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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 (list; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A030612 A371921 A327528 * A370645 A340596 A340654
KEYWORD
nonn
AUTHOR
Gionata Neri, Nov 26 2015
STATUS
approved

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Last modified April 23 22:36 EDT 2024. Contains 371917 sequences. (Running on oeis4.)