: abcd, (ab)cd, a(bc)d, ab(cd), (ab)(cd), a(bcd), a(b(cd)), a((bc)d), (abc)d, (a(bc))d, ((ab)c)d.
Sum over partitions formula: a(3) = 2*a(0)*a(2) + 1*a(1)^2 + 3*(a(0)^2)*a(1) + 1*a(0)^4 = 6 + 1 + 3 + 1 = 11.
a(4) = 45 since the top row of Q^3 = (11, 14, 12, 8, 0, 0, 0, ...); (11 + 14 + 12 + 8) = 45.
		

References

  • D. Arques and A. Giorgetti, Une bijection géometrique entre une famille d'hypercartes et une famille de polygones énumérées par la série de Schroeder, Discrete Math., 217 (2000), 17-32.
  • Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
  • N. Bernasconi et al., On properties of random dissections and triangulations, Combinatorica, 30 (6) (2010), 627-654.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 618.
  • Peter J. Cameron, Some treelike objects. Quart. J. Math. Oxford Ser. (2) 38 (1987), no. 150, 155--183. MR0891613 (89a:05009). See p. 155, also p. 179, line -9. - N. J. A. Sloane, Apr 18 2014
  • C. Coker, A family of eigensequences, Discrete Math. 282 (2004), 249-250.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 57.
  • D. E. Davenport, L. W. Shapiro and L. C. Woodson, The Double Riordan Group, The Electronic Journal of Combinatorics, 18(2) (2012), #P33. - From N. J. A. Sloane, May 11 2012
  • Emeric Deutsch, A bijective proof of an equation linking the Schroeder numbers, large and small, Discrete Math., 241 (2001), 235-240.
  • Tomislav Doslic and Darko Veljan, Logarithmic behavior of some combinatorial sequences. Discrete Math. 308 (2008), no. 11, 2182--2212. MR2404544 (2009j:05019) - From N. J. A. Sloane, May 01 2012
  • Michael Drmota, Anna de Mier, and Marc Noy, Extremal statistics on non-crossing configurations. Discrete Math. 327 (2014), 103--117. MR3192420. See Eq. (2). - N. J. A. Sloane, May 18 2014
  • I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
  • I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
  • P. Flajolet and M. Noy, Analytic combinatorics of non-crossing permutations, Discrete Math., 204 (1999), 203-229, Section 3.1.
  • D. Foata and D. Zeilberger, A classic proof of a recurrence for a very classical sequence, J. Comb Thy A 80 380-384 1997.
  • Wolfgang Gatterbauer and Dan Suciu, Dissociation and propagation for approximate lifted inference with standard relational database management systems, The VLDB Journal, February 2017, Volume 26, Issue 1, pp. 5-30; DOI 10.1007/s00778-016-0434-5
  • Ivan Geffner and Marc Noy, Counting Outerplanar Maps, Electronic Journal of Combinatorics 24(2) (2017), #P2.3.
  • D. Gouyou-Beauchamps and B. Vauquelin, Deux propriétés combinatoires des nombres de Schroeder, Theor. Inform. Appl., 22 (1988), 361-388.
  • N. S. S. Gu, N. Y. Li and T. Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
  • P.-Y. Huang, S.-C. Liu, and Y.-N. Yeh, Congruences of Finite Summations of the Coefficients in certain Generating Functions, The Electronic Journal of Combinatorics, 21 (2014), #P2.45.
  • M. Klazar, On numbers of Davenport-Schinzel sequences, Discr. Math., 185 (1998), 77-87.
  • D. E. Knuth, The Art of Computer Programming, Vol. 1, various sections (e.g. p. 534 of 2nd ed.).
  • D. E. Knuth, The Art of Computer Programming, Vol. 1, (p. 539 of 3rd ed.).
  • D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.2.1.6, Problem 66, p. 479.
  • J. S. Lew, Polynomial enumeration of multidimensional lattices, Math. Systems Theory, 12 (1979), 253-270.
  • Ana Marco and J.-J. Martinez, A total positivity property of the Marchenko-Pastur Law, Electronic Journal of Linear Algebra, 30 (2015), #7.
  • T. S. Motzkin, Relations between hypersurface cross ratios and a combinatorial formula for partitions of a polygon, for permanent preponderance and for non-associative products, Bull. Amer. Math. Soc., 54 (1948), 352-360.
  • L. Ozsvart, Counting ordered graphs that avoid certain subgraphs, Discr. Math., 339 (2016), 1871-1877.
  • R. C. Read, On general dissections of a polygon, Aequat. Mathem. 18 (1978) 370-388, Table 6
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 168.
  • E. Schroeder, Vier combinatorische Probleme, Zeit. f. Math. Phys., 15 (1870), 361-376.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see page 178; see page 239, Exercise 6.39b.
  • H. N. V. Temperley and D. G. Rogers, A note on Baxter's generalization of the Temperley-Lieb operators, pp. 324-328 of Combinatorial Mathematics (Canberra, 1977), Lect. Notes Math. 686, 1978.
  • I. Vardi, Computational Recreations in Mathematica. Addison-Wesley, Redwood City, CA, 1991, p. 198.
  • Sheng-Liang Yang and Mei-yang Jiang, The m-Schröder paths and m-Schröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.

Crossrefs

See A000081, A000108, A001190, A001699, for other ways to count parentheses.
Row sums of A033282, A033877, A086810, A126216.
Right-hand column 1 of convolution triangle A011117.
Column 1 of A336573. Column 0 of A104219.
The sequences listed in Yang-Jiang's Table 1 appear to be A006318, this sequence, A027307, A034015, A144097, A243675, A260332, A243676. - N. J. A. Sloane, Mar 28 2021
Cf. A006318 (Schroeder numbers).

