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.

User: Georg Muntingh

Georg Muntingh's wiki page.

Georg Muntingh has authored 3 sequences.

A172004 Let y = y(u,v) be implicitly defined by g(u,v,y(u,v)) = 0. Read as a triangle by rows k = 1,2,..., the sequence represents the number of terms a(i,k-i) in the expansion of the partial derivatives d^k y/du^i dv^{k-i} in terms of partial derivatives of g.

Original entry on oeis.org

1, 1, 3, 4, 3, 9, 15, 15, 9, 24, 47, 59, 47, 24, 61, 136, 195, 195, 136, 61, 145, 360, 580, 663, 580, 360, 145, 333, 904, 1586, 2032, 2032, 1586, 904, 333, 732, 2152, 4077, 5684, 6350, 5684, 4077, 2152, 732, 1565, 4927, 9948, 14938, 18123, 18123, 14938, 9948
Offset: 1

Author

Georg Muntingh, Jan 22 2010

Keywords

Comments

The sequence starts with a(1,0),a(0,1),a(2,0),a(1,1),a(0,2),a(3,0),...
The subsequences a(1,0),a(2,0),a(3,0),... and a(0,1),a(0,2),a(0,3),... coincide with the sequence A003262, which is the corresponding sequence for univariate implicit functions.

Examples

			The formulas dy/du = -g_u/g_y,
d^2y/du^2 = -g_yy g_u^2/g_y^3 + 2g_uy g_u/g_y^2 - g_uu/g_y,
d^2y/dudv = -2g_yy g_u g_v / g_y^3 + g_uy g_v/g_y^2 + g_vy g_u/g_y^3 - g_uv/g_y
imply that a(1,0) = 1, a(2,0) = 3, and a(1,1) = 4.
		

Crossrefs

Cf. A003262, which is the univariate variant of this sequence.
Cf. A172003, which is the analogous sequence for implicit divided differences, and A162326 for its univariate variant.

Programs

  • Sage
    # Upon executing the following code in Sage 4.2 (using Singular as a backend), it
    # computes the number of terms a(n1,n2) and stores it in the entry A[n1][n2] of the
    # double list A.
    N = 9
    E1 = N
    E2 = N
    p = [[[0 for i1 in range(E1+1)] for i2 in range(E2+1)] for j in range(E1 + E2)]
    q = [[[0 for i1 in range(E1+1)] for i2 in range(E2+1)] for j in range(E1 + E2)]
    for m in range(1, E1 + E2):
        for d in range(1, m+1):
            quotient, remainder = divmod(m, d)
            if remainder == 0:
                for i1 in range(quotient + 1 + 1):
                    for i2 in range(quotient + 1 - i1 + 1):
                        if d*i1 <= E1 and d*i2 <= E2:
                            q[m][i1*d][i2*d] += 1/d
    for i1 in range(E1 + 1):
        for i2 in range(E2 + 1):
            p[0][i1][i2] = 1
    for n in range(1, E1 + E2):
        for s in range(n+1):
            for k1 in range(E1+1):
                for k2 in range(E2+1):
                    for i1 in range(k1 + 1):
                        for i2 in range(k2 + 1):
                            p[n][k1][k2] += 1/n * s * q[s][k1-i1][k2-i2] * p[n-s][i1][i2]
    A = [[ p[n1+n2-1][n1][n2] for n1 in range(E1+1)] for n2 in range(E2+1)]

Formula

Let E = N^3 \ {(0,0,0), (0,0,1)} be a set of triples of natural numbers. The number of terms a(m,n) is the coefficient of u^m * v^n * y^{m+n-1} in Product_{(r,s,t) in E} (1 - u^r * v^s * y^{r+s+t-1})^{-1}.

A172003 Let y = y(u,v) be implicitly defined by g(u,v,y(u,v)) = 0. Read as a triangle by rows, the sequence represents the number of terms a(i,k-i) in the expansion of the bivariate divided difference [u_0,...,u_i; v_0,...,v_{k-i}]y in terms of trivariate divided differences of g.

Original entry on oeis.org

1, 1, 3, 5, 3, 13, 33, 33, 13, 71, 245, 351, 245, 71, 441, 1921, 3597, 3597, 1921, 441, 2955, 15525, 35931, 46709, 35931, 15525, 2955, 20805, 127905, 352665, 563821, 563821, 352665, 127905, 20805, 151695, 1067925, 3417975, 6483285, 7963151
Offset: 1

Author

Georg Muntingh, Jan 22 2010

Keywords

Comments

The sequence starts with a(1,0),a(0,1),a(2,0),a(1,1),a(0,2),a(3,0),...

