login
A206442
Number of distinct irreducible factors of the polynomial p(n,x) defined at A206284.
5
0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 0, 3, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 3, 1, 1, 1, 2, 2, 0, 3, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 1, 1, 2, 1, 4, 2, 3, 1, 1, 1, 2
OFFSET
1,14
COMMENTS
The factorization is over the ring of polynomials having integer coefficients.
From Robert Israel, Oct 09 2016: (Start)
a(n) = 0 iff n is a power of 2.
a(n) <= A061395(n)-1 for n > 1. (End)
LINKS
EXAMPLE
From Antti Karttunen, Oct 09 2016: (Start)
For n = 1, the corresponding polynomial is zero-polynomial, thus a(1) = 0.
For n = 2, the corresponding polynomial is constant 1, thus a(2) = 0.
For n = 3 = prime(2), the corresponding polynomial is x, thus a(3) = 1.
For n = 11 = prime(5), the corresponding polynomial is x^4 which factorizes as (x)(x)(x)(x), thus a(11) = 1. (Only distinct factors are counted by this sequence).
For n = 14 = prime(4) * prime(1), the corresponding polynomial is x^3 + 1, which factorizes as (x + 1)(x^2 - x + 1), thus a(14) = 2.
For n = 33 = prime(5) * prime(2), the corresponding polynomial is x^4 + x, which factorizes as x(x+1)(x^2 - x + 1), thus a(33) = 3.
For n = 90 = prime(3) * prime(2)^2 * prime(1), the corresponding polynomial is x^2 + 2x + 1, which factorizes as (x + 1)^2, thus a(90) = 1.
For n = 93 = prime(11) * prime(2), the corresponding polynomial is x^10 + x, which factorizes as x(x+1)(x^2 - x + 1)(x^6 - x^3 + 1), thus a(93) = 4.
For n = 177 = prime(17) * prime(2), the corresponding polynomial is x^16 + x, which factorizes as x(x + 1)(x^2 - x + 1)(x^4 - x^3 + x^2 - x + 1)(x^8 + x^7 - x^5 - x^4 - x^3 + x + 1), thus a(177) = 5.
(End)
MAPLE
P:= n -> add(f[2]*x^(numtheory:-pi(f[1])-1), f = ifactors(n)[2]):
seq(nops(factors(P(n))[2]), n=1..200); # Robert Israel, Oct 09 2016
MATHEMATICA
b[n_] := Table[x^k, {k, 0, n}];
f[n_] := f[n] = FactorInteger[n]; z = 1000;
t[n_, m_, k_] := If[PrimeQ[f[n][[m, 1]]] && f[n][[m, 1]] == Prime[k], f[n][[m, 2]], 0];
u = Table[Apply[Plus,
Table[Table[t[n, m, k], {k, 1, PrimePi[n]}], {m, 1,
Length[f[n]]}]], {n, 1, z}];
p[n_, x_] := u[[n]].b[-1 + Length[u[[n]]]]
TableForm[Table[{n, FactorInteger[n],
p[n, x], -1 + Length[FactorList[p[n, x]]]},
{n, 1, z/4}]]
Table[-1 + Length[FactorList[p[n, x]]], {n, 1, z/4}]
(* A206442 *)
PROG
(PARI)
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f)};
pfps(n) = if(1==n, 0, if(!(n%2), 1 + pfps(n/2), 'x*pfps(A064989(n))));
A206442 = n -> if(!bitand(n, (n-1)), 0, #(factor(pfps(n))~));
\\ Alternatively, one may use the version of pfps given by Charles R Greathouse IV in A277322:
pfps(n)=my(f=factor(n)); sum(i=1, #f~, f[i, 2] * 'x^(primepi(f[i, 1])-1));
\\ In which case this version of the "main function" should suffice:
A206442 = n -> if(1==n, 0, #(factor(pfps(n))~));
\\ Antti Karttunen, Oct 09 2016
CROSSREFS
Cf. also A277322 (gives the number of irreducible polynomial factors with multiplicity).
Sequence in context: A294624 A290531 A376629 * A137581 A156311 A270740
KEYWORD
nonn
AUTHOR
Clark Kimberling, Feb 07 2012
EXTENSIONS
Example section rewritten by Antti Karttunen, Oct 09 2016
STATUS
approved