Programs

  • Haskell
    a001003 = last . a144944_row  -- Reinhard Zumkeller, May 11 2013
    
  • Magma
    R:=PowerSeriesRing(Rationals(), 50);
    Coefficients(R!( (1+x -Sqrt(1-6*x+x^2) )/(4*x) )); // G. C. Greubel, Oct 27 2024
  • Maple
    t1 := (1/(4*x))*(1+x-sqrt(1-6*x+x^2)); series(t1,x,40);
    invtr:= proc(p) local b; b:= proc(n) option remember; local i; `if`(n<1, 1, add(b(n-i) *p(i-1), i=1..n+1)) end end: a:='a': f:= (invtr@@2)(a): a:= proc(n) if n<0 then 1 else f(n-1) fi end: seq(a(n), n=0..30); # Alois P. Heinz, Apr 01 2009
    # Computes n -> [a[0],a[1],..,a[n]]
    A001003_list := proc(n) local j,a,w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w] := a[w-1]+2*add(a[j]*a[w-j-1],j=1..w-1) od;
    convert(a,list) end: A001003_list(100); # Peter Luschny, May 17 2011
  • Mathematica
    Table[Length[Flatten[Nest[ #/.a_Integer:> Join[Range[0, a + 1], Range[a, 0, -1]] &, {0}, n]]], {n, 0, 10}]; Sch[ 0 ] = Sch[ 1 ] = 1; Sch[ n_Integer ] := Sch[ n ] = (3(2n - 1)Sch[ n - 1 ] - (n - 2)Sch[ n - 2 ])/(n + 1); Array[ Sch, 24, 0]
    (* Second program: *)
    a[n_] := Hypergeometric2F1[-n + 1, n + 2, 2, -1]; a[0] = 1; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Nov 09 2011, after Vladeta Jovovic *)
    a[ n_] := SeriesCoefficient[ (1 + x - Sqrt[1 - 6 x + x^2]) / (4 x), {x, 0, n}]; (* Michael Somos, Aug 26 2015 *)
    Table[(KroneckerDelta[n] - GegenbauerC[n+1, -1/2, 3])/4, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 25 2015 *)
    a[n_] := -LegendreP[n, -1, 2, 3] I / Sqrt[2]; a[0] = 1;
    Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Feb 16 2019 *)
    a[1]:=1; a[2]:=1; a[n_]:=a[n] = a[n-1]+2 Sum[a[k] a[n-k], {k,2,n-1}]; Map[a, Range[24]] (* Oliver Seipel, Nov 03 2024, after Schröder 1870 *)
    CoefficientList[InverseSeries[Series[x/(Series[(1 - x)/(1 - 2  x), {x, 0, 24}]), {x, 0, 24}]]/x, x] (* Mats Granvik, Jun 30 2025 *)
  • PARI
    {a(n) = if( n<1, n==0, sum( k=0, n, 2^k * binomial(n, k) * binomial(n, k-1) ) / (2*n))}; /* Michael Somos, Mar 31 2007 */
    
  • PARI
    {a(n) = my(A); if( n<1, n==0, n--; A = x * O(x^n); n! * simplify( polcoef( exp(3*x + A) * besseli(1, 2*x * quadgen(8) + A), n)))}; /* Michael Somos, Mar 31 2007 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoef( serreverse( (x - 2*x^2) / (1 - x) + x * O(x^n)), n))}; /* Michael Somos, Mar 31 2007 */
    
  • PARI
    N=30; x='x+O('x^N); Vec( (1+x-(1-6*x+x^2)^(1/2))/(4*x) ) \\ Hugo Pfoertner, Nov 19 2018
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [a(0), a(1), ..., a(n)]
    def A001003_list(n):
        a = [0 for i in range(n)]
        a[0] = 1
        for w in range(1, n):
            s = 0
            for j in range(1, w):
                s += a[j]*a[w-j-1]
            a[w] = a[w-1]+2*s
        return a
    # Peter Luschny, May 17 2011
    
  • Python
    from gmpy2 import divexact
    A001003 = [1, 1]
    for n in range(3,10**3):
        A001003.append(divexact(A001003[-1]*(6*n-9)-(n-3)*A001003[-2],n))
    # Chai Wah Wu, Sep 01 2014
    
  • Sage
    # Generalized algorithm of L. Seidel
    def A001003_list(n) :
        D = [0]*(n+1); D[1] = 1/2
        b = True; h = 2; R = [1]
        for i in range(2*n-2) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1;
            else :
                for k in range(1,h, 1) : D[k] += D[k-1]
                R.append(D[h-1]);
            b = not b
        return R
    A001003_list(24) # Peter Luschny, Jun 02 2012
    