Examples

			The subsequences a(1,0),a(2,0),a(3,0),... and a(0,1),a(0,2),a(0,3),... coincide with the sequence A162326.
For (m,n) = (1,1), one expresses [u_0,u_1;v_0,v_1]y as a sum of 5 terms,
[01;01]y =
- [0;0;(0,0),(1,0),(1,1)]g * [01;0;(1,0)]g * [1;01;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [0;0;(0,0),(1,0)]g * [1;0;(1,0),(1,1)]g )
+ [01;0;(1,0),(1,1)]g * [1;01;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [1;0;(1,0),(1,1)]g )
- [01;01;(1,1)]g / [0;0;(0,0),(1,1)]g
- [0;0;(0,0),(0,1),(1,1)]g * [0;01;(0,1)]g * [01;1;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [0;0;(0,0),(0,1)]g * [0;1;(0,1),(1,1)]g )
+ [0;01;(0,1),(1,1)]g * [01;1;(1,1)]g /
( [0;0;(0,0),(1,1)]g * [0;1;(0,1),(1,1)]g ),
where the numbers refer to the indices of the corresponding variable, e.g.
[1;01;(1,1)]g = [u_1;v_0,v_1;y(u_1,v_1)]g.
		

Crossrefs

Cf. A162326, which is the univariate variant of this sequence.
Cf. A172004, which is the analogous sequence for implicit derivatives, and A003262 for its univariate variant.

Programs

  • Sage
    R. = PolynomialRing(ZZ,3)
    def P(n1,n2,q):
        E = cartesian_product([list(range(n1+1)), list(range(n2+1)), list(range(n1+n2+1))])
        E = [(i1,i2,j) for (i1,i2,j) in E if (i1,i2,j) != (0,0,0) and
             (i1,i2,j) != (0,0,1) and i1 + i2 + j <= n1 + n2 and
             2*(i1 + i2) + j - 1 <= 2*(n1+n2) - q]
        return R.sum(X1^s1 * X2^s2 * Y^(s1+s2+t-1) for s1,s2,t in E)
    n1, n2 = 4, 4
    L = [[0 for _ in range(n1 + 1)]] * (n2 + 1)
    h = 1 + sum(((P(n1,n2,q))^q)/q for q in range(1,2*(n1+n2)))
    for k1 in range(n1+1):
        for k2 in range(k1+1):
            if (k1, k2) != (0, 0):
                print(k1, k2, h.coefficient({X1:k1, X2:k2, Y:k1+k2-1}))

Formula

Let E = N^3 \ {(0,0,0), (0,0,1)} be a set of triples of natural numbers. The number of terms a(m,n) is the coefficient of u^m * v^n * y^{m+n-1} of the generating function
- log(1 - Sum_{(r,s,t) in E} u^r * v^s * y^{r+s+t-1})
= Sum_{q >= 1} (Sum_{(r,s,t) in E} u^r * v^s * y^{r+s+t-1})^q / q.

A162326 Let a(0) = a(1) = 1, and n*a(n) = 2*(-7+5*n)*a(n-1) + 9*(2-n)*a(n-2) for n >= 2.

Original entry on oeis.org

1, 1, 3, 13, 71, 441, 2955, 20805, 151695, 1135345, 8671763, 67320573, 529626839, 4213228969, 33833367963, 273892683573, 2232832964895, 18314495896545, 151037687326755, 1251606057754605, 10416531069771111, 87029307323766681
Offset: 0

Author

Georg Muntingh, Jul 01 2009

Keywords

Comments

Let y = y(x) be implicitly defined by g(x,y(x)) = 0, with dg/dy not identically zero. For n >= 1, the sequence a(n) is the number of terms in the expansion of the divided difference [x0,...,xn]y in terms of bivariate divided differences of g.
(1 + 3*x + 13*x^2 + 71*x^3 + ...) = (1 + 4*x + 20*x^2 + 116*x^3 + ...) * 1/(1 + x + 4*x^2 + 20*x^3 + 116*x^4 + ...); where A082298 = (1, 4, 20, 116, 740, ...). - Gary W. Adamson, Nov 17 2011
The shifted sequence 1,3,13,71,... is the binomial transform of A151374. - Georg Muntingh, Jul 19 2012
a(n+1) is the number of Schröder paths of semilength n in which the (2,0)-steps come in 3 colors and with no peaks at level 1. - José Luis Ramírez Ramírez, Mar 31 2013
Define an infinite triangle by T(n,0)=1 and the other cells by T(n,k) = Sum_{c=0..k-1} T(n,c) + Sum_{r=k..n-1} T(r,k), the sum of the cells to the left and above a cell. The column k=1 contains A000079, the column k=2 essentially A001792. Then T(n,n)=a(n) on the diagonal. - J. M. Bergot, May 22 2013

Examples

			Write [0...n]y for [x0,...,xn]y and [0...s,0...t]g for [x0,...,xs;y0,...,yt]g.
For n = 1 one finds 1 term, [01]y = -[01;1]g/[0;01]g.
For n = 2 one finds 3 terms, [012]y = -[012;2]g/[0;02]g + ([01;12]g[12;2]g)/([0;02]g[1;12]g) - ([0;012]g[01;1]g[12;2]g)/([0;02]g[0;01]g[1;12]g).
		

