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.

A284308 Number A(n,k) of singular vector tuples for a general k-dimensional {n}^k tensor; square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 6, 3, 1, 1, 24, 37, 4, 1, 1, 120, 997, 240, 5, 1, 1, 720, 44121, 51264, 1621, 6, 1, 1, 5040, 2882071, 23096640, 2940841, 11256, 7, 1, 1, 40320, 260415373, 18754813440, 14346274601, 180296088, 79717, 8, 1, 1, 362880, 31088448777, 24874143759360, 153480509680141, 9859397817600, 11559133741, 572928, 9, 1
Offset: 1

Views

Author

Alois P. Heinz, Mar 24 2017

Keywords

Examples

			Square array A(n,k) begins:
  1, 1,     1,         1,             1,                   1, ...
  1, 2,     6,        24,           120,                 720, ...
  1, 3,    37,       997,         44121,             2882071, ...
  1, 4,   240,     51264,      23096640,         18754813440, ...
  1, 5,  1621,   2940841,   14346274601,     153480509680141, ...
  1, 6, 11256, 180296088, 9859397817600, 1435747717722810960, ...
		

Crossrefs

Rows n=1-3 give: A000012, A000142, A274308.
Main diagonal gives A284309.

A271905 Number of singular vector tuples for a general n X n X n tensor.

Original entry on oeis.org

1, 6, 37, 240, 1621, 11256, 79717, 572928, 4164841, 30553116, 225817021, 1679454816, 12556853401, 94313192616, 711189994357, 5381592930816, 40848410792017, 310909645663332, 2372280474687277, 18141232682656320, 139010366280363601, 1067160872528170536, 8206301850166625797, 63203453697218605440
Offset: 1

Views

Author

Doron Zeilberger, Apr 21 2016

Keywords

References

  • Bernd Sturmfels, Eigenvectors of Tensors, Colloquium Talk, Rutgers University, Apr 22 2016.

Crossrefs

See A272551 for the n X n X n X n version.
Column k=3 of A284308.
Cf. A274308.

Programs

  • Mathematica
    a[1] = 1;
    a[n_] := Module[{a, b, c, s}, s = Series[(((a + b)^n - c^n)((a + c)^n - b^n)((b + c)^n - a^n))/((a + b - c)(a + c - b)(b + c - a)), {a, 0, n}, {b, 0, n}, {c, 0, n}] // Normal // Expand; Cases[List @@ s, k_Integer a^(n-1) b^(n-1) c^(n-1)] /. (a|b|c) -> 1 // First];
    Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 24}] (* Jean-François Alcover, Aug 18 2018 *)

Formula

From Eq. (1.3) of Ottaviani-Friedland (2014), a(n) is the coefficient of (abc)^(n-1) in the polynomial
{((a+b)^n-c^n)*((a+c)^n-b^n)*((b+c)^n-a^n)} / {(a+b-c)*(a+c-b)*(b+c-a)}.
a(n) satisfies the following fifth-order recurrence equation with polynomial coefficients:
72*(n + 2)*(245*n^4 + 3094*n^3 + 14447*n^2 + 29474*n + 22100)*(n + 1)^2*a(n) - (n + 2)*(21805*n^6 + 330981*n^5 + 2012733*n^4 + 6230951*n^3 + 10263446*n^2 + 8425060*n + 2639760)*a(n + 1) + (-10279296 - 13230*n^7 - 249641*n^6 - 29331496*n - 22847777*n^3 - 1998705*n^5 - 8785333*n^4 - 35069178*n^2)*a(n + 2) + (16026528 + 21560*n^7 + 413637*n^6 + 3343917*n^5 + 14735333*n^4 + 38132651*n^3 + 57777574*n^2 + 47273504*n)*a(n + 3) - (4410*n^6 + 70147*n^5 + 452903*n^4 + 1516515*n^3 + 2769127*n^2 + 2601986*n + 975888)*(n + 4)*a(n + 4) + (n + 5)*(n + 4)*(n + 3)*(245*n^4 + 2114*n^3 + 6635*n^2 + 8882*n + 4224)*a(n + 5) = 0
with initial conditions
[a(1), ..., a(5)] = [1, 6, 37, 240, 1621]
and asymptotically
a(n) ~ (2/(sqrt(3)*Pi))*8^n/n.

