A203614 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 equal to zero.

105, 140, 231, 627, 748, 750, 897, 935, 1470, 1581, 1729, 2205, 2465, 2625, 2967, 3404, 3549, 4123, 4301, 4715, 5452, 5487, 6256, 7623, 7685, 7881, 9009, 9717, 10707, 10829, 10988, 11319, 11339, 13310, 14993, 15470, 16377, 17353, 17457, 17901, 20213, 20915

Offset: 1

Author

Paolo P. Lava, Jan 05 2012

Keywords

• nonn

Comments

Prime numbers are excluded because are banal solutions: in fact for them min(pi)=max(pi)=pi and then the area is zero.

Any squarefree number with an odd number of prime factors which are symmetrically distributed around the central one is part of the sequence. For instance with n=53295 the prime factors are 3, 5, 11, 17, 19 and 3+8=11=19-8, 5+6=11=17-6.

Examples

			n=140. Prime factors: 2, 2, 5, 7: min(pi)=2, max(pi)=7. Polynomial to integrate from 2 to 7: (x-2)^2*(x-5)*(x-7)=x^4-16*x^3+87*x^2-188x+140. The resulting area is equal to zero.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A203612, A203613.

Programs

Maple

with(numtheory);
P:=proc(i)
local a,b,c,d,k,m,m1,m2,n;
for k from 1 to i do
a:=ifactors(k)[2]; b:=nops(a); c:=op(a); d:=1;
if b>1 then
   m1:=c[1,1]; m2:=0;
   for n from 1 to b do
     for m from 1 to c[n][2] do d:=d*(x-c[n][1]); od;
     if c[n,1]m2 then m2:=c[n,1]; fi;
   od;
   if int(d,x=m1..m2)=0 then print(k); fi;
fi;
od;
end:
P(500000);