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.

A365673 Array A(n, k) read by ascending antidiagonals. Polygonal number weighted generalized Catalan sequences.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 4, 1, 1, 1, 4, 15, 8, 1, 1, 1, 5, 34, 105, 16, 1, 1, 1, 6, 61, 496, 945, 32, 1, 1, 1, 7, 96, 1385, 11056, 10395, 64, 1, 1, 1, 8, 139, 2976, 50521, 349504, 135135, 128, 1, 1, 1, 9, 190, 5473, 151416, 2702765, 14873104, 2027025, 256, 1
Offset: 0

Views

Author

Peter Luschny, Sep 30 2023

Keywords

Comments

Using polygonal numbers as weights, a recursion for triangles is defined, whose main diagonals represents a family of sequences, which include, among others, the powers of 2, the double factorial of odd numbers, the reduced tangent numbers, and the Euler numbers.
Apart from the edge cases k = 0 and k = n the recursion is T(n, k) = w(n, k) * T(n, k - 1) + T(n - 1, k). T(n, 0) = 1 and T(n, n) = T(n, n-1) if n > 0.
The weights w(n, k) identical to 1 yield the recursion of the Catalan triangle A009766 (with main diagonal the Catalan numbers). Here the polygonal numbers are used as weights in the form w(n, k) = p(s, n - k + 1), where the parameter s is the number of sides of the polygon and p(s, n) = ((s-2) * n^2 - (s-4) * n) / 2, see A317302.

Examples

			Array A(n, k) starts:                            (polygon|diagonal|triangle)
[0] 1, 1, 1,   1,     1,       1,         1, ...  A258837  A000012
[1] 1, 1, 2,   4,     8,      16,        32, ...  A080956  A011782
[2] 1, 1, 3,  15,   105,     945,     10395, ...  A001477  A001147  A001498
[3] 1, 1, 4,  34,   496,   11056,    349504, ...  A000217  A002105  A365674
[4] 1, 1, 5,  61,  1385,   50521,   2702765, ...  A000290  A000364  A060058
[5] 1, 1, 6,  96,  2976,  151416,  11449296, ...  A000326  A126151  A366138
[6] 1, 1, 7, 139,  5473,  357721,  34988647, ...  A000384  A126156  A365672
[7] 1, 1, 8, 190,  9080,  725320,  87067520, ...  A000566  A366150  A366149
[8] 1, 1, 9, 249, 14001, 1322001, 188106489, ...  A000567
           A054556                         A366137

Links

Crossrefs

Poly weights: A258837, A080956, A001477, A000217, A000290, A000326, A000384.
Rows: A000012, A011782, A001147, A002105, A000364, A126151, A126156, A366150.
Triangles: A001498, A365674, A060058, A366138, A365672, A366149.
Cf. A009766, A366137 (central diagonal), A317302 (table of polygonal numbers).
Cf. A112934, A303943, A305532, A305533.

Programs

  • Maple
    poly := (s, n) -> ((s - 2) * n^2 - (s - 4) * n) / 2:
T := proc(s, n, k) option remember; if k = 0 then 1 else if k = n then T(s, n, k-1) else poly(s, n - k + 1) * T(s, n, k - 1) + T(s, n - 1, k) fi fi end:
for n from 0 to 8 do A := (n, k) -> T(n, k, k): seq(A(n, k), k = 0..9) od;
# Alternative, using continued fractions:
A := proc(p, L) local CF, poly, k, m, P, ser;
   poly := (s, n) -> ((s - 2)*n^2 - (s - 4)*n)/2;
   CF := 1 + x;
   for k from 1 to L do
       m := L - k + 1;
       P := poly(p, m);
       CF := 1/(1 - P*x*CF)
   od;
   ser := series(CF, x, L);
   seq(coeff(ser, x, m), m = 0..L-1)
end:
for p from 0 to 8 do lprint(A(p, 8)) od;
  • Mathematica
    poly[s_, n_] := ((s - 2) * n^2 - (s - 4) * n) / 2;
T[s_, n_, k_] := T[s, n, k] = If[k == 0, 1, If[k == n, T[s, n, k - 1], poly[s, n - k + 1] * T[s, n, k - 1] + T[s, n - 1, k]]];
A[n_, k_] := T[n, k, k];
Table[A[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 27 2023, from first Maple program *)
  • PARI
    A(p, n) = {
       my(CF = 1 + x,
           poly(s, n) = ((s - 2)*n^2 - (s - 4)*n)/2,
           m, P
       );
       for(k = 1, n,
           m = n - k + 1;
           P = poly(p, m);
           CF = 1/(1 - P*x*CF)
        );
        Vec(CF + O(x^(n)))
}
for(p = 0, 8, print(A(p, 8)))
\\  Michel Marcus and Peter Luschny, Oct 02 2023
  • Python
    from functools import cache
@cache
def T(s, n, k):
    if k == 0: return 1
    if k == n: return T(s, n, k - 1)
    p = (n - k + 1) * ((s - 2) * (n - k + 1) - (s - 4)) // 2
    return p * T(s, n, k - 1) + T(s, n - 1, k)
def A(n, k): return T(n, k, k)
for n in range(9): print([A(n, k) for k in range(9)])