Formula

D-finite with recurrence: (n+1) * a(n) = (6*n-3) * a(n-1) - (n-2) * a(n-2) if n>1. a(0) = a(1) = 1.
a(n) = 3*a(n-1) + 2*A065096(n-2) (n>2). If g(x) = 1 + 3*x + 11*x^2 + 45*x^3 + ... + a(n)*x^n + ..., then g(x) = 1 + 3(x*g(x)) + 2(x*g(x))^2, g(x)^2 = 1 + 6*x + 31*x^2 + 156*x^3 + ... + A065096(n)*x^n + ... - Paul D. Hanna, Jun 10 2002
a(n+1) = -a(n) + 2*Sum_{k=1..n} a(k)*a(n+1-k). - Philippe Deléham, Jan 27 2004
a(n-2) = (1/(n-1))*Sum_{k=0..n-3} binomial(n-1,k+1)*binomial(n-3,k)*2^(n-3-k) for n >= 3 [G. Polya, Elemente de Math., 12 (1957), p. 115.] - N. J. A. Sloane, Jun 13 2015
G.f.: (1 + x - sqrt(1 - 6*x + x^2) )/(4*x) = 2/(1 + x + sqrt(1 - 6*x + x^2)).
a(n) ~ W*(3+sqrt(8))^n*n^(-3/2) where W = (1/4)*sqrt((sqrt(18)-4)/Pi) [See Knuth I, p. 534, or Perez. Note that the formula on line 3, page 475 of Flajolet and Sedgewick seems to be wrong - it has to be multiplied by 2^(1/4).] - N. J. A. Sloane, Apr 10 2011
The correct asymptotic for this sequence is a(n) ~ W*(3+sqrt(8))^n*n^(-3/2), where W = (1+sqrt(2))/(2*2^(3/4)*sqrt(Pi)) = 0.404947065905750651736243... Result in book by D. Knuth (p. 539 of 3rd edition, exercise 12) is for sequence b(n), but a(n) = b(n+1)/2. Therefore is asymptotic a(n) ~ b(n) * (3+sqrt(8))/2. - Vaclav Kotesovec, Sep 09 2012
The Hankel transform of this sequence gives A006125 = 1, 1, 2, 8, 64, 1024, ...; example: det([1, 1, 3, 11; 1, 3, 11, 45; 3, 11, 45, 197; 11, 45, 197, 903]) = 2^6 = 64. - Philippe Deléham, Mar 02 2004
a(n+1) = Sum_{k=0..floor((n-1)/2)} 2^k * 3^(n-1-2k) * binomial(n-1, 2k) * Catalan(k). This formula counts colored Dyck paths by the same parameter by which Touchard's identity counts ordinary Dyck paths: number of DDUs (U=up step, D=down step). See also Gouyou-Beauchamps reference. - David Callan, Mar 14 2004
From Paul Barry, May 24 2005: (Start)
a(n) = (1/(n+1))*Sum_{k=0..n} C(n+1, k)*C(2n-k, n)*(-1)^k*2^(n-k) [with offset 0].
a(n) = (1/(n+1))*Sum_{k=0..n} C(n+1, k+1)*C(n+k, k)*(-1)^(n-k)*2^k [with offset 0].
a(n) = Sum_{k=0..n} (1/(k+1))*C(n, k)*C(n+k, k)*(-1)^(n-k)*2^k [with offset 0].
a(n) = Sum_{k=0..n} A088617(n, k)*(-1)^(n-k)*2^k [with offset 0]. (End)
E.g.f. of a(n+1) is exp(3*x)*BesselI(1, 2*sqrt(2)*x)/(sqrt(2)*x). - Vladeta Jovovic, Mar 31 2004
Reversion of (x-2*x^2)/(1-x) is g.f. offset 1.
For n>=1, a(n) = Sum_{k=0..n} 2^k*N(n, k) where N(n, k) = (1/n)*C(n, k)*C(n, k+1) are the Narayana numbers (A001263). - Benoit Cloitre, May 10 2003 [This formula counts colored Dyck paths by number of peaks, which is easy to see because the Narayana numbers count Dyck paths by number of peaks and the number of peaks determines the number of interior ascent vertices.]
a(n) = Sum_{k=0..n} A088617(n, k)*2^k*(-1)^(n-k). - Philippe Deléham, Jan 21 2004
For n > 0, a(n) = (1/(n+1)) * Sum_{k = 0 .. n-1} binomial(2*n-k, n) * binomial(n-1, k). This formula counts colored Dyck paths (as above) by number of white vertices. - David Callan, Mar 14 2004
a(n-1) = (d^(n-1)/dx^(n-1))((1-x)/(1-2*x))^n/n!|_{x=0}. (For a proof see the comment on the unsigned row sums of triangle A111785.)
From Wolfdieter Lang, Sep 12 2005: (Start)
a(n) = (1/n)*Sum_{k=1..n} binomial(n, k)*binomial(n+k, k-1).
a(n) = hypergeom([1-n, n+2], [2], -1), n>=1. (End)
a(n) = hypergeom([1-n, -n], [2], 2) for n>=0. - Peter Luschny, Sep 22 2014
a(m+n+1) = Sum_{k>=0} A110440(m, k)*A110440(n, k)*2^k = A110440(m+n, 0). - Philippe Deléham, Sep 14 2005
Sum over partitions formula (reference Schroeder paper p. 362, eq. (1) II). Number the partitions of n according to Abramowitz-Stegun pp. 831-832 (see reference under A105805) with k=1..p(n)= A000041(n). For n>=1: a(n-1) = Sum_{k=2..p(n)} A048996(n,k)*a(1)^e(k, 1)*a(1)^e(k, 2)*...*a(n-2)^e(k, n-1) if the k-th partition of n in the mentioned order is written as (1^e(k, 1), 2^e(k, 2), ..., (n-1)e(k, n-1)). Note that the first (k=1) partition (n^1) has to be omitted. - Wolfdieter Lang, Aug 23 2005
Starting (1, 3, 11, 45, ...), = row sums of triangle A126216 = A001263 * [1, 2, 4, 8, 16, ...]. - Gary W. Adamson, Nov 30 2007
From Paul Barry, May 15 2009: (Start)
G.f.: 1/(1+x-2x/(1+x-2x/(1+x-2x/(1+x-2x/(1-.... (continued fraction).
G.f.: 1/(1-x/(1-x-x/(1-x-x/(1-x-x/(1-... (continued fraction).
G.f.: 1/(1-x-2x^2/(1-3x-2x^2/(1-3x-2x^2/(1-... (continued fraction). (End)
G.f.: 1 / (1 - x / (1 - 2*x / (1 - x / (1 - 2*x / ... )))). - Michael Somos, May 19 2013
a(n) = (LegendreP(n+1,3)-3*LegendreP(n,3))/(4*n) for n>0. - Mark van Hoeij, Jul 12 2010 [This formula is mentioned in S.-J. Kettle's 1982 letter - see link. N. J. A. Sloane, Jun 13 2015]
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) = upper left term in M^n, where M is the production matrix:
1, 1, 0, 0, 0, 0, ...
2, 2, 2, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
2, 2, 2, 2, 2, 0, ...
1, 1, 1, 1, 1, 1, ...
... (End)
From Gary W. Adamson, Aug 23 2011: (Start)
a(n) is the sum of top row terms of Q^(n-1), where Q is the infinite square production matrix:
1, 2, 0, 0, 0, ...
1, 1, 2, 0, 0, ...
1, 1, 1, 2, 0, ...
1, 1, 1, 1, 2, ...
... (End)
Let h(t) = (1-t)^2/(2*(1-t)^2-1) = 1/(1-(2*t+3*t^2+4*t^3+...)), an o.g.f. for A003480, then for A001003 a(n) = (1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=1. (Cf. A086810.) - Tom Copeland, Sep 06 2011
A006318(n) = 2*a(n) if n>0. - Michael Somos, Mar 31 2007
BINOMIAL transform is A118376 with offset 0. REVERT transform is A153881. INVERT transform is A006318. INVERT transform of A114710. HANKEL transform is A139685. PSUM transform is A104858. - Michael Somos, May 19 2013
G.f.: 1 + x/(Q(0) - x) where Q(k) = 1 + k*(1-x) - x - x*(k+1)*(k+2)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n) = A144944(n,n) = A186826(n,0). - Reinhard Zumkeller, May 11 2013
a(n)=(-1)^n*LegendreP(n,-1,-3)/sqrt(2), n > 0, LegendreP(n,a,b) is the Legendre function. - Karol A. Penson, Jul 06 2013
Integral representation as n-th moment of a positive weight function W(x) = W_a(x) + W_c(x), where W_a(x) = Dirac(x)/2, is the discrete (atomic) part, and W_c(x) = sqrt(8-(x-3)^2)/(4*Pi*x) is the continuous part of W(x) defined on (3 sqrt(8),3+sqrt(8)): a(n) = int( x^n*W_a(x), x=-eps..eps ) + int( x^n*W_c(x), x = 3-sqrt(8)..3+sqrt(8) ), for any eps>0, n>=0. W_c(x) is unimodal, of bounded variation and W_c(3-sqrt(8)) = W_c(3+sqrt(8)) = 0. Note that the position of the Dirac peak (x=0) lies outside support of W_c(x). - Karol A. Penson and Wojciech Mlotkowski, Aug 05 2013
G.f.: 1 + x/G(x) with G(x) = 1 - 3*x - 2*x^2/G(x) (continued fraction). - Nikolaos Pantelidis, Dec 17 2022

