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 1-2 of 2 results.

A064094 Triangle composed of generalized Catalan numbers.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 3, 1, 1, 1, 14, 13, 4, 1, 1, 1, 42, 67, 25, 5, 1, 1, 1, 132, 381, 190, 41, 6, 1, 1, 1, 429, 2307, 1606, 413, 61, 7, 1, 1, 1, 1430, 14589, 14506, 4641, 766, 85, 8, 1, 1, 1, 4862, 95235, 137089, 55797, 10746, 1279, 113, 9, 1, 1
Offset: 0

Views

Author

Wolfdieter Lang, Sep 13 2001

Keywords

Comments

The column m sequence (without leading zeros and the first 1) appears in the Derrida et al. 1992 reference as Z_{N}=Y_{N}(N+1), N >=0, for alpha = m, beta = 1 (or alpha = 1, beta = m). In the Derrida et al. 1993 reference the formula in eq. (39) gives Z_{N}(alpha,beta)/(alpha*beta)^N for N>=1.

Examples

			Triangle begins:
  1;
  1,    1;
  1,    1,     1;
  1,    2,     1,     1;
  1,    5,     3,     1,    1;
  1,   14,    13,     4,    1,   1;
  1,   42,    67,    25,    5,   1,   1;
  1,  132,   381,   190,   41,   6,   1,   1;
  1,  429,  2307,  1606,  413,  61,   7,   1,   1;
  1, 1430, 14589, 14506, 4641, 766,  85,   8,   1,   1;

Links

Crossrefs

Diagonals: A000012, A000012, A000027, A001844, A064096, A064302, A064303, A064304, A064305.
Columns (without leading zeros): A000012 (k=0), A000108 (k=1), A064062 (k=2), A064063 (k=3), A064087 (k=4), A064088 (k=5), A064089 (k=6), A064090 (k=7), A064091 (k=8), A064092 (k=9), A064093 (k=10).
Cf. A064095 (row sums).

Programs

  • Magma
    function A064094(n,k)
  if k eq 0 or k eq n then return 1;
  else return (&+[(n-k-j)*Binomial(n-k-1+j, j)*k^j: j in [0..n-k-1]])/(n-k);
  end if;
end function;
[A064094(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 27 2024
  • Mathematica
    T[n_, 0] = 1; T[n_, 1] := CatalanNumber[n - 1]; T[n_, n_] = 1; T[n_, m_] := (1/(1 - m))^(n - m)*(1 - m*Sum[ CatalanNumber[k]*(m*(1 - m))^k, {k, 0, n - m - 1}]); Table[ T[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013 *)
  • SageMath
    def A064094(n,k):
    if (k==0 or k==n): return 1
    else: return sum((n-k-j)*binomial(n-k-1+j,j)*k^j for j in range(n-k))//(n-k)
flatten([[A064094(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 27 2024

Formula

G.f. for column m: (x^m)/(1-x*c(m*x)) = (x^m)*((m-1)+m*x*c(m*x))/(m-1+x) with the g.f. c(x) of Catalan numbers A000108.
T(n, m) = Sum_{j=0..n-m-1} (n-m-j)*binomial(n-m-1+j, j)*(m^j)/(n-m) or T(n, m) = (1/(1-m))^(n-m)*(1 - m*Sum_{j=0..n-m-1} C(j)*(m*(1-m))^j ), for n - m >= 1, T(n, n) = 1, T(n, m) = 0 if nA000108(k) (Catalan).

A323206 A(n, k) = hypergeometric([-k, k+1], [-k-1], n), square array read by ascending antidiagonals for n,k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 13, 14, 1, 1, 5, 25, 67, 42, 1, 1, 6, 41, 190, 381, 132, 1, 1, 7, 61, 413, 1606, 2307, 429, 1, 1, 8, 85, 766, 4641, 14506, 14589, 1430, 1, 1, 9, 113, 1279, 10746, 55797, 137089, 95235, 4862, 1
Offset: 0

Views

Author

Peter Luschny, Feb 21 2019

Keywords

Comments

Conjecture: A(n, k) is odd if and only if n is even or (n is odd and k + 2 = 2^j for some j > 0).

Examples

			Array starts:
    [n\k 0  1    2     3       4        5         6           7  ...]
    [0]  1, 1,   1,    1,      1,       1,        1,          1, ... A000012
    [1]  1, 2,   5,   14,     42,     132,      429,       1430, ... A000108
    [2]  1, 3,  13,   67,    381,    2307,    14589,      95235, ... A064062
    [3]  1, 4,  25,  190,   1606,   14506,   137089,    1338790, ... A064063
    [4]  1, 5,  41,  413,   4641,   55797,   702297,    9137549, ... A064087
    [5]  1, 6,  61,  766,  10746,  161376,  2537781,   41260086, ... A064088
    [6]  1, 7,  85, 1279,  21517,  387607,  7312789,  142648495, ... A064089
    [7]  1, 8, 113, 1982,  38886,  817062, 17981769,  409186310, ... A064090
    [8]  1, 9, 145, 2905,  65121, 1563561, 39322929, 1022586105, ... A064091
         A001844 A064096 A064302  A064303   A064304   A064305  diag: A323209
.
Seen as a triangle (by reading ascending antidiagonals):
                               1
                              1, 1
                            1, 2, 1
                           1, 3, 5, 1
                        1, 4, 13, 14, 1
                      1, 5, 25, 67, 42, 1
                   1, 6, 41, 190, 381, 132, 1

Links

Crossrefs

Rows: A000108, A064062, A064063, A064087, A064088, A064089, A064090, A064091.
Columns: A001844, A064096, A064302, A064303, A064304, A064305.
Diagonals: A323209 (main), A323208 (sup main), A323217 (sub main).
Sums of antidiagonals: A323207
Cf. A064094, A009766, A238762.

