A070014 Ceiling of number of prime factors of n divided by the number of n's distinct prime factors.
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
Keywords
Examples
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.
Links
Crossrefs
Programs
-
Mathematica
Table[Ceiling[PrimeOmega[n]/PrimeNu[n]], {n, 2, 106}] (* Michael De Vlieger, Jul 12 2017 *)
-
PARI
v=[]; for(n=2,150,v=concat(v,ceil(bigomega(n)/omega(n)))); v
-
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
-
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
Formula
a(n) = ceiling(bigomega(n)/omega(n)) for n>=2.
Comments