A209616 Sum of positive Dyson's ranks of all partitions of n.

Original entry on oeis.org

0, 1, 2, 4, 7, 12, 18, 29, 42, 63, 89, 128, 176, 246, 333, 453, 603, 807, 1058, 1393, 1807, 2346, 3011, 3867, 4915, 6248, 7879, 9926, 12421, 15529, 19297, 23954, 29585, 36486, 44802, 54937, 67096, 81831, 99459, 120700, 146026, 176410, 212512, 255636, 306734
Offset: 1

Views

Author

Omar E. Pol, Mar 10 2012

Keywords

Comments

The Dyson's rank of a partition is the largest part minus the number of parts.

Examples

			For n = 5 we have:
--------------------------
Partitions        Dyson's
of 5               rank
--------------------------
5               5 - 1 =  4
4+1             4 - 2 =  2
3+2             3 - 2 =  1
3+1+1           3 - 3 =  0
2+2+1           2 - 3 = -1
2+1+1+1         2 - 4 = -2
1+1+1+1+1       1 - 5 = -4
--------------------------
The sum of positive Dyson's ranks of all partitions of 5 is 4+2+1 = 7 so a(5) = 7.
		

Crossrefs

Column 1 of triangle A208482.

Programs

  • Maple
    # Maple program based on Theorem 1 of Andrews-Chan-Kim:
    M:=101;
    qinf:=mul(1-q^i,i=1..M);
    qinf:=series(qinf,q,M);
    R1:=add((-1)^(n+1)*q^(n*(3*n+1)/2)/(1-q^n),n=1..M);
    R1:=series(R1/qinf,q,M);
    seriestolist(%); # N. J. A. Sloane, Sep 04 2012
  • Mathematica
    M = 101;
    qinf = Product[1-q^i, {i, 1, M}];
    qinf = Series[qinf, {q, 0, M}];
    R1 = Sum[(-1)^(n+1) q^(n(3n+1)/2)/(1-q^n), {n, 1, M}];
    R1 = Series[R1/qinf, {q, 0, M}];
    CoefficientList[R1, q] // Rest (* Jean-François Alcover, Aug 18 2018, translated from Maple *)
  • PARI
    my(N=50, x='x+O('x^N)); concat(0, Vec(1/prod(k=1, N, 1-x^k)*sum(k=1, N, (-1)^(k-1)*x^(k*(3*k+1)/2)/(1-x^k)))) \\ Seiichi Manyama, May 21 2023

