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.

Showing 1-3 of 3 results.

A118340 Pendular triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + T(n-1,k), for n>=k>0, with T(n,0) = 1 and T(n,n) = 0^n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 9, 5, 1, 0, 1, 5, 15, 20, 6, 1, 0, 1, 6, 22, 48, 28, 7, 1, 0, 1, 7, 30, 85, 113, 37, 8, 1, 0, 1, 8, 39, 132, 282, 169, 47, 9, 1, 0, 1, 9, 49, 190, 519, 688, 237, 58, 10, 1, 0, 1, 10, 60, 260, 837, 1762, 1074, 318, 70, 11, 1, 0
Offset: 0

Views

Author

Paul D. Hanna, Apr 25 2006

Keywords

Comments

Definitions. A pendular triangle is a triangle in which row n is generated from the pendular sums of row n-1. Pendular sums of a row are partial sums taken in back-and-forth order, starting with the leftmost term, jumping to the rightmost term, back to the leftmost unused term, then forward to the rightmost unused term, etc.
In each pass, the partial sum is placed in the new row directly under the term most recently used in the sum. Continue in this way until all the terms of the prior row have been used and then complete the new row by appending a zero at the end. Pendular sums are so named because the process resembles a swinging pendulum that slows down on each pass until it eventually comes to rest in the center.
In the simplest case, pendular triangles obey the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + p*T(n-1,k), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n, for some fixed number p.
In which case the g.f. G=G(x) of the central terms satisfies: G = 1 - p*x*G + p*x*G^2 + x*G^3.
More generally, a pendular triangle is defined by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + Sum_{j>=1} p(j)*T(n-1,k-1+j), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n.
Remarkably, the g.f. G=G(x) of the central terms satisfies: G = 1 + x*G^3 + Sum_{j>=1} p(j)*x^j*[G^(2*j) - G^(2*j-1)].
Further, the g.f. of the m-th lower semi-diagonal equals G(x)^(m+1) for m>=0, where the m-th semi-diagonal consists of those terms located at m rows below the central terms.
For variants of pendular triangles, the main diagonal may be nonzero, but then the g.f.s of the semi-diagonals are more complex.

Examples

			Row 6 equals the pendular sums of row 5:
  [1,  4,  9,  5,  1,  0], where the sums proceed as follows:
  [1, __, __, __, __, __], T(6,0) = T(5,0) = 1;
  [1, __, __, __, __,  1], T(6,5) = T(6,0) + T(5,5) = 1 + 0 = 1;
  [1,  5, __, __, __,  1], T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
  [1,  5, __, __,  6,  1], T(6,4) = T(6,1) + T(5,4) = 5 + 1 = 6;
  [1,  5, 15, __,  6,  1], T(6,2) = T(6,4) + T(5,2) = 6 + 9 = 15;
  [1,  5, 15, 20,  6,  1], T(6,3) = T(6,2) + T(5,3) = 15 + 5 = 20;
  [1,  5, 15, 20,  6,  1, 0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  4,   1,    0;
  1,  4,  9,   5,    1,    0;
  1,  5, 15,  20,    6,    1,    0;
  1,  6, 22,  48,   28,    7,    1,    0;
  1,  7, 30,  85,  113,   37,    8,    1,   0;
  1,  8, 39, 132,  282,  169,   47,    9,   1,  0;
  1,  9, 49, 190,  519,  688,  237,   58,  10,  1,  0;
  1, 10, 60, 260,  837, 1762, 1074,  318,  70, 11,  1, 0;
  1, 11, 72, 343, 1250, 3330, 4404, 1568, 413, 83, 12, 1, 0; ...
Central terms are T(2*n,n) = A108447(n);
semi-diagonals form successive self-convolutions of the central terms:
T(2*n+1,n) = A118341(n) = [A108447^2](n),
T(2*n+2,n) = A118342(n) = [A108447^3](n).

Links

Crossrefs

Cf. A108447 (central terms), A118341, A118343.
variants: A118344 (Catalan), A118362 (ternary), A118350, A118355.
Cf. A167763 (p=0), this sequence (p=1), A118345 (p=2), A118350 (p=3).

