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.

A120062 Number of triangles with integer sides a <= b <= c having integer inradius n.

Original entry on oeis.org

1, 5, 13, 18, 15, 45, 24, 45, 51, 52, 26, 139, 31, 80, 110, 89, 33, 184, 34, 145, 185, 103, 42, 312, 65, 96, 140, 225, 36, 379, 46, 169, 211, 116, 173, 498, 38, 123, 210, 328, 44, 560, 60, 280, 382, 134, 64, 592, 116, 228, 230, 271, 47, 452, 229, 510, 276, 134, 54
Offset: 1

Views

Author

Hugo Pfoertner, Jun 11 2006

Keywords

Comments

It is conjectured that the longest possible side c of a triangle with integer sides and inradius n is given by A057721(n) = n^4 + 3*n^2 + 1.
For n >= 1, a(n) >= 1 because triangle (a, b, c) = (n^2 + 2, n^4 + 2*n^2 + 1, n^4 + 3*n^2 + 1) has inradius n. - David W. Wilson, Jun 17 2006
Previous name was "Number of triangles with integer sides a<=bA362669); so, now effectively, a(10) = 52. - Bernard Schott, Apr 24 2023

Examples

			a(1)=1: {3,4,5} is the only triangle with integer sides and inradius 1.
a(2)=5: {5,12,13}, {6,8,10}, {6,25,29}, {7,15,20}, {9,10,17} are the only triangles with integer sides and inradius 2.
a(4)=A120252(1)+A120252(2)+A120252(4)=1+4+13 because 1, 2 and 4 are the factors of 4. The 1 primitive triangle with inradius n=1 is (3,4,5). The 4 primitive triangles with n=2 are (5,12,13), (9,10,17), (7,15,20), (6,25,29). The 13 primitive triangles with n=4 are (13,14,15), (15,15,24), (11,25,30), (15,26,37), (10,35,39), (9,40,41), (33,34,65), (25,51,74), (9,75,78), (11,90,97), (21,85,104), (19,153,170), (18,289,305). (Primitive means GCD(a, b, c, n)=1.)
		

Crossrefs

Cf. A078644 [Pythagorean triangles with inradius n], A057721 [n^4+3*n^2+1].
Let S(n) be the set of triangles with integer sides a<=b<=c and inradius n. Then:
A120062(n) gives number of triangles in S(n).
A120261(n) gives number of triangles in S(n) with gcd(a, b, c) = 1.
A120252(n) gives number of triangles in S(n) with gcd(a, b, c, n) = 1.
A005408(n) = 2n+1 gives shortest short side a of triangles in S(n).
A120064(n) gives shortest middle side b of triangles in S(n).
A120063(n) gives shortest long side c of triangles in S(n).
A120570(n) gives shortest perimeter of triangles in S(n).
A120572(n) gives smallest area of triangles in S(n).
A058331(n) = 2n^2+1 gives longest short side a of triangles in S(n).
A082044(n) = n^4+2n^2+1 gives longest middle side b of triangles in S(n).
A057721(n) = n^4+3n^2+1 gives longest long side c of triangles in S(n).
A120571(n) = 2n^4+6n^2+4 gives longest perimeter of triangles in S(n).
A120573(n) = gives largest area of triangles in S(n).
Cf. A120252 [primitive triangles with integer inradius], A120063 [minimum of longest sides], A057721 [maximum of longest sides], A120064 [minimum of middle sides], A082044 [maximum of middle sides], A005408 [minimum of shortest sides], A058331 [maximum of shortest sides], A007237 [number of triangles with integer sides and area = n times perimeter].

Programs

  • Mathematica
    (* See link above. *)

Formula

The even-numbered terms are given by a(2*n)=A007237(n).
a(n) = Sum_{k|n} A120252(k).

Extensions

More terms from Graeme McRae and Hugo Pfoertner, Jun 12 2006
Name corrected by Bernard Schott, Apr 24 2023

A289156 Largest area of triangles with integer sides and area = n times perimeter.

Original entry on oeis.org

60, 1224, 8436, 34320, 103020, 254040, 546084, 1060896, 1907100, 3224040, 5185620, 8004144, 11934156, 17276280, 24381060, 33652800, 45553404, 60606216, 79399860, 102592080, 130913580, 165171864, 206255076, 255135840, 312875100, 380625960, 459637524, 551258736
Offset: 1

Views

Author

Zhining Yang, Jun 26 2017

Keywords

Examples

			For n = 4, a(4) = 34320 means for the largest triangles (a,b,c) = (66,4225,4289), the area is 34320 which is 4 times the perimeter 8580.
		

Crossrefs

Programs

  • Mathematica
    Table[4 n (2 n^2 + 1) (4 n^2 + 1), {n, 27}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {60, 1224, 8436, 34320, 103020, 254040}, 27] (* or *) Rest@ CoefficientList[Series[12 x (5 + 72 x + 166 x^2 + 72 x^3 + 5 x^4)/(1 - x)^6, {x, 0, 27}], x] (* Michael De Vlieger, Jul 03 2017 *)
  • PARI
    Vec(12*x*(5 + 72*x + 166*x^2 + 72*x^3 + 5*x^4)/(1 - x)^6 + O(x^30)) \\ Colin Barker, Jun 28 2017

Formula

From Colin Barker, Jun 28 2017: (Start)
G.f.: 12*x*(5 + 72*x + 166*x^2 + 72*x^3 + 5*x^4)/(1 - x)^6.
a(n) = 4*n*(2*n^2 + 1)*(4*n^2 + 1).
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>6. (End)
a(n) = A120573(2*n). - Ray Chandler, Jul 27 2017
From Elmo R. Oliveira, Sep 01 2025: (Start)
E.g.f.: 4*exp(x)*x*(15 + 138*x + 206*x^2 + 80*x^3 + 8*x^4).
a(n) = 12*A005900(n)*A053755(n) = A053755(n)*A007900(n)/2. (End)
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