A203613 - OEIS

%I #10 Mar 30 2012 18:53:49

%S 55,85,91,115,133,145,187,195,204,205,217,235,247,253,259,265,275,285,

%T 295,301,319,351,355,357,385,391,403,415,425,427,445,451,465,469,476,

%U 481,483,493,505,511,517,535,553,555,559,565,575,583,589,595,609,621,637

%N For any number n take the polynomial formed by the product of the terms (x-pi), where pi’s are the prime factors of n. Then calculate the area between the minimum and the maximum value of the prime factors. This sequence lists the numbers for which the area is a negative integer.

%H Paolo P. Lava, <a href="/A203613/b203613.txt">Table of n, a(n) for n = 1..10000</a>

%e n=217. Prime factors: 7, 31: min(pi)=7, max(pi)=31. Polynomial: (x-7)*(x-31)=x^2-38*x+217. Integral: x^3/3-19*x^2+217*x. The area from x=7 to x=31 is -2304.

%e n=53151. Prime factors: 3, 7, 2531: min(pi)=3, max(pi)=2531. Polynomial: (x-7)*(x-349)*(x-409)=x^3-2541*x^2+25331*x-53151. Integral: x^4/4-847*x^3+25331/2*x^2-53151*x. The area from x=3 to x=2531 is - 3392739409920.

%p with(numtheory);

%p P:=proc(i)

%p local a,b,c,d,k,m,m1,m2,n,p;

%p for k from 1 to i do

%p a:=ifactors(k)[2]; b:=nops(a); c:=op(a); d:=1;

%p if b>1 then

%p    m1:=c[1,1]; m2:=0;

%p    for n from 1 to b do

%p      for m from 1 to c[n][2] do d:=d*(x-c[n][1]); od;

%p      if c[n,1]<m1 then m1:=c[n,1]; fi; if c[n,1]>m2 then m2:=c[n,1]; fi;

%p    od;

%p    p:=int(d,x=m1..m2); if (trunc(p)=p and p<0) then print(k); fi;

%p fi;

%p od;

%p end:

%p P(500000);

%Y Cf. A203612, A203614.

%K nonn

%O 1,1

%A _Paolo P. Lava_, Jan 05 2012