Formula

a(n) = A115995(n) - A195012(n). - Omar E. Pol, Apr 06 2012
G.f.: (1/Product_{k>=1} (1-x^k)) * Sum_{k>=1} (-1)^(k-1) * x^(k*(3*k+1)/2) / (1-x^k). - Seiichi Manyama, May 21 2023
a(n) ~ log(2) * exp(Pi*sqrt(2*n/3)) / (Pi*2^(3/2)*sqrt(n)). - Vaclav Kotesovec, Jul 06 2025

Extensions

More terms from Alois P. Heinz, Mar 10 2012

A208478 Triangle read by rows: T(n,k) = number of partitions of n with positive k-th rank.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 3, 2, 1, 5, 2, 4, 4, 2, 1, 6, 3, 5, 6, 4, 2, 1, 10, 5, 7, 9, 7, 4, 2, 1, 13, 7, 9, 11, 11, 7, 4, 2, 1, 19, 11, 12, 15, 16, 12, 7, 4, 2, 1, 25, 16, 15, 19, 22, 18, 12, 7, 4, 2, 1, 35, 24, 20, 26, 29, 27, 19, 12, 7, 4, 2, 1
Offset: 1

Views

Author

Omar E. Pol, Mar 07 2012

Keywords

Comments

We define the k-th rank of a partition as the k-th part minus the number of parts >= k. Every partition of n has n ranks. This is a generalization of the Dyson's rank of a partition which is the largest part minus the number of parts. Since the first part of a partition is also the largest part of the same partition so the Dyson's rank of a partition is the case for k = 1.
The sum of the k-th ranks of all partitions of n is equal to zero.
Also T(n,k) = number of partitions of n with negative k-th rank.
It appears that reversed rows converge to A000070, the same as A208482. - Omar E. Pol, Mar 11 2012
From Omar E. Pol, Dec 12 2019: (Start)
1) The k-th part of a partition of n is also the number of parts >= k of its conjugate partition.
2) The k-th rank of a partitions is also the number of parts >= k of its conjugate partition minus the number of parts >= k.
For example: for n = 9 consider the partition [5, 3, 1]. The first part is 5, so the conjugate partition has five parts >= 1. The second part is 3, so the conjugate partition has three parts >= 2. The third part is 1, so the conjugate partition has only one part >= 3. The mentioned conjugate partition is [3, 2, 2, 1, 1]. And conversely, consider the partition [3, 2, 2, 1, 1]. The first part is 3, so the conjugate partition has three parts >= 1. The second part is 2, so the conjugate partition has two parts >= 2. the Third part is 2, so the conjugate partition has two parts >= 3, and so on. In this case the conjugate partition is [5, 3, 1].
3) The difference between the k-th part and the (k+1)-st part of the partition of n is also the number of k's in its conjugate partition. For example: consider the partition [5, 3, 1]. The difference between the first and the second part is 5 - 3 = 2, equals the number of 1's in its conjugate partition. The difference between the second and the third part is 3 - 1 = 2, equals the number of 2's in its conjugate partition. The difference between the third and the fourth (virtual) part is 1 - 0 = 1, equals the number of 3's in its conjugate partition [3, 2, 2, 1, 1]. And conversely, consider the partition [3, 2, 2, 1, 1]. The difference between the first and the second part is 3 - 2 = 1, equals the number of 1's in its conjugate partition. The difference between the second and the third part is 2 - 2 = 0, equals the number of 2's in its conjugate partition. The difference between the third and the fourth part is 2 - 1 = 1, equals the number of 3's in its conjugate partition, and so on.
4) The list of n ranks of a partition of n equals the list of n ranks multiplied by -1 of its conjugate partition. For example the nine ranks of the partition [5, 3, 1] of 9 are [2, 1, -1, -1, -1, -1, 0, 0, 0], and the nine ranks of its conjugate partition [3, 2, 2, 1, 1] are [-2, -1, 1, 1, 1, 1, 0, 0, 0].
For a list of partitions of the positive integers ordered by its k-th ranks see A330370. (End)

Examples

			For n = 4 the partitions of 4 and the four types of ranks of the partitions of 4 are
----------------------------------------------------------
Partitions    First      Second       Third      Fourth
of 4          rank        rank        rank        rank
----------------------------------------------------------
4           4-1 =  3    0-1 = -1    0-1 = -1    0-1 = -1
3+1         3-2 =  1    1-1 =  0    0-1 = -1    0-0 =  0
2+2         2-2 =  0    2-2 =  0    0-0 =  0    0-0 =  0
2+1+1       2-3 = -1    1-1 =  0    1-0 =  1    0-0 =  0
1+1+1+1     1-4 = -3    1-0 =  1    1-0 =  1    1-0 =  1
----------------------------------------------------------
The number of partitions of 4 with positive k-th ranks are 2, 1, 2, 1 so row 4 lists 2, 1, 2, 1.
Triangle begins:
   0;
   1,  1;
   1,  1,  1;
   2,  1,  2,  1;
   3,  1,  3,  2,  1;
   5,  2,  4,  4,  2,  1;
   6,  3,  5,  6,  4,  2,  1;
  10,  5,  7,  9,  7,  4,  2,  1;
  13,  7,  9, 11, 11,  7,  4,  2,  1;
  19, 11, 12, 15, 16, 12,  7,  4,  2,  1;
  25, 16, 15, 19, 22, 18, 12,  7,  4,  2,  1;
  35, 24, 20, 26, 29, 27, 19, 12,  7,  4,  2,  1;
  ...
		

