|
|
A051904
|
|
Minimal exponent in prime factorization of n.
|
|
62
|
|
|
0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
The asymptotic mean of this sequence is 1 (Niven, 1969). - Amiram Eldar, Jul 10 2020
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
For n = 72 = 2^3*3^2, a(72) = min(exponents) = min(3,2) = 2.
For n = 72, using alternative definition: rad(72) = 6; and 6^2 = 36 divides 72 but no higher power of 6 divides 72, so a(72) = 2.
For n = 432, rad(432) = 6 and 6^3 = 216 divides 432 but no higher power of 6 divides 432, therefore a(432) = 3. - David James Sycamore, Sep 08 2023
|
|
MAPLE
|
a := proc (n) if n = 1 then 0 else min(seq(op(2, op(j, op(2, ifactors(n)))), j = 1 .. nops(op(2, ifactors(n))))) end if end proc: seq(a(n), n = 1 .. 100); # Emeric Deutsch, May 20 2015
|
|
MATHEMATICA
|
Table[If[n == 1, 0, Min @@ Last /@ FactorInteger[n]], {n, 100}] (* Ray Chandler, Jan 24 2006 *)
|
|
PROG
|
(Haskell)
a051904 1 = 0
(Python)
from sympy import factorint
def a(n):
f = factorint(n)
l = [f[p] for p in f]
return 0 if n == 1 else min(l)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|