

A334763


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


0



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, 63, 68, 72, 77, 80, 84, 87, 93, 99, 103, 104, 107, 110, 115, 118, 123, 126, 131, 134, 138, 139, 142, 146, 149, 153, 158, 168, 176, 182, 185, 190, 194, 200, 204, 208
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OFFSET

2,1


COMMENTS

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 ngons 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.
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).


LINKS

Table of n, a(n) for n=2..61.
Eric Weisstein's World of Mathematics, Circumradius.
Eric Weisstein's World of Mathematics, Ulam Sequence.
Wikipedia, Ulam number.


FORMULA

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(sa)(sb)(sc)) and where s = (a+b+c)/2.


EXAMPLE

a(2)=3 because a triangle with sides 2,3,4 has area = (1/4)*sqrt((2+3+4)(2+34)(23+4)(2+3+4)) = 2.904... and circumradius = 2*3*4/(4A) = 2.065...


MATHEMATICA

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(sa)(sb)(sc)]; R=a*b*c/(4 A); AppendTo[lst, Ceiling@R], {n, 2, 100}]; lst


CROSSREFS

Cf. A002858, A331676.
Sequence in context: A298006 A026438 A026442 * A307712 A048869 A039051
Adjacent sequences: A334760 A334761 A334762 * A334764 A334765 A334766


KEYWORD

nonn


AUTHOR

Frank M Jackson, May 10 2020


STATUS

approved