Crossrefs

Cf. A172003, which is a generalization to bivariate implicit functions.
Cf. A003262, which is the analogous sequence for implicit derivatives, and A172004 for its generalization to bivariate implicit functions.

Programs

  • Magma
    m:=20; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (5-Sqrt((1-9*x)/(1-x)))/4 )); // G. C. Greubel, Feb 07 2019
    
  • Magma
    a:=[1,3]; for n in [3..21] do Append(~a,(2*(-7+5*n)*a[n-1] + 9*(2-n)*a[n-2]) div n); end for ; [1] cat a; // Marius A. Burtea, Jan 20 2020
    
  • Mathematica
    CoefficientList[Series[(5-Sqrt[(1-9*x)/(1-x)])/4, {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(binomial(n,k)*2^(n-k-1)*binomial(2*n-2*k-2,n-k-1),k,0,n)/n; /* Vladimir Kruchinin, Mar 13 2016 */
    
  • PARI
    a(n) = if(n<2, 1, (2*(-7+5*n)*a(n-1) + 9*(2-n)*a(n-2))/n);
    vector(25, n, a(n-1)) \\ Altug Alkan, Oct 06 2015
    
  • PARI
    my(x='x+O('x^20)); Vec((5-sqrt((1-9*x)/(1-x)))/4) \\ G. C. Greubel, Feb 07 2019
    
  • Python
    L = [1, 1]
    for n in range(2,22):
        L.append( ((-14 + 10*n)*L[-1] + (18-9*n)*L[-2])//n )
    print(L)
    # Georg Muntingh, Jul 19 2012
    
  • Sage
    ((5-sqrt((1-9*x)/(1-x)))/4).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 07 2019

Formula

Let E = N x N \ {(0,0), (0,1)} be a set of pairs of natural numbers. The number of terms a(n) is the coefficient of x^n*y^{n-1} of the generating function 1 - log(1 - Sum_{(s,t) in E} x^s*y^{s+t-1}) = 1 + Sum_{q >= 1} (Sum_{(s,t) in E} x^s*y^{s+t-1})^q / q.
From Georg Muntingh, Jul 19 2012: (Start)
a(n) = 2F1(1/2,1-n;2;-8), where 2F1 is the Gauss hypergeometric series.
G.f.: (5 - sqrt( (1-9*x)/(1-x) ))/4.
Quadratic recurrence relation: a(n) = 1 + 2*Sum_{m=1..n-1} a(m)*a(n-m).
(End)
a(n) ~ 3^(2*n+1)/(16*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
a(n) = Sum_{k=0..n} (binomial(n,k)*2^(n-k-1)*binomial(2*n-2*k-2,n-k-1))/n, n>0, a(0)=1. - Vladimir Kruchinin, Mar 13 2016
From Peter Bala, Jan 19 2020: (Start)
a(n+1) = Sum_{k = 0..n} 2^k*C(n,k)*Catalan(k).
a(n+1) = (2/Pi) * Integral_{x = -1..1} (1 + 8*x^2)^n*sqrt(1 - x^2) dx.
O.g.f.: 1 + x/(1 - x)*c(2*x/(1-x)), where c(x) is the o.g.f. for A000108. (End)
Conjecture: a(n) = t_n for n > 0 with a(0) = 1 where we start with vector v of fixed length m with elements v_i = 1, then set t = v and for i=1..m-1, for j=i+1..m apply [v_i, v_j] := [v_i + 2*v_j, 2*v_i + v_j] (here square brackets mean that instead of sequentially assigning v_i and then v_j, we reserve their values (for example, as A = v_i, B = v_j) and then assign them in any order) and t_{i+1} := v_{i+1} (after ending each cycle for j). It also looks like that if we change 2*v_i to z*v_i it gives us a(n+1) = Sum_{k=0..n} A090981(n, k)*2^(n-k) for n >= 0. - Mikhail Kurkov, Aug 14 2024

Extensions

Edited by Georg Muntingh, Jan 22 2010