Programs

  • Maple
    # The function ballot is defined in A238762.
A := (n, k) -> add(ballot(2*j, 2*k)*n^j, j=0..k):
for n from 0 to 6 do seq(A(n, k), k=0..9) od;
# Or by recurrence:
A := proc(n, k) option remember;
if n = 1 then return `if`(k = 0, 1, (4*k + 2)*A(1, k-1)/(k + 2)) fi:
if k < 2 then return [1, n+1][k+1] fi; n*(4*k - 2);
((%*(n - 1) - k - 1)*A(n, k-1) + %*A(n, k-2))/((n - 1)*(k + 1)) end:
for n from 0 to 6 do seq(A(n, k), k=0..9) od;
# Alternative:
Arow := proc(n, len) # Function REVERT is in Sloane's 'Transforms'.
[seq(1 + n*k, k=0..len-1)]; REVERT(%); seq((-1)^k*%[k+1], k=0..len-1) end:
for n from 0 to 8 do Arow(n, 8) od;
  • Mathematica
    A[n_, k_] := Hypergeometric2F1[-k, k + 1, -k - 1, n];
Table[A[n, k], {n, 0, 8}, {k, 0, 8}]
(* Alternative: *)
prev[f_, n_] := InverseSeries[Series[-x f, {x, 0, n}]]/(-x);
f[n_, x_] := (1 + (n - 1) x)/((1 - x)^2);
For[n = 0, n < 9, n++, Print[CoefficientList[prev[f[n, x], 8], x]]]
(* Continued fraction: *)
num[k_, n_] := If[k < 2, 1, If[k == 2, -x, -n x]];
cf[n_, len_] := ContinuedFractionK[num[k, n], 1, {k, len + 2}];
Arow[n_, len_] := Rest[CoefficientList[Series[cf[n, len], {x, 0, len}], x]];
For[n = 0, n < 9, n++, Print[Arow[n, 8]]]
  • PARI
    {A(n,k) = polcoeff((1/x)*serreverse(x*((1+(n-1)*(-x))/((1-(-x))^2)+x*O(x^k))), k)}
for(n=0, 8, for(k=0, 8, print1(A(n, k), ", ")); print())
  • Sage
    # Valid for n > 0.
def genCatalan(n): return SR(1/(x- x^2*(1 - sqrt(1 - 4*x*n))/(2*x*n)))
for n in (1..8): print(genCatalan(n).series(x).list())
# Alternative:
def pseudo_reversion(g, invsign=false):
    if invsign: g = g.subs(x=-x)
    g = g.shift(1)
    g = g.reverse()
    g = g.shift(-1)
    return g
R. = PowerSeriesRing(ZZ)
for n in (0..6):
    f = (1+(n-1)*x)/((1-x)^2)
    s = pseudo_reversion(f, true)
    print(s.list())

Formula

A(n, k) = [x^k] 1/(x - x^2*C(n*x)) if n > 0 and C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function of the Catalan numbers A000108.
A(n, k) = Sum_{j=0..k} (binomial(2*k-j, k) - binomial(2*k-j, k+1))*n^(k-j).
A(n, k) = Sum_{j=0..k} binomial(k + j, k)*(1 - j/(k + 1))*n^j (cf. A009766).
A(n, k) = 1 + Sum_{j=0..k-1} ((1+j)*binomial(2*k-j, k+1)/(k-j))*n^(k-j).
A(n, k) = (1/(2*Pi))*Integral_{x=0..4*n} (sqrt(x*(4*n-x))*x^k)/(1+(n-1)*x), n>0.
A(n, k) ~ ((4*n)^k/(Pi^(1/2)*k^(3/2)))*(1+1/(2*n-1))^2.
If we shift the series f with constant term 1 to the right, invert it with respect to composition and shift the result back to the left then we call this the 'pseudo reversion' of f, prev(f). Row n of the array gives the coefficients of the pseudo reversion of f = (1 + (n - 1)*x)/((1 - x)^2) with an additional inversion of sign. Note that f is not revertible. See also the Sage implementation below.
A(n, k) = [x^k] prev((1 + (n - 1)*(-x))/(1 - (-x))^2).
A(n, k) = [x^(k+1)] cf(n, x) where cf(n, x) = K_{i>=1} c(i)/b(i) in the notation of Gauß with b(i) = 1, c(1) = 1, c(2) = -x and c(i) = -n*x for i > 2.
For a recurrence see the Maple section.
Showing 1-2 of 2 results.

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