A277322 a(n) = number of irreducible polynomial factors (counted with multiplicity) in the polynomial (with nonnegative integral coefficients) constructed from the prime factorization of n.

0, 0, 1, 0, 2, 1, 3, 0, 1, 1, 4, 1, 5, 2, 2, 0, 6, 1, 7, 1, 2, 1, 8, 1, 2, 2, 1, 1, 9, 1, 10, 0, 3, 2, 3, 1, 11, 2, 2, 1, 12, 1, 13, 1, 2, 1, 14, 1, 3, 1, 3, 1, 15, 1, 3, 1, 3, 3, 16, 1, 17, 2, 2, 0, 4, 1, 18, 1, 3, 1, 19, 1, 20, 2, 2, 1, 4, 2, 21, 1, 1, 2, 22, 2, 3, 2, 2, 1, 23, 2, 4, 1, 4, 2, 4, 1, 24, 1, 2, 1, 25, 1, 26, 1, 2

Offset: 1

Author

Antti Karttunen, Oct 09 2016

Keywords

• nonn

Comments

Let p(n,x) be the completely additive polynomial-valued function such that p(prime(n),x) = x^(n-1) as defined by Clark Kimberling in A206284. Then this sequence is the number of irreducible factors in p(n,x), counted with multiplicity.

Examples

			For n = 7 = prime(4), the corresponding polynomial is x^3, which factorizes as (x)(x)(x), thus a(7) = 3.
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 = 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) = 2.
pfps(660) = pfps(2^2*3*5*11) = pfps(2^2) + pfps(3) + pfps(5) + pfps(11) = 2 + x + x^2 + x^4 which is irreducible, so a(660) = 1.
For n = 30030 = Product_{i=1..6} prime(i), the corresponding polynomial is x^5 + x^4 + x^3 + x^2 + x + 1, which factorizes as (x+1)(x^2 - x + 1)(x^2 + x + 1), thus a(30030) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..30033

Crossrefs

Cf. A206442 (gives the number of irreducible polynomial factors without multiplicity), A206284 (positions of 1's, i.e., irreducible polynomials).

Cf. A000040, A007188, A064989, A260443, A277013.

Programs

PARI

allocatemem(2^29);
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))));
A277322 = n -> if(!bitand(n,(n-1)), 0, vecsum(factor(pfps(n))[,2]));
for(n=1, 121121, write("b277322.txt", n, " ", A277322(n)));

PARI

pfps(n)=my(f=factor(n)); sum(i=1,#f~, f[i,2] * 'x^(primepi(f[i,1])-1))
A277322(n) = if(1==n, 0, vecsum(factor(pfps(n))[, 2])); \\ Charles R Greathouse IV, test for one added by Antti Karttunen, Oct 09 2016

Formula

a(2^n) = 0. [By an explicit convention.]

a(A000040(n)) = n-1.

a(A007188(n)) = n.

a(A260443(n)) = A277013(n).