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.

A229926 Integer areas of the integer-sided triangles T(n) defined by the property: a(0) = 6 ; for n > 0, a(n) is the area A where the smallest side of T(n) is the greatest side of T(n-1).

Original entry on oeis.org

6, 12, 24, 48, 96, 192, 384, 768, 1080, 1080, 3888, 4320, 15360, 69120, 69120, 248832, 349920, 349920, 1259712, 342144, 7226112, 10782720, 17031168, 18095616, 19226592, 21660210, 30270240, 44706816, 81544320, 128798208
Offset: 0

Views

Author

Michel Lagneau, Oct 03 2013

Keywords

Comments

Subsequence of A188158.
The sequence of the common sides is {5, 6, 10, 12, 20, 24, 40, 48, 51, 90, 108, 208, 384, 408, 720, 864, 918, 1620, 1944, 3880, 4656, 6240, 6336, ...}
a(n) = 6*2^n for n = 0, 1, 2,..., 7, and then this property disappears.
The area is given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) where the semiperimeter s = (a + b + c)/2.
The following table gives the first values (n, A, a, b, c) where a <= b <= c are the integer sides of the triangles.
+----+------+-----+-----+-----+
| n | A | a | b | c |
+----+------+-----+-----+-----+
| 0 | 6 | 3 | 4 | 5 |
| 1 | 12 | 5 | 5 | 6 |
| 2 | 24 | 6 | 8 | 10 |
| 3 | 48 | 10 | 10 | 12 |
| 4 | 96 | 12 | 16 | 20 |
| 5 | 192 | 20 | 20 | 24 |
| 6 | 384 | 24 | 32 | 40 |
| 7 | 768 | 40 | 40 | 48 |
| 8 | 1080 | 48 | 51 | 51 |
| 9 | 1080 | 51 | 51 | 90 |
| 10 | 3888 | 90 | 90 | 108 |
| 11 | 4320 | 108 | 116 | 208 |
+----+------+-----+-----+-----+

Crossrefs

Cf. A188158.

Programs

  • Maple
    with(numtheory):nn:=15000:a:=5: printf ( "%d %d %d %d %d \n",1,6,3,4,a):
        for n from 2 to 40 do:
           ii:=0:
          for b from a to nn while(ii=0) do:
            for c from b to nn while(ii=0)  do:
              p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c):
              if x>0
              then
              x0:= sqrt(x):
              else
              fi:
               if x0=floor(x0)
               then
               ii:=1:printf ( "%d %d %d %d %d \n",n,x0,a,b,c):
               a:=max(b,c):
               else
               fi:
              od:
            od:
           od: