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A203614
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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.
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9
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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Paolo P. Lava, Table of n, a(n) for n = 1..1000
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EXAMPLE
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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.
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MAPLE
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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]<m1 then m1:=c[n, 1]; fi; 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);
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CROSSREFS
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Cf. A203612, A203613.
Sequence in context: A115935 A069702 A239589 * A252069 A133509 A013590
Adjacent sequences: A203611 A203612 A203613 * A203615 A203616 A203617
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KEYWORD
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nonn
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AUTHOR
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Paolo P. Lava, Jan 05 2012
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STATUS
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approved
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