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.

A050270 Largest value b for Diophantine 1-doubles (a,b) ordered by smallest b.

Original entry on oeis.org

3, 4, 5, 6, 7, 8, 8, 8, 9, 10, 11, 12, 12, 12, 13, 14, 15, 15, 15, 16, 16, 16, 17, 18, 19, 20, 20, 20, 21, 21, 21, 22, 23, 24, 24, 24, 24, 24, 24, 24, 25, 26, 27, 28, 28, 28, 29, 30, 30, 30, 31, 32, 32, 32, 33, 33, 33, 34, 35, 35, 35, 36, 36, 36, 37, 38, 39, 39, 39, 40, 40, 40
Offset: 1

Views

Author

Keywords

Crossrefs

Cf. A050269.

A176097 Degree of the hyperdeterminant of the cubic format (k+1) X (k+1) X (k+1).

Original entry on oeis.org

1, 4, 36, 272, 2150, 16992, 134848, 1072192, 8536914, 68036600, 542607560, 4329671040, 34561892560, 275979195520, 2204266118400, 17609217372416, 140698273234634, 1124340854572296, 8985828520591912, 71822662173752800
Offset: 0

Views

Author

Benjamin J. Young, Apr 08 2010

Keywords

Examples

			For k=1, the hyperdeterminant of the matrix (a_ijk) (for 0 <= i,j,k <= 1) is (a_000 * a_111)^2 + (a001 * a110)^2 + (a_010 * a_101)^2 + (a_011 * a_100)^2 -2(a_000 * a_001 * a_110 * a_111 + a_000 * a_010 * a_101 * a_111 + a_000 * a_011 * a_100 * a_111 + a_001 * a_010 * a_101 * a_110 + a_001 * a_011 * a_110 * a_100 + a_010 * a_011 * a_101 * a_100) + 4(a_000 * a_011 * a_101 * a_110 + a_001 * a_010 * a_100 * a_111) (see Gelfand, Kapranov & Zelevinsky, pp. 2 and 448.) [Corrected by _Petros Hadjicostas_, Sep 12 2019]
		

References

  • I. M. Gelfand, M. M. Kapranov and A. V. Zelevinsky, Discriminants, Resultants and Multidimensional Determinants, Birkhauser, 2008, p. 456 (Ch. 14, Corollary 2.9).

Crossrefs

Programs

  • Maple
    a:= k-> add((j+k+1)! /(j!)^3 /(k-2*j)! *2^(k-2*j), j=0..floor(k/2)): seq(a(n), n=0..20);
    # Second program:
    a := proc(n) option remember; if n = 0 then return 1 elif n = 1 then return 4 fi;
    (a(n-1)*(21*n^3-10*n^2-9*n+6)+a(n-2)*(24*n^3+16*n^2))/((3*n-1)*n^2) end:
    seq(a(n), n=0..19); # Peter Luschny, Sep 12 2019
  • Mathematica
    Table[Sum[(j + n + 1)!*2^(n - 2*j)/(j!^3*(n - 2*j)!), {j, 0, n/2}], {n, 0, 20}] (* Vaclav Kotesovec, Sep 12 2019 *)

Formula

a(n) = Sum_{j = 0..n/2} ( (j+n+1)! * 2^(n-2j) )/((j!)^3 * (n-2j)!).
a(n) = (n+1)^2*(8*A000172(n)-A000172(n+1))/6. - Mark van Hoeij, Jul 02 2010
G.f.: hypergeom([-1/3, 1/3],[1],27*x^2/(1-2*x)^3)*(1-2*x)/((x+1)^2*(1-8*x)). - Mark van Hoeij, Apr 11 2014
a(n) ~ 8^(n+1) / (Pi * 3^(3/2)). - Vaclav Kotesovec, Sep 12 2019
a(n) = (a(n-1)*(21*n^3 - 10*n^2 - 9*n + 6) + a(n-2)*(24*n^3 + 16*n^2))/((3*n - 1)*n^2) for n >= 2. - Peter Luschny, Sep 12 2019
Showing 1-2 of 2 results.