[go: up one dir, main page]

login
A070014
Ceiling of number of prime factors of n divided by the number of n's distinct prime factors.
6
1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 3, 4, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 3, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1
OFFSET
2,3
COMMENTS
a(n) is the ceiling of the average of the exponents in the prime factorization of n.
FORMULA
a(n) = ceiling(bigomega(n)/omega(n)) for n>=2.
EXAMPLE
a(12) = 2 because 12 = 2^2 * 3^1 and ceiling(bigomega(12)/omega(12)) = ceiling((2+1)/2) = 2. a(36) = 2 because 36 = 2^2 * 3^2 and ceiling(bigomega(36)/omega(36)) = ceiling((2+2)/2) = 2. a(60) = 2 because 60 = 2^2 * 3^1 * 5^1 and ceiling(bigomega(60)/omega(60)) = ceiling((2+1+1)/3) = 2. 36 is in A067340. 12 and 60 are in A070011.
MATHEMATICA
Table[Ceiling[PrimeOmega[n]/PrimeNu[n]], {n, 2, 106}] (* Michael De Vlieger, Jul 12 2017 *)
PROG
(PARI) v=[]; for(n=2, 150, v=concat(v, ceil(bigomega(n)/omega(n)))); v
(Scheme) (define (A070014 n) (let ((a (A001222 n)) (b (A001221 n))) (if (zero? (modulo a b)) (/ a b) (+ 1 (/ (- a (modulo a b)) b))))) ;; Antti Karttunen, Jul 12 2017
(Python)
from sympy import primefactors, ceiling
def bigomega(n): return 0 if n==1 else bigomega(n//primefactors(n)[0]) + 1
def omega(n): return len(primefactors(n))
def a(n): return ceiling(bigomega(n)/omega(n))
print([a(n) for n in range(2, 51)]) # Indranil Ghosh, Jul 13 2017
CROSSREFS
Cf. A001221 (omega(n)), A001222 (bigomega(n)), A067340 (ratio is integer before ceil is applied), A070011 (ratio is not an integer), A070012 (floor of ratio), A070013 (ratio rounded), A046660 (bigomega(n)-omega(n)), A088529, A088530.
Sequence in context: A371148 A088388 A070013 * A051903 A324912 A157754
KEYWORD
nonn
AUTHOR
Rick L. Shepherd, Apr 11 2002
STATUS
approved