Crossrefs

Extensions

More terms from Alois P. Heinz, Mar 11 2012

A208482 Triangle read by rows: T(n,k) = sum of positive k-th ranks of all partitions of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 4, 1, 2, 1, 7, 1, 3, 2, 1, 12, 2, 5, 4, 2, 1, 18, 3, 6, 6, 4, 2, 1, 29, 6, 9, 10, 7, 4, 2, 1, 42, 9, 11, 13, 11, 7, 4, 2, 1, 63, 16, 15, 19, 17, 12, 7, 4, 2, 1, 89, 24, 18, 25, 24, 18, 12, 7, 4, 2, 1, 128, 39, 24, 36, 34, 28, 19, 12, 7, 4, 2, 1
Offset: 1

Views

Author

Omar E. Pol, Mar 07 2012

Keywords

Comments

For the definition of the k-th rank see A208478.
It appears that the sum of the k-th ranks of all partitions of n is equal to zero.
It appears that reversed rows converge to A000070, the same as A208478. - Omar E. Pol, Mar 10 2012

Examples

			For n = 4 the partitions of 4 and the four types of ranks of the partitions of 4 are
----------------------------------------------------------
Partitions    First      Second       Third      Fourth
of 4          rank        rank        rank        rank
----------------------------------------------------------
4           4-1 =  3    0-1 = -1    0-1 = -1    0-1 = -1
3+1         3-2 =  1    1-1 =  0    0-1 = -1    0-0 =  0
2+2         2-2 =  0    2-2 =  0    0-0 =  0    0-0 =  0
2+1+1       2-3 = -1    1-1 =  0    1-0 =  1    0-0 =  0
1+1+1+1     1-4 = -3    1-0 =  1    1-0 =  1    1-0 =  1
----------------------------------------------------------
The sums of positive k-th ranks of the partitions of 4 are 4, 1, 2, 1 so row 4 lists 4, 1, 2, 1.
Triangle begins:
0;
1,    1;
2,    1,  1;
4,    1,  2,  1;
7,    1,  3,  2,  1;
12,   2,  5,  4,  2,  1;
18,   3,  6,  6,  4,  2,  1;
29,   6,  9, 10,  7,  4,  2,  1;
42,   9, 11, 13, 11,  7,  4,  2,  1;
63,  16, 15, 19, 17, 12,  7,  4,  2,  1;
89,  24, 18, 25, 24, 18, 12,  7,  4,  2,  1;
128, 39, 24, 36, 34, 28, 19, 12,  7,  4,  2,  1;
		

Crossrefs

Extensions

Terms a(1)-a(22) confirmed and additional terms added by John W. Layman, Mar 10 2012

A194547 Triangle read by rows: T(n,k) = Dyson's rank of the k-th partition of n, with partitions in lexicographic order.

Original entry on oeis.org

0, -1, 1, -2, 0, 2, -3, -1, 1, 0, 3, -4, -2, 0, -1, 2, 1, 4, -5, -3, -1, -2, 1, 0, 3, -1, 2, 1, 5, -6, -4, -2, -3, 0, -1, 2, -2, 1, 0, 4, 0, 3, 2, 6, -7, -5, -3, -4, -1, -2, 1, -3, 0, -1, 3, -1, 2, 1, 5, -2, 1, 0, 4, 3, 2, 7, -8, -6, -4, -5, -2, -3, 0, -4, -1
Offset: 1

Views

Author

Omar E. Pol, Dec 10 2011

Keywords

Comments

Row n has length A000041(n). The sum of row n is equal to zero.

Examples

			Written as a triangle:
  0;
  -1,1;
  -2,0,2;
  -3,-1,1,0,3;
  -4,-2,0,-1,2,1,4;
  -5,-3,-1,-2,1,0,3,-1,2,1,5;
  -6,-4,-2,-3,0,-1,2,-2,1,0,4,0,3,2,6;
  -7,-5,-3,-4,-1,-2,1,-3,0,-1,3,-1,2,1,5,-2,1,0,4,3,2,7;
		

Crossrefs

Programs

  • Maple
    T:= proc(n) local b, l;
          b:= proc(n, i, t)
                if n=0 then l:=l, i-t
              elif i>n then
              else b(n-i, i, t+1); b(n, i+1, t)
                fi
              end;
          l:= NULL; b(n, 1, 0); l
        end:
    seq(T(n), n=1..10);  # Alois P. Heinz, Dec 22 2011
  • Mathematica
    T[n_] := Module[{b, l}, b[n0_, i_, t_] := If [n0==0, l = Append[l, i-t], If[i>n0, , b[n0-i, i, t+1]; b[n0, i+1, t]]]; l = {}; b[n, 1, 0]; l];
    Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, Mar 05 2021, after Alois P. Heinz *)

Formula

a(n) = A194546(n) - A193173(n).

Extensions

More terms from Alois P. Heinz, Dec 22 2011

A111786 Array used to obtain the complete symmetric function in n variables in terms of the elementary symmetric functions; irregular triangle T(n,k), read by rows, with n >= 1 and 1 <= k <= A000041(n).

Original entry on oeis.org

1, -1, 1, 1, -2, 1, -1, 2, 1, -3, 1, 1, -2, -2, 3, 3, -4, 1, -1, 2, 2, 1, -3, -6, -1, 4, 6, -5, 1, 1, -2, -2, -2, 3, 6, 3, 3, -4, -12, -4, 5, 10, -6, 1, -1, 2, 2, 2, 1, -3, -6, -6, -3, -3, 4, 12, 6, 12, 1, -5, -20, -10, 6, 15, -7, 1, 1, -2, -2, -2, -2, 3, 6, 6, 3, 3, 6, 1, -4, -12, -12, -12, -12, -4, 5, 20, 10, 30, 5, -6, -30, -20, 7, 21, -8, 1, -1
Offset: 1

