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 1-5 of 5 results.

A188158 Area A of the triangles such that A and the sides are integers.

Original entry on oeis.org

6, 12, 24, 30, 36, 42, 48, 54, 60, 66, 72, 84, 90, 96, 108, 114, 120, 126, 132, 144, 150, 156, 168, 180, 192, 198, 204, 210, 216, 234, 240, 252, 264, 270, 288, 294, 300, 306, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 432, 456, 462, 468, 480, 486, 504, 510, 522, 528
Offset: 1

Views

Author

Michel Lagneau, Mar 22 2011

Keywords

Comments

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. A given area often corresponds to more than one triangle; for example, a(9) = 60 for the triangles (a,b,c) = (6,25,29), (8,17,15), (13,13,10) and (13,13,24).
If only primitive integer triangles (that is, the lengths of the sides are coprime) are considered, then the possible areas are 6 times the terms in A083875. - T. D. Noe, Mar 23 2011

Examples

			a(3) = 24 because the area of the triangle whose sides are 4, 15, 13 is given by sqrt(p(p-4)(p-15)(p-13)) = 24, where p = (4 + 15 + 13)/2 = 16.

Links

Crossrefs

Cf. A007237, A009112, A024153, A024365, A051516, A051584, A051585, A055592, A055593, A055594, A055595.

Programs

  • Maple
    # storage of areas in T(i)
T:=array(1..4000):nn:=100:k:=1:for a from 1
  to nn do: for b from 1 to nn do: for c from 1 to nn do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c):   if x>0 then x1:=abs(x):s:=sqrt(x1) :else fi:if s=floor(s) then T[k]:=s:k:=k+1:else
  fi:od:od:od:
# sort of T(i)
for jj from 1 to k-1 do: ii:=jj:for k1 from  ii+1 to k-1 do:if T[ii]>T[k1] then ii:=k1:else fi:od: m:=T[jj]:T[jj]:=T[ii]:T[ii]:=m:od:liste:=convert(T,set):print(liste):
# second program:
isA188158 := proc(A::integer)
    local Asqr, s,a,b,c ;
    Asqr := A^2 ;
    for s in numtheory[divisors](Asqr) do
        if s^2> A then
        for a from 1 to s-1 do
            if modp(Asqr,s-a) = 0 then
                for b from a to s-1 do
                    c := 2*s-a-b ;
                    if s*(s-a)*(s-b)*(s-c) = Asqr then
                        return true ;
                    end if;
                end do:
            end if;
        end do:
        end if;
    end do:
    false ;
end proc:
for n from 3 to 600 do
    if isA188158(n) then
        printf("%d,\n",n) ;
    end if;
end do: # R. J. Mathar, May 02 2018
  • Mathematica
    nn = 528; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst] (* T. D. Noe, Mar 23 2011 *)

A051584 Number of integer-sided triangles of area n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 4
Offset: 1

Views

Author

David W. Wilson

Keywords

Comments

If integer-sided triangle has integer area, area is divisible by 6.

Links

Crossrefs

Cf. A051585, A051586.

Programs

  • PARI
    A051585(n) = sum(z=sqrtint(sqrtint(192*n^2)-1)+1, sqrtint(9*(64*n^2+5)\20), sum(y=z\2+1, z, my(t=(y*z)^2-(12*n)^2, x); if(issquare(t, &t), (issquare(y^2+z^2-2*t, &x) && x<=y) + (t && issquare(y^2+z^2+2*t, &x) && x<=y), 0))); \\ From A051585 by Charles R Greathouse IV
A051584(n) = if((n%6),0,A051585(n/6)); \\ Antti Karttunen, Aug 23 2019

Extensions

Data section extended up to a(120) by Antti Karttunen, Aug 23 2019

A051586 There are exactly n integer-sided triangles of area a(n).

Original entry on oeis.org

6, 12, 126, 60, 240, 210, 1080, 336, 1260, 924, 2016, 2640, 5280, 11424, 420, 2520, 840, 5544, 3696, 14280, 3360, 25200, 1680, 6720, 10920, 26880, 15960, 137280, 23100, 43680, 64680, 21840, 32760, 24024, 13860, 7560, 68040, 49140, 262080, 60480, 73920, 133056, 207900
Offset: 1

Views

Author

David W. Wilson

Keywords

Comments

If integer-sided triangle has integer area, area is divisible by 6.

Links

Crossrefs

Cf. A051584, A051585.

Extensions

More terms from Giovanni Resta, Nov 04 2019

A227003 Number of primitive Heronian triangles with area 6n.

