|
|
A277707
|
|
a(n) = index of the least prime divisor of n which has an odd exponent, or 0 if n is a perfect square.
|
|
4
|
|
|
0, 1, 2, 0, 3, 1, 4, 1, 0, 1, 5, 2, 6, 1, 2, 0, 7, 1, 8, 3, 2, 1, 9, 1, 0, 1, 2, 4, 10, 1, 11, 1, 2, 1, 3, 0, 12, 1, 2, 1, 13, 1, 14, 5, 3, 1, 15, 2, 0, 1, 2, 6, 16, 1, 3, 1, 2, 1, 17, 2, 18, 1, 4, 0, 3, 1, 19, 7, 2, 1, 20, 1, 21, 1, 2, 8, 4, 1, 22, 3, 0, 1, 23, 2, 3, 1, 2, 1, 24, 1, 4, 9, 2, 1, 3, 1, 25, 1, 5, 0, 26, 1, 27, 1, 2
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
For n = 8 = 2*2*2 = prime(1)^3, the exponent of the least (and the only) prime factor 2 is 3, an odd number, thus a(8) = 1 as 2 = prime(1).
|
|
PROG
|
(Scheme, two implementations)
(PARI) a(n) = my(f = factor(core(n))); if (!#f~, 0, primepi(vecmin(f[, 1]))); \\ Michel Marcus, Oct 30 2016
(Python)
from sympy import primepi, isprime, primefactors
from sympy.ntheory.factor_ import core
def a049084(n): return primepi(n)*(1*isprime(n))
def a055396(n): return 0 if n==1 else a049084(min(primefactors(n)))
|
|
CROSSREFS
|
Cf. A000290 (after its initial zero-term gives the positions of zeros in this sequence).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|