

A264857


Number of values of k, n included, such that the distinct prime divisors of k are the same as those of n and Omega(k) = Omega(n), where Omega(n) is the number of prime divisors of n with multiplicity (A001222); a(1) = 1.


1



1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 1, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 2, 1, 1, 1, 5, 1, 2, 2, 3
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OFFSET

1,12


COMMENTS

a(n) > 1 if 1 < omega(n) < Omega(n), where omega(n) is the number of distinct prime divisors of n (A001221).
a(n) = 1 if n is a squarefree number (A005117) or a prime power p^e where p is a prime and e >= 2 (A246547).


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = C(A001222(n)1, A001221(n)1).  Robert Israel, Nov 27 2015


EXAMPLE

The numbers k with the same prime divisors 2 and 3 are (6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...) and those for which A001222(k) = 3 are (12, 18), so a(12) = a(18) = 2.
The numbers k with the same prime divisor 5 are (5, 25, 125, 625, 3125, ...) and those for which A001222(k) = 4 are (625), so a(625) = 1.


MAPLE

seq(binomial(numtheory:bigomega(n)1, nops(numtheory:factorset(n))1) , n=1..200); # Robert Israel, Nov 27 2015


MATHEMATICA

Table[Binomial[PrimeOmega@ n  1, PrimeNu@ n  1], {n, 100}] (* Michael De Vlieger, Nov 28 2015 *)


PROG

(PARI) a(n) = if(n==1, 1, binomial(bigomega(n)1, omega(n)1)) \\ Altug Alkan, Dec 04 2015


CROSSREFS

Cf. A001221, A001222.
Sequence in context: A276088 A030612 A327528 * A303837 A286520 A320105
Adjacent sequences: A264854 A264855 A264856 * A264858 A264859 A264860


KEYWORD

nonn


AUTHOR

Gionata Neri, Nov 26 2015


STATUS

approved



