%I #6 Oct 01 2013 17:58:26
%S 4,6,9,12,14,18,19,25,30,32,38,41,43,48,54,59,61,67,71,73,79,83,90,97,
%T 101,103,107,109,114,128,131,137,139,149,151,157,163,167,173,179,181,
%U 191,193,197,199,212,223,227,229,233,239,241,251,257,263,269,271,277
%N Floor of the slanted side of a right trapezoid formed by 3 consecutive primes.
%C It is easy to prove that for any 3 consecutive primes p1,p2,p3, a(n) always lies between p2 and p3.
%e For the first 3 prime numbers, go up 2, go right 3 and go down 5. Connecting the figure to form a right trapezoid we have the slanted side = sqrt(18). The integer part of this is 4, the first term of the sequence.
%o (PARI) g(n) = { for(x=1,n,p1=prime(x);p2=prime(x+1);p3=prime(x+2);y=p3p1; print1(floor(sqrt(p2^2+y^2)),",")) )
%K easy,nonn
%O 1,1
%A _Cino Hilliard_, Sep 20 2006
