A206442 Number of distinct irreducible factors of the polynomial p(n,x) defined at A206284.
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
Keywords
Examples
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)
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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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
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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 *) -
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
Extensions
Example section rewritten by Antti Karttunen, Oct 09 2016
Comments