A334763 - OEIS

%I #10 Jun 13 2020 01:45:05

%S 3,4,5,6,7,9,10,14,15,19,21,24,26,29,31,34,37,40,43,45,48,52,55,58,60,

%T 63,68,72,77,80,84,87,93,99,103,104,107,110,115,118,123,126,131,134,

%U 138,139,142,146,149,153,158,168,176,182,185,190,194,200,204,208

%N Ceiling of circumradius of triangle whose sides are consecutive Ulam numbers (A002858).

%C It has been proved that three consecutive Ulam numbers U(n) for n > 1 satisfy the triangle inequality. See Wikipedia link below. Consequently it is possible to create n-gons using n consecutive Ulam numbers. The sequence starts at offset 2 because using the first Ulam number generates a triangle with sides (1,2,3) that is degenerate with infinite circumradius.

%C Conjecture: Triangles whose sides are consecutive Ulam numbers are acute apart from (1,2,3), (2,3,4), (3,4,6), (4,6,8), (6,8,11) and (16,18,26).

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Circumradius.html">Circumradius</a>.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/UlamSequence.html">Ulam Sequence</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ulam_number">Ulam number</a>.

%F Circumradius of a triangle with sides a, b, c is given by R = a*b*c/(4A) where the Area A is given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) and where s = (a+b+c)/2.

%e a(2)=3 because a triangle with sides 2,3,4 has area = (1/4)*sqrt((2+3+4)(2+3-4)(2-3+4)(-2+3+4)) = 2.904... and circumradius = 2*3*4/(4A) = 2.065...

%t lst1=ReadList["https://oeis.org/A002858/b002858.txt", {Number, Number}]; lst={}; Do[{a, b, c}={lst1[[n]][[2]], lst1[[n+1]][[2]], lst1[[n+2]][[2]]}; s=(a+b+c)/2; A=Sqrt[s(s-a)(s-b)(s-c)]; R=a*b*c/(4 A); AppendTo[lst, Ceiling@R], {n, 2, 100}]; lst

%Y Cf. A002858, A331676.

%K nonn

%O 2,1

%A _Frank M Jackson_, May 10 2020