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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. 9

%I #12 Mar 30 2012 18:53:50

%S 105,140,231,627,748,750,897,935,1470,1581,1729,2205,2465,2625,2967,

%T 3404,3549,4123,4301,4715,5452,5487,6256,7623,7685,7881,9009,9717,

%U 10707,10829,10988,11319,11339,13310,14993,15470,16377,17353,17457,17901,20213,20915

%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 equal to zero.

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

%C 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.

%H Paolo P. Lava, <a href="/A203614/b203614.txt">Table of n, a(n) for n = 1..1000</a>

%e 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.

%p with(numtheory);

%p P:=proc(i)

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

%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 if int(d,x=m1..m2)=0 then print(k); fi;

%p fi;

%p od;

%p end:

%p P(500000);

%Y Cf. A203612, A203613.

%K nonn

%O 1,1

%A _Paolo P. Lava_, Jan 05 2012

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