Views

Author

Wolfdieter Lang, Aug 23 2005

Keywords

Comments

The unsigned numbers give A048996. They are not listed on pp. 831-832 of Abramowitz and Stegun (reference given in A103921). One could call these numbers M_0 (like M_1, M_2, M_3 given in A036038, A036039, A036040, resp.).
The sequence of row lengths is A000041(n) (partition numbers).
The sign is (-1)^(n + m(n,k)) with m(n,k) the number of parts of the k-th partition of n taken in the mentioned order. For m(n,k), see A036043.
The row sum is 1 for n = 1, and 0 otherwise. The unsigned row sum is 2^(n-1) = A000079(n-1) for n >= 1.
The complete symmetric polynomial is also h(n; a[1],...,a[n]) = Det(A_n) with the matrix elements of the n X n matrix A_n given by A_n(k, k+1) = 1 for 1 <= k < n, A(k, m) = a[k-m+1] for n >= k >= m >= 1, and 0 otherwise. [For an explanation of this statement, see the example for n = 4 below. See also p. 3 in MacMahon (1960).]

Examples

			Triangle T(n,k) (with rows n >= 1 and columns k >= 1) begins as follows:
   1;
  -1,  1;
   1, -2,  1;
  -1,  2,  1, -3,  1;
   1, -2, -2,  3,  3, -4,  1;
  -1,  2,  2,  1, -3, -6, -1, 4, 6, -5, 1,
   ...
h(4; a[1],...,a[4])= -1*a[4] + 2*a[1]*a[3] + 1*a[2]^2 - 3*a[1]^2*a[2] + a[1]^4.
Consider variables x_1, x_2, x_3, x_4, and let a[1] = Sum_i x_i, a[2] = Sum_{i,j} x_i*x_j, a[3] = Sum_{i,j,k} x_i*x_j*x_k, and a[4] = x1*x2*x3*x4, where in all the sums no term is repeated twice.
Then h(4; a[1],...,a[4]) = Sum_i x_i^4 + Sum_{i,j} x_i^3*x_j + Sum_{i,j} x_i^2*x_j^2 + Sum_{i,j,k} x_i^2*x_j*x_k + Sum_{i,j,k,m} x_i*x_j*x_k*x_m, where again in all the sums no term is repeated twice. Thus, indeed, h is the complete symmetric polynomial in four variables x_1, x_2, x_3, x_4.
		

References

  • V. Krishnamurthy, Combinatorics, Ellis Horwood, Chichester, 1986, p. 55, eqs. (48) and (50).

Crossrefs

Formula

The complete symmetric row polynomials h(n; a[1], ..., a[n]):= sum k over partitions of n of T(n, k)* A[k], with A[k] := a[1]^e(k, 1) * a[2]^e(k, 2) * ... * a[n]^e(k, n) is the k-th partition of n, in Abramowitz-Stegun order (see A105805 for this reference), is [1^e(k, 1), 2^e(k, 2), ..., n^e(k, n)], for k = 1..p(n), where p(n) = A000041(n) (partition numbers).
G.f.: A(x) = 1/(1 + Sum_{j = 1..infinity} (-1)^j * a[j]).
T(n, k) is the coefficient of x^n and a[1]^e(k, 1) * a[2]^e(k, 2) * ... * a[n]^e(k, n) in A(x) if the k-th partition of n, counted using the Abramowitz-Stegun order, is [1^e(k, 1), 2^e(k, 2), ..., n^e(k, n)] with e(k, j) >= 0 (and if e(k, j) = 0 then j^0 is not recorded).
T(n, k) = (-1)^(n + m(n, k)) * m(n, k)!/(Product_{j = 1..n} e(k, j)!), where m(n, k) := Sum_{j = 1..n} e(k, j), with [1^e(k, 1), 2^e(k, 2), ..., n^e(k, n)] being the k-th partition of n in the mentioned order. Here m(n, k) is the number of parts of the k-th partition of n. For m(n,k), see A036043.

Extensions

Various sections edited by Petros Hadjicostas, Dec 15 2019

A195822 Triangle read by rows in which row n lists the Dyson's ranks of all partitions of n that do not contain 1 as a part, in nonincreasing order.

Original entry on oeis.org

0, 1, 2, 3, 0, 4, 1, 5, 2, 1, -1, 6, 3, 2, 0, 7, 4, 3, 2, 1, 0, -2, 8, 5, 4, 3, 2, 1, 0, -1, 9, 6, 5, 4, 3, 3, 2, 1, 1, 0, -1, -3, 10, 7, 6, 5, 4, 4, 3, 2, 2, 1, 1, 0, -1, -2, 11, 8, 7, 6, 5, 5, 4, 4, 3, 3, 2, 2, 2, 1, 1, 0, 0, -1, -1, -2, -4
Offset: 1

Views

Author

Omar E. Pol, Nov 06 2011

Keywords

Comments

The sum of row n is equal to A000041(n-1), if n >= 2. Proof: Dyson defined the rank of a partition as the largest part minus the number of parts. On the other hand the total number of parts in all partitions of n equals the sum of largest parts of all partitions of n (Cf. A006128), hence the sum of the ranks of all partitions of n is equal to zero. Let p(n) be the number of partitions of n. If we now add an part equal to 1 in each partition of n we obtain the partitions of n+1 that contain 1 as a part. The sum of the ranks of these partitions is p(n)*(-1) because the largest parts are the same but now there is an additional part in each partition. On the other hand the sum of the ranks of all partitions of n+1 is equal to zero, hence the sum of the ranks of all partitions of n+1 that do not contain 1 as a part is equal to p(n).

Examples

			Triangle begins:
0;
1;
2;
3, 0;
4, 1;
5, 2, 1, -1;
6, 3, 2, 0;
7, 4, 3, 2, 1, 0, -2;
8, 5, 4, 3, 2, 1, 0, -1;
9, 6, 5, 4, 3, 3, 2, 1, 1, 0, -1, -3;
10, 7, 6, 5, 4, 4, 3, 2, 2, 1, 1, 0, -1, -2;
		

Crossrefs

Row n has length A002865(n), n >= 2.