Programs

  • Magma
    function T(n,k,p)
  if k lt 0 or n lt k then return 0;
  elif k eq 0 then return 1;
  elif k eq n then return 0;
  elif n gt 2*k then return T(n,n-k,p) + T(n-1,k,p);
  else return T(n,n-k-1,p) + p*T(n-1,k,p);
  end if;
  return T;
end function;
[T(n,k,1): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2021
  • Mathematica
    T[n_, k_, p_]:= T[n,k,p] = If[nG. C. Greubel, Feb 17 2021 *)
  • PARI
    {T(n,k) = if(n2*k, T(n-1,k) + T(n,n-k), T(n-1,k) + T(n,n-1-k)))))}
for(n=0,12, for(k=0,n, print1(T(n,k),", "));print(""))
  • Sage
    @CachedFunction
def T(n, k, p):
    if (k<0 or n2*k): return T(n,n-k,p) + T(n-1,k,p)
    else: return T(n, n-k-1, p) + p*T(n-1, k, p)
flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 17 2021

Formula

T(2*n+m,n) = [A108447^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A108447; compare semi-diagonals to the diagonals of convolution triangle A118343.

A118341 Self-convolution square of A108447.

Original entry on oeis.org

1, 2, 9, 48, 282, 1762, 11488, 77270, 532239, 3735488, 26617976, 192061278, 1400453568, 10303466638, 76391703591, 570195719792, 4281169118106, 32312317899352, 245016567557504, 1865677590201192, 14259825593908356
Offset: 0

Views

Author

Paul D. Hanna, Apr 25 2006

Keywords

Crossrefs

Cf. A108447, A118342, A118343.

Programs

  • PARI
    {a(n)=polcoeff((serreverse(x*(1-x+sqrt((1-x)*(1-5*x)+x*O(x^n)))/2/(1-x))/x)^2,n)}

A263917 Riordan array (f(x)^3, f(x)), where 1 + x*f^3(x)/(1 - x*f(x)) = f(x).

Original entry on oeis.org

1, 3, 1, 15, 4, 1, 85, 22, 5, 1, 519, 132, 30, 6, 1, 3330, 837, 190, 39, 7, 1, 22135, 5516, 1250, 260, 49, 8, 1, 151089, 37404, 8461, 1773, 343, 60, 9, 1, 1052805, 259280, 58550, 12324, 2422, 440, 72, 10, 1, 7458236, 1829018, 412375, 87045, 17283, 3214, 552, 85, 11, 1
Offset: 0

Views

Author

Peter Bala, Oct 29 2015

Keywords

Comments

Riordan arrays of the form (f(x)^(m+1), f(x)), where f(x) satisfies 1 + x*f^(m+1)(x)/(1 - x*f(x)) = f(x) include (modulo differences of offset) the Motzkin triangle A091836 (m = -1), the Catalan triangle A033184 (m = 0) and the Schroder triangle A091370 (m = 1). This is the case m = 2. See A263918 for the case m = 3.
The coefficients of the power series solution of the equation 1 + x*f^(m+1)(x)/(1 - x*f(x)) = f(x) appear to be given by [x^0] f(x) = 1 and [x^n] f(x) = 1/n * Sum_{k = 1..n} binomial(n,k)*binomial(n + m*k, k - 1) for n >= 1.
This triangle appears in Novelli et al., Figure 8, p. 24, where a combinatorial interpretation is given in terms of trees.

Examples

			Triangle begins:
       1
       3     1
      15     4     1
      85    22     5    1
     519   132    30    6   1
    3330   837   190   39   7  1
   22135  5516  1250  260  49  8 1
  151089 37404  8461 1773 343 60 9 1

Links

Crossrefs

Cf. A108447 (row sums), A118342 (column 0).
Cf. A033184, A091370, A091836, A263918.

Programs

  • Maple
    # For the function TreesByArityOfTheRoot_Row(m, n) see A263918.
A263917_row := n -> TreesByArityOfTheRoot_Row(2,n):
seq(A263917_row(n), n=0..9); # Peter Luschny, Oct 31 2015
  • Mathematica
    rows = 9;
f[] = 1; Do[f[x] = 1 + x*f[x]*(f[x]^2 + f[x] - 1) + O[x]^(rows+1) // Normal, {rows+1}];
coes = CoefficientList[f[x]^3/(1 - x*t*f[x]) + O[x]^(rows+1), x];
row[n_] := CoefficientList[coes[[n+1]], t];
Table[row[n], {n, 0, rows}] // Flatten (* Jean-François Alcover, Jul 19 2018 *)

Formula

O.g.f. f^3(x)/(1 - x*t*f(x)), where f(x) = 1 + x + 4*x^2 + 20*x^3 + 113*x^4 + ... satisfies 1 + x*f^3(x)/(1 - x*f(x)) = f(x);
f(x) is the o.g.f. for A108447.
First column o.g.f f(x)^3 is the o.g.f. for A118342.
f(x) - 1 is the g.f. for the row sums of the array.
Showing 1-3 of 3 results.

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