A303943 Expansion of 1/(1 - x/(1 - 1^2*x/(1 - 2^2*x/(1 - 3^2*x/(1 - 4^2*x/(1 - ...)))))), a continued fraction.

Original entry on oeis.org

1, 1, 2, 8, 76, 1540, 53684, 2812148, 205054036, 19805016628, 2444724910292, 375282530128052, 70102075181928148, 15655136160745164340, 4118456236678107528404, 1260512820941791994429876, 444069171743010266366969044, 178408825363590577961830752052
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 04 2018

Keywords

Comments

Invert transform of Euler (or secant) numbers (A000364), shifted right one place.

Links

Crossrefs

Cf. A000290, A000364, A112934, A305532.

Programs

  • Maple
    b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
a:= proc(n) option remember; `if`(n<1, 1,
      add(a(n-i)*b((i-1)*2, 0), i=1..n))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 13 2018
# Alternative:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)
else (n - k)^2 * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..17);  # Peter Luschny, Oct 02 2023
  • Mathematica
    nmax = 17; CoefficientList[Series[1/(1 - x/(1 + ContinuedFractionK[-k^2 x, 1, {k, 1, nmax}])), {x, 0, nmax}], x]
nmax = 17; CoefficientList[Series[1/(1 - x Sum[Abs[EulerE[2 k]] x^k, {k, 0, nmax}]), {x, 0, nmax}], x]
a[0] = 1; a[n_] := a[n] = Sum[Abs[EulerE[2 (k - 1)]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 17}]

Formula

a(n) ~ 2^(4*n - 1) * n^(2*n - 3/2) / (Pi^(2*n - 3/2) * exp(2*n)). - Vaclav Kotesovec, Jun 08 2019

A305533 Expansion of 1/(1 - x/(1 - 1*x/(1 - 3*x/(1 - 6*x/(1 - 10*x/(1 - ... - (k*(k + 1)/2)*x/(1 - ...))))))), a continued fraction.

Original entry on oeis.org

1, 1, 2, 7, 47, 592, 12287, 374857, 15639302, 851542747, 58536120467, 4953497262712, 505784457870707, 61300510121162077, 8698776162350603222, 1428545280744850604767, 268795232754158224790687, 57445320930331531152213232, 13837791987711934467999437927
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 04 2018

Keywords

Comments

Invert transform of reduced tangent numbers (A002105).

Links

Crossrefs

Cf. A000217, A002105, A305532.

Programs

  • Maple
    T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)
else (((k - n - 1)*(k - n)) / 2) * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Oct 01 2023
  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 - x/(1 + ContinuedFractionK[-k (k + 1)/2 x, 1, {k, 1, nmax}])), {x, 0, nmax}], x]
nmax = 18; CoefficientList[Series[1/(1 - Sum[PolyGamma[2 k - 1, 1/2]/(2^(k - 2) Pi^(2 k)) x^k, {k, 1, nmax}]), {x, 0, nmax}], x]
a[0] = 1; a[n_] := a[n] = Sum[2^k (2^(2 k) - 1) Abs[BernoulliB[2 k]]/k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 18}]

Formula

a(n) ~ 2^(3*n + 2) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n - 1/2)). - Vaclav Kotesovec, Jun 08 2019
Showing 1-3 of 3 results.

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