A330368 Irregular triangle read by rows in which row n lists the ranks of the partitions of n in nonincreasing order.

Original entry on oeis.org

0, 1, -1, 2, 0, -2, 3, 1, 0, -1, -3, 4, 2, 1, 0, -1, -2, -4, 5, 3, 2, 1, 1, 0, -1, -1, -2, -3, -5, 6, 4, 3, 2, 2, 1, 0, 0, 0, -1, -2, -2, -3, -4, -6, 7, 5, 4, 3, 3, 2, 2, 1, 1, 1, 0, 0, -1, -1, -1, -2, -2, -3, -3, -4, -5, -7, 8, 6, 5, 4, 4, 3, 3, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0, -1, -1, -1, -2, -2, -2, -3, -3, -4, -4, -5, -6, -8
Offset: 1

Views

Author

Omar E. Pol, Dec 12 2019

Keywords

Comments

The rank of a partition is the largest part minus the number of parts.
For more about this ordering, see A330370.
First differs from A105805 at a(49) = T(7,5).

Examples

			Triangle begins:
                              0;
                            1, -1;
                          2,  0, -2;
                      3,  1,  0, -1, -3;
                  4,  2,  1,  0, -1, -2, -4;
          5,  3,  2,  1,  1,  0, -1, -1, -2, -3, -5;
  6,  4,  3,  2,  2,  1,  0,  0,  0, -1, -2, -2, -3, -4, -6;
...
		

Crossrefs

Row n has length A000041(n).
Row sums give A000004.

Extensions

Edited by N. J. A. Sloane, Sep 15 2020

A179864 The unrestricted partition statistic defined by A049085(n,k)+A036043(n,k)- 1.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 4, 3, 4, 4, 5, 5, 4, 5, 4, 5, 5, 6, 6, 5, 4, 6, 5, 4, 6, 5, 6, 6, 7, 7, 6, 5, 7, 6, 5, 5, 7, 6, 5, 7, 6, 7, 7, 8, 8, 7, 6, 5, 8, 7, 6, 6, 5, 8, 7, 6, 6, 5, 8, 7, 6, 8, 7, 8, 8, 9, 9, 8, 7, 6, 9, 8, 7, 6, 7, 6, 5, 9, 8, 7, 7, 6, 6, 9, 8, 7, 7, 6, 9, 8, 7, 9, 8, 9, 9, 10, 10, 9, 8, 7, 6, 10, 9, 8
Offset: 1

Views

Author

Alford Arnold, Aug 02 2010

Keywords

Examples

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

Crossrefs

Cf. A000041 (row lengths), A179862 (row sums), A105805 (the rank statistic)

A330374 Triangle read by rows: T(n,k) is the number of partitions of n whose absolute value of Dyson's rank is equal to k, with 0 <= k < n.

Original entry on oeis.org

1, 0, 2, 1, 0, 2, 1, 2, 0, 2, 1, 2, 2, 0, 2, 1, 4, 2, 2, 0, 2, 3, 2, 4, 2, 2, 0, 2, 2, 6, 4, 4, 2, 2, 0, 2, 4, 6, 6, 4, 4, 2, 2, 0, 2, 4, 10, 6, 8, 4, 4, 2, 2, 0, 2, 6, 10, 12, 6, 8, 4, 4, 2, 2, 0, 2, 7, 16, 12, 12, 8, 8, 4, 4, 2, 2, 0, 2, 11, 16, 18, 14, 12, 8, 8, 4, 4, 2, 2, 0, 2, 11, 26, 20, 20, 14, 14
Offset: 1

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Author

Omar E. Pol, Dec 18 2019

Keywords

Comments

The rank of a partition is the largest part minus the number of parts.
Since the largest part of a partition equals the number of parts of its conjugate partition, so the rank of a partition also is equal to the difference between the number of parts of its conjugate partition and the number of parts of the partition.

Examples

			Triangle begins:
--------------------------------------------------------------------
  n \ k   0   1   2   3   4   5   6   7   8   9  10  11  12  13  14
--------------------------------------------------------------------
[ 1]      1;
[ 2]      0,  2;
[ 3]      1,  0,  2;
[ 4]      1,  2,  0,  2;
[ 5]      1,  2,  2,  0,  2;
[ 6]      1,  4,  2,  2,  0,  2;
[ 7]      3,  2,  4,  2,  2,  0,  2;
[ 8]      2,  6,  4,  4,  2,  2,  0,  2;
[ 9]      4,  6,  6,  4,  4,  2,  2,  0,  2;
[10]      4, 10,  6,  8,  4,  4,  2,  2,  0,  2;
[11]      6, 10, 12,  6,  8,  4,  4,  2,  2,  0,  2;
[12]      7, 16, 12, 12,  8,  8,  4,  4,  2,  2,  0,  2;
[13]     11, 16, 18, 14, 12,  8,  8,  4,  4,  2,  2,  0,  2;
[14]     11, 26, 20, 20, 14, 14,  8,  8,  4,  4,  2,  2,  0,  2;
[15]     16, 28, 30, 22, 22, 14, 14,  8,  8,  4,  4,  2,  2,  0,  2;
...
		

Crossrefs

Row sums give A000041, n >= 1.
Leading diagonal gives A040000.
Second diagonal gives A000004.
Column k=0 is A047993.

Formula

T(n,k) = A063995(n,k)*A040000(k), 0 <= k < n.
Showing 1-10 of 10 results.