A322677 -id:A322677 - cp's OEIS frontend

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).

0, 0, 0, 0, 1, 0, 0, 8, 2, 0, 0, 49, 24, 3, 0, 0, 288, 242, 48, 4, 0, 0, 1681, 2400, 675, 80, 5, 0, 0, 9800, 23762, 9408, 1444, 120, 6, 0, 0, 57121, 235224, 131043, 25920, 2645, 168, 7, 0, 0, 332928, 2328482, 1825200, 465124, 58080, 4374, 224, 8, 0

Offset: 0

Seiichi Manyama, Dec 23 2018

			Square array begins:
   0, 0,   0,    0,      0,       0,        0, ...
   0, 1,   8,   49,    288,    1681,     9800, ...
   0, 2,  24,  242,   2400,   23762,   235224, ...
   0, 3,  48,  675,   9408,  131043,  1825200, ...
   0, 4,  80, 1444,  25920,  465124,  8346320, ...
   0, 5, 120, 2645,  58080, 1275125, 27994680, ...
   0, 6, 168, 4374, 113568, 2948406, 76545000, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-5 give A000004, A001477, A033996, A322675, A322677, A322745.
Rows 0-9 give A000004, A001108, A132596, A007654(n+1), A132584, A322707, A322708, A322709, A132592, A132593.
Main diagonal gives A322746.
Cf. A173175 (A(n,2*n)), A322790.

Unprotect[Power]; 0^0 := 1; Protect[Power]; Table[(-1 + Sum[Binomial[2 k, 2 j] (# + 1)^(k - j)*#^j, {j, 0, k}])/2 &[n - k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Jan 01 2019 *)
nmax = 9; row[n_] := LinearRecurrence[{4n+3, -4n-3, 1}, {0, n, 4n(n+1)}, nmax+1]; T = Array[row, nmax+1, 0]; A[n_, k_] := T[[n+1, k+1]];
Table[A[n-k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jan 06 2019 *)

def ncr(n, r)
  return 1 if r == 0
  (n - r + 1..n).inject(:*) / (1..r).inject(:*)
end
def A(k, n)
  (0..n).map{|i| (0..k).inject(-1){|s, j| s + ncr(2 * k, 2 * j) * (i + 1) ** (k - j) * i ** j} / 2}
end
def A322699(n)
  a = []
  (0..n).each{|i| a << A(i, n - i)}
  ary = []
  (0..n).each{|i|
    (0..i).each{|j|
      ary << a[i - j][j]
    }
  }
  ary
end
p A322699(10)

sqrt(A(n,k)+1) + sqrt(A(n,k)) = (sqrt(n+1) + sqrt(n))^k.
sqrt(A(n,k)+1) - sqrt(A(n,k)) = (sqrt(n+1) - sqrt(n))^k.
A(n,0) = 0, A(n,1) = n and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) + 2*n for k > 1.
A(n,k) = (T_{k}(2*n+1) - 1)/2 where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = (2*n+1-1)/2 = n.

A180324 Vassiliev invariant of fourth order for the torus knots.

Original entry on oeis.org

0, 3, 25, 98, 270, 605, 1183, 2100, 3468, 5415, 8085, 11638, 16250, 22113, 29435, 38440, 49368, 62475, 78033, 96330, 117670, 142373, 170775, 203228, 240100, 281775, 328653, 381150, 439698, 504745, 576755, 656208, 743600, 839443, 944265, 1058610, 1183038, 1318125

Offset: 0

Sergey Allenov, Jan 18 2011

a(n) is the Vassiliev invariant of fourth order for the torus knots. a(n) can be calculated as the number of attachments of the two arrow diagrams in the arrow diagram of the torus knot. Arrow diagram of the torus knot is 2n+1 intersecting arrows with mixing ends.
Antidiagonal sums of the convolution array A213847. - Clark Kimberling, Jul 05 2012
First differences of the terms produced by convolving the odd and even triangular numbers, with  n>0. The sequence begins 0, 3, 28, 126, 396, 1001, 2184, 4284, 7752, 13167, 21252..starting at n=1 and has the formula (4*n^5 - 5*n^3 + 30*n)/30. - J. M. Bergot, Sep 09 2016

			a(1) = 1*2*3^2/6 = 3.
a(2) = 2*(2+1)*(2*2+1)^2/6 = 5^2 = 25.

S. V. Allenov, Explicit formulas for Vassil'ev invariants of the fourth order for knots, Journal of Mathematical Sciences, New York: Springer, Vol. 157, No. 3 (2009), pp. 413-423.
Michael Polyak and Oleg Viro, Gauss diagram formulas for Vassiliev invariants, Int. Math. Res. Notices, No. 11 (1994), pp. 445-453.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000330, A185096, A213847, A322677, A339483.

a:=n->(1/6)*n*(n+1)*(2*n+1)^2;
a:=n->binomial(2*n+2, 4)+binomial(2*n+2, 3)/2;

Table[Binomial[2n+2,4]+Binomial[2n+2,3]/2,{n,0,40}] (* Harvey P. Dale, Sep 18 2018 *)
Table[Sum[x^2 + y^2, {x, -g, g}, {y, -g, g}], {g, 0, 33}]/4 (* Horst H. Manninger, Jun 19 2025 *)

