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%I #11 Feb 18 2019 02:10:57
%S 5,11,23,59,71,89,211,239,269,349,389,419,431,467,479,521,571,577,647,
%T 863,983,1087,1213,1223,1733,1747,1759,1933,1949,1973,2131,2297,2411,
%U 2521,2659,2879,2909,2999,3011,3191,3203,3209,3391,3467,3469,3517,3559
%N A triangle with three consecutive primes as sides has an area that is a prime after rounding. The sequence gives the first of the three consecutive primes.
%F Simply use the formula for the area of a triangle given the three sides.
%e For sides 5,7,11 the formula gives 12.96 and with rounding this becomes 13, a prime.
%p s:=proc(n) local a,b,c,p,A: a:=ithprime(n): b:=ithprime(n+1): c:=ithprime(n+2): p:=(a+b+c)/2: A:=sqrt(p*(p-a)*(p-b)*(p-c)): if isprime(round(A))=true then a else fi end: seq(s(n),n=1..700); # _Emeric Deutsch_, May 25 2007
%p Digits := 60 : isA106171 := proc(p) local q,r,s,area ; if isprime(p) then q := nextprime(p) ; r := nextprime(q) ; s := (p+q+r)/2 ; area := round(sqrt(s*(s-p)*(s-q)*(s-r))) ; RETURN(isprime(area)) ; else false ; fi ; end: for n from 1 to 900 do p := ithprime(n) : if isA106171(p) then printf("%d,",p) ; fi ; od : # _R. J. Mathar_, Jun 08 2007
%K nonn
%O 1,1
%A _J. M. Bergot_, May 19 2007
%E More terms from _Emeric Deutsch_ and _R. J. Mathar_, May 25 2007
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