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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.
9
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
1,3
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 >= ...
LINKS
FORMULA
a(2^n) = 0. [By an explicit convention.]
Other identities. For all n >= 1:
A284011(n) = a(A260443(n)).
EXAMPLE
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.
PROG
(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])); }
CROSSREFS
Cf. A046523, A206284 (positions of 2's), A206442, A277322, A284011, A284012.
Cf. also A260443, A278233, A278243.
Sequence in context: A163123 A194346 A328598 * A278082 A327442 A068773
KEYWORD
nonn
AUTHOR
Antti Karttunen, Mar 20 2017
STATUS
approved