PARI
```
a(n) = n*(n+1)*(2*n+1)^2/6
```

a(n) = (n*(n+1)*(2*n+1)^2)/6.
a(n) = C(2*n+2,4) + C(2*n+2,3)/2.
a(n) = (2*n+1)*A000330(n).
a(n) = 3*A000330(n)^2/A000217(n).
a(n) = (A000330(1) + A000330(2) + ... + A000330(2*n-1) + A000330(2*n))/2.
G.f.: x*(3+x)*(1+3*x)/(1-x)^5. - Colin Barker, Mar 17 2012
Sum_{n>=1} 1/a(n) = 30 - 3*Pi^2. - Amiram Eldar, Jun 20 2025
From Elmo R. Oliveira, Aug 20 2025: (Start)
E.g.f.: exp(x)*x*(2 + x)*(9 + 24*x + 4*x^2)/6.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A322677(n)/96 = A185096(n)/12 = A339483(n)/3. (End)

A339483 Number of regular polygons that can be drawn with vertices on a centered hexagonal grid with side length n.

Original entry on oeis.org

0, 9, 75, 294, 810, 1815, 3549, 6300, 10404, 16245, 24255, 34914, 48750, 66339, 88305, 115320, 148104, 187425, 234099, 288990, 353010, 427119, 512325, 609684, 720300, 845325, 985959, 1143450, 1319094, 1514235, 1730265, 1968624, 2230800, 2518329, 2832795

Offset: 0

Peter Kagey, Dec 06 2020

The only regular polygons that can be drawn with vertices on the centered hexagonal grid are equilateral triangles and regular hexagons.

			There are a(2) = 75 regular polygons that can be drawn on the centered hexagonal grid with side length 2: A000537(2) = 9 regular hexagons and A008893(n) = 66 equilateral triangles.
The nine hexagons are:
    * . *       . * .       * * .
   . . . .     * . . *     * . * .
  * . . . *   . . . . .   . * * . .
   . . . .     * . . *     . . . .
    * . *       . * .       . . .
      1           1           7
which are marked with the number of ways to draw the hexagons up to translation.
The 66 equilateral triangles are:
    * . .       * . .       * . .       * . *       * . .       . . .
   * * . .     . . * .     . . . .     . . . .     . . . .     * . . *
  . . . . .   . * . . .   . . . * .   . . * . .   . . . . *   . . . . .
   . . . .     . . . .     * . . .     . . . .     . . . .     . . . .
    . . .       . . .       . . .       . . .       * . .       . * .
     24          14          12          12           2           2
which are marked with the number of ways to draw the triangles up to translation and dihedral action of the hexagon.

Peter Kagey, Table of n, a(n) for n = 0..10000
Burkard Polster, What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented, Mathologer video (2020).
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000537 (regular hexagons), A008893 (equilateral triangles).
Cf. A338323 (cubic grid).
Cf. A003215, A180324, A185096, A322677, A348670.

a[n_] := n*(n+1)*(2*n+1)^2/2; Array[a, 35, 0] (* Amiram Eldar, Jun 20 2025 *)

a(n) = A000537(n) + A008893(n).
a(n) = (1/2)*(n+1)*n*(2*n+1)^2.
a(n) = 3*A180324(n).
Sum_{n>=1} 1/a(n) = 10 - Pi^2 (A348670). - Amiram Eldar, Jun 20 2025
From Elmo R. Oliveira, Aug 20 2025: (Start)
G.f.: -3*x*(x + 3)*(3*x + 1)/(x - 1)^5.
E.g.f.: exp(x)*x*(2 + x)*(9 + 24*x + 4*x^2)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A185096(n)/4 = A322677(n)/32. (End)

A322675 a(n) = n * (4*n + 3)^2.

Original entry on oeis.org

0, 49, 242, 675, 1444, 2645, 4374, 6727, 9800, 13689, 18490, 24299, 31212, 39325, 48734, 59535, 71824, 85697, 101250, 118579, 137780, 158949, 182182, 207575, 235224, 265225, 297674, 332667, 370300, 410669, 453870, 499999, 549152, 601425, 656914, 715715, 777924, 843637

Offset: 0

Seiichi Manyama, Dec 23 2018

			(sqrt(2) - sqrt(1))^3 = 5*sqrt(2) - 7 = sqrt(50) - sqrt(49). So a(1) = 49.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Column 3 of A322699.

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^k: A033996(n) (k=2), this sequence (k=3), A322677 (k=4), A322745 (k=5).

PARI
```
{a(n) = n*(4*n+3)^2}
```

concat(0, Vec(x*(49 + 46*x + x^2) / (1 - x)^4 + O(x^40))) \\ Colin Barker, Dec 23 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^3.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^3.
Sum_{n>=1} 1/a(n) = 8/27 + 2*c/3 + Pi/18 - Pi^2/12 - log(2)/3 = 0.027956857336446942649782759291008857522041405948099294509008..., where c is the Catalan constant A006752. - Vaclav Kotesovec, Dec 23 2018
From Colin Barker, Dec 23 2018: (Start)
G.f.: x*(49 + 46*x + x^2) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3.
(End)

cp's OEIS Frontend

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j).

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A180324 Vassiliev invariant of fourth order for the torus knots.

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A339483 Number of regular polygons that can be drawn with vertices on a centered hexagonal grid with side length n.

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A322675 a(n) = n * (4*n + 3)^2.

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A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).