A206719 Number of distinct irreducible factors of the polynomial p(n,x) defined at A206073.
0, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 1, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 2, 3, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 3, 1, 2, 3, 1, 2, 1, 3, 1, 3, 1, 2, 3, 2, 1, 3, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 1, 4, 1, 2, 2, 2, 2, 2, 1, 3, 2
Offset: 1
Keywords
Examples
p(1,n) = 1, so a(1)=0 p(2,n) = x, so a(2)=1 p(6,n) = x(1+x), so a(6)=2 p(18,n) = x(x+1)(1-x+x^2), so a(18)=3 p(90,n) = x(1+x)(1+x^2)(1-x+x^2), so a(90)=4
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
Programs
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Mathematica
t = Table[IntegerDigits[n, 2], {n, 1, 1000}]; b[n_] := Reverse[Table[x^k, {k, 0, n}]] p[n_, x_] := p[n, x] = t[[n]].b[-1 + Length[t[[n]]]] TableForm[Table[{n, p[n, x], FactorList[p[n, x]], -1 + Length[FactorList[p[n, x]]]}, {n, 1, 9}]] Table[Length[FactorList[p[n, x]]], {n, 1, 120}]
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PARI
A206719(n) = { my(f = factor(Pol(binary(n)))); (#f~); }; \\ Antti Karttunen, Dec 16 2017
Comments