A321711 Triangle T(n,k) read by rows: coefficients of polynomials P_n(t) defined in Formula section.

Original entry on oeis.org

1, 1, 0, 3, 0, 0, 11, 9, 0, 1, 53, 120, 60, 40, 9, 309, 1410, 1800, 1590, 885, 216, 2119, 16560, 39960, 55120, 52065, 29016, 7570, 16687, 202755, 801780, 1696555, 2433165, 2300403, 1326850, 357435, 148329, 2624496, 15606360, 48387024, 99650670, 141429456, 135382464, 79738800, 22040361, 1468457, 36080100, 304274880, 1323453180, 3760709526, 7493549868, 10570597800, 10199809980, 6103007505, 1721632024
Offset: 0

Views

Author

Gheorghe Coserea, Nov 27 2018

Keywords

Examples

			For n=3 we have s1 = z1 + z2 + z3, s2 = z1^2 + z2^2 + z3^2, s12 = z1*z2 + z1*z3 + z2*z3, f1 = z1^2 + z2^2 + z3^2 + t*z2*z3 + z1*(z2 + z3), f2 = z1^2 + z2^2 + z3^2 + t*z1*z3 + z2*(z1 + z3), f3 = z1^2 + z2^2 + z3^2 + t*z1*z2 + z3*(z1 + z2), [(z1*z2*z3)^2] f1*f2*f3 = 11 + 9*t + t^3, therefore P_3(t) = 11 + 9*t + t^3.
A(x;t) = 1 + x + 3*x^2 + (11 + 9*t + t^3)*x^3 + (53 + 120*t + 60*t^2 + 40*t^3 + 9*t^4)*x^4 + ...
Triangle starts:
n\k [0]    [1]     [2]     [3]      [4]      [5]      [6]      [7]
[0] 1;
[1] 1;     0;
[2] 3;     0;      0;
[3] 11,    9,      0,      1;
[4] 53,    120,    60,     40,      9;
[5] 309,   1410,   1800,   1590,    885,     216;
[6] 2119,  16560,  39960,  55120,   52065,   29016,   7570;
[7] 16687, 202755, 801780, 1696555, 2433165, 2300403, 1326850, 357435;
[8] ...
		

Crossrefs

Programs

  • PARI
    P(n, t='t) = {
      my(z=vector(n, k, eval(Str("z", k))),
         s1=sum(k=1, #z, z[k]), s2=sum(k=1, #z, z[k]^2), s12=(s1^2 - s2)/2,
         f=vector(n, k, s2 + t*(s12 - z[k]*(s1 - z[k])) + z[k]*(s1 - z[k])), g=1);
      for (i=1, n, g *= f[i]; for(j=1, n, g=substpol(g, z[j]^3, 0)));
      for (k=1, n, g=polcoef(g, 2, z[k]));
      g;
    };
    seq(N) = concat([[1], [1, 0], [3, 0, 0]], apply(n->Vecrev(P(n,'t)), [3..N]));
    concat(seq(9))

Formula

Let z1..zn be n variables and s1 = Sum_{k=1..n} zk, s2 = Sum_{k=1..n} zk^2, s12 = (s1^2 - s2)/2, fk = s2 + t*(s12 - zk*(s1 - zk)) + zk*(s1 - zk) for k=1..n; we define P_n(t) = [(z1..zn)^2] Product_{k=1..n} fk.
A000255(n) = T(n,0).
A007107(n) = T(n,n).
A000681(n) = Sum_{k=0..n} T(n,k).
A274308(n) = Sum_{k=0..n} T(n,k)*2^k.
Showing 1-3 of 3 results.