Original entry on oeis.org

1, 2, 0, 1, 1, 2, 1, 0, 0, 4, 1, 1, 0, 4, 2, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 2, 0, 3, 0, 2, 0, 0, 1, 1, 6, 1, 0, 0, 1, 1, 0, 3, 0, 2, 1, 0, 0, 1, 0, 2, 1, 0, 0, 0, 4, 4, 0, 0, 0, 3, 0, 0, 0, 0, 1, 3, 0, 1, 0, 15, 0, 0, 0, 0, 0, 3, 2, 1, 0, 1, 0, 0, 0, 3, 2, 0, 1, 2, 0, 0
Offset: 1

Views

Author

Frank M Jackson, Jun 26 2013

Keywords

Comments

The Mathematica program captures all primitive Heronian areas up to 540 by searching through integer triangles with a longest side ranging from 3 to at least 484. This upper limit for the longest side is determined by observing that the shortest side of a Heronian triangle is >= 3 and the smallest area of an integer triangle with longest side z and shortest side 3 is generated by the integer triple (3, z-2, z).

Examples

			a(10) = 4 as there are 4 primitive Heronian triangles with area 60. The triples are (10,13,13), (8,15,17), (13,13,24), (6,25,29).

Crossrefs

Cf. A051585, A083875.

Programs

  • Mathematica
    nn=540; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s]&&GCD[a, b, c]==1, area2=s(s-a)(s-b)(s-c); If[area2>0&&IntegerQ[Sqrt[area2]], AppendTo[lst, Sqrt[area2]]]], {a, 3, nn}, {b, a}, {c, b}]; lst1=Sort@lst/6; Table[Length@Select[lst1, n==# &], {n, 1, nn/6}] (* using T. D. Noe's program A083875 *)
  • PARI
    a(n)=sum(z=sqrtint(sqrtint(192*n^2)-1)+1,sqrtint(9*(64*n^2+5)\20), sum(y=z\2+1,z, my(t=(y*z)^2-(12*n)^2,x,g=gcd(y,z)); if(issquare(t,&t), (issquare(y^2+z^2-2*t,&x) && gcd(x,g)==1 && x<=y) + (t && issquare(y^2+z^2+2*t,&x) && gcd(x,g)==1 && x<=y), 0))) \\ Charles R Greathouse IV, Jun 27 2013

A301384 Number of integer-sided triangles of area A188158(n).

Original entry on oeis.org

1, 2, 2, 1, 2, 1, 2, 1, 4, 1, 1, 4, 2, 2, 2, 1, 4, 3, 1, 2, 1, 2, 4, 2, 2, 1, 1, 6, 3, 1, 5, 3, 3, 2, 2, 1, 4, 1, 2, 4, 8, 5, 1, 2, 1, 3, 1, 15, 2, 4, 2, 1, 5, 1, 6, 2, 1, 3, 4, 3, 1, 2, 2, 1, 2, 4, 5, 1, 5, 4, 1, 2, 3, 5, 1, 1, 1, 6, 2, 4, 2, 1, 2, 1, 17, 4, 1
Offset: 1

Views

Author

Michel Lagneau, Mar 20 2018

Keywords

Comments

Nonzero terms of A051584.

Examples

			a(9) = 4 because A188158(9) = 60 corresponding to 4 triangles (a, b, c) = (6, 25, 29), (8, 15, 17), (10, 13, 13) and (13, 13, 24) of area 60.

Links

Crossrefs

Cf. A188158, A024153 (distinct side lengths), A051516, A051584, A051585.

Programs

  • Maple
    A301384 := proc(A::integer)
    local Asqr, s,a,b,c,sol ;
    sol := 0 ;
    Asqr := A^2 ;
    for s in numtheory[divisors](Asqr) do
        if s^2> A then
        for a from 1 to s-1 do
            if modp(Asqr,s-a) = 0 then
                for b from a to s-1 do
                    c := 2*s-a-b ;
                    if s*(s-a)*(s-b)*(s-c) = Asqr and c >= b then
                        sol := sol+1 ;
                    end if;
                end do:
            end if;
        end do:
        end if;
    end do:
    sol ;
end proc:
for n from 3 to 600 do
    a301384 := A301384(n) ;
    if a301384 > 0 then
        printf("%d,\n",a301384) ;
    end if;
end do: # R. J. Mathar, May 02 2018
  • Mathematica
    nn=1000;lst={};lst2={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s(s-a)(s-b)(s-c);If[0
Showing 1-5 of 5 results.

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