A284010 a(n) = least natural number with the same prime signature polynomial p(n,x) has when it is factored over Z. Polynomial p(n,x) has only nonnegative integer coefficients that are encoded in the prime factorization of n.

0, 0, 2, 0, 4, 2, 8, 0, 2, 2, 16, 2, 32, 6, 6, 0, 64, 2, 128, 2, 6, 2, 256, 2, 4, 6, 2, 2, 512, 2, 1024, 0, 30, 6, 12, 2, 2048, 6, 6, 2, 4096, 2, 8192, 2, 6, 2, 16384, 2, 8, 2, 30, 2, 32768, 2, 12, 2, 30, 30, 65536, 2, 131072, 6, 6, 0, 60, 2, 262144, 2, 30, 2, 524288, 2, 1048576, 6, 6, 2, 24, 6, 2097152, 2, 2, 6, 4194304, 6, 12, 6, 6, 2, 8388608, 4, 24, 2, 210

Offset: 1

Author

Antti Karttunen, Mar 20 2017

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. To compute a(n), factor p(n,x) over Z and collect the exponents of its irreducible polynomial factors using them as exponents of primes (in Z) as 2^e1 * 3^e2 * 5^e3 * ..., with e1 >= e2 >= e3 >= ...

Examples

			For n = 7 = prime(4), the corresponding polynomial is x^3, which factorizes as (x)(x)(x), thus a(7) = 2^3 = 8.
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) = 2^1 * 3^1 = 6.
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) = 2^2 = 4.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8192

Crossrefs

Cf. A046523, A206284 (positions of 2's), A206442, A277322, A284011, A284012.

Cf. also A260443, A278233, A278243.

Programs

PARI

\\ After Charles R Greathouse IV's code in A046523 and A277322:
pfps(n) = { my(f=factor(n)); sum(i=1, #f~, f[i, 2] * 'x^(primepi(f[i, 1])-1)); };
A284010(n) = { if(!bitand(n, (n-1)), 0, my(p=0, f=vecsort(factor(pfps(n))[, 2], ,4)); prod(i=1, #f, (p=nextprime(p+1))^f[i])); }

Formula

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

Other identities. For all n >= 1:

A284011(n) = a(A260443(n)).