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.

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A103772 Larger of two sides in a (k,k,k-1)-integer-sided triangle with integer area.

Original entry on oeis.org

1, 17, 241, 3361, 46817, 652081, 9082321, 126500417, 1761923521, 24540428881, 341804080817, 4760716702561, 66308229755041, 923554499868017, 12863454768397201, 179164812257692801, 2495443916839302017, 34757050023492535441, 484103256412056194161
Offset: 1

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Author

Zak Seidov, Feb 23 2005

Keywords

Comments

Corresponding areas are 0, 120, 25080, 4890480, 949077360, 184120982760, ...
Values of (x^2 + y^2)/2, where the pair (x, y) satisfies x^2 - 3*y^2 = -2, i.e., a(n) = {(A001834(n))^2 + (A001835(n))^2}/2 = {(A001834(n))^2 + A046184(n)}/2. - Lekraj Beedassy, Jul 13 2006
The heights of these triangles are given in A028230. (A028230(n), A045899(n), A103772(n)) forms a primitive Pythagorean triple.
Shortest side of (k,k+2,k+3) triangle such that median to longest side is integral. Sequence of such medians is A028230. - James R. Buddenhagen, Nov 22 2013
Numbers n such that (n+1)*(3n-1) is a square. - James R. Buddenhagen, Nov 22 2013

Crossrefs

Programs

  • Magma
    I:=[1,17]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+4: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016
  • Mathematica
    a[1] = 1; a[2] = 17; a[3] = 241; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n] - 1, {n, 17}] (* Robert G. Wilson v, Mar 24 2005 *)
    LinearRecurrence[{15,-15,1},{1,17,241},20] (* Harvey P. Dale, Jan 02 2016 *)
    RecurrenceTable[{a[1] == 1, a[2] == 17, a[n] == 14 a[n-1] - a[n-2] + 4}, a, {n, 20}] (* Vincenzo Librandi, Mar 05 2016 *)
  • PARI
    Vec(x*(1+x)^2/((1-x)*(1-14*x+x^2)) + O(x^25)) \\ Colin Barker, Mar 05 2016
    

Formula

a(n) = (4*A001570(n+1) - 1)/3, n > 0. - Ralf Stephan, May 20 2007
a(n) = A052530(n-1)*A052530(n) + 1. - Johannes Boot, May 21 2011
G.f.: x*(1+x)^2/((1-x)*(1-14*x+x^2)). - Colin Barker, Apr 09 2012
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3); a(1)=1, a(2)=17, a(3)=241. - Harvey P. Dale, Jan 02 2016
a(n) = (-1+(7-4*sqrt(3))^n*(2+sqrt(3))-(-2+sqrt(3))*(7+4*sqrt(3))^n)/3. - Colin Barker, Mar 05 2016
a(n) = 14*a(n-1) - a(n-2) + 4. - Vincenzo Librandi, Mar 05 2016
a(n) = A001353(n)^2 + A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020

Extensions

More terms from Robert G. Wilson v, Mar 24 2005

A051047 For n > 5, a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3); initial terms are 1, 3, 8, 120, 1680.

Original entry on oeis.org

1, 3, 8, 120, 1680, 23408, 326040, 4541160, 63250208, 880961760, 12270214440, 170902040408, 2380358351280, 33154114877520, 461777249934008, 6431727384198600, 89582406128846400, 1247721958419651008, 17378525011746267720, 242051628206028097080
Offset: 1

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Author

Keywords

Comments

The recurrence gives an infinite sequence of polynomials S={x,x+2,c_1(x),c_2(x),...} such that the product of any two consecutive polynomials, increased by 1, is the square of a polynomial - see the Jones reference.

Crossrefs

Cf. A051048. Essentially the same as A045899.

Programs

  • Magma
    I:=[1,3,8,120,1680]; [n le 5 select I[n] else 14*Self(n-1)-Self(n-2)+8: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016
  • Mathematica
    With[{x = 1},
    Join[{x, x + 2},
    RecurrenceTable[{c[-1] == c[0] == 0,
    c[k] == (4 x^2 + 8 x + 2) c[k - 1] - c[k - 2] + 4 (x + 1)}, c, {k, 1, 12}]]]
    LinearRecurrence[{15, -15, 1}, {1, 3, 8, 120, 1680}, 22] (* Charles R Greathouse IV, Oct 31 2011 *)
    Join[{1, 3}, RecurrenceTable[{a[1] == 8, a[2] == 120, a[n] == 14 a[n-1] - a[n-2] + 8}, a, {n, 20}]] (* Vincenzo Librandi, Mar 05 2016 *)
  • PARI
    Vec((3*x^4-44*x^3+22*x^2+12*x-1)/(x^3-15*x^2+15*x-1)+O(x^99)) \\ Charles R Greathouse IV, Oct 31 2011
    

Formula

G.f.: x*(3*x^4-44*x^3+22*x^2+12*x-1) / (x^3-15*x^2+15*x-1).
For n>4, a(n) = 14*a(n-1)-a(n-2)+8. - Vincenzo Librandi, Mar 05 2016

Extensions

Entry revised by N. J. A. Sloane, Oct 25 2009, following correspondence with Eric Weisstein

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
Previous Showing 11-13 of 13 results.