OFFSET
1,6
COMMENTS
The zeros of this sequences are the powerful numbers (A001694). There are no arbitrarily long subsequences with a given upper bound; for example, every sequence of 4 values includes one divisible by 2 but not 4, so there are no more than 3 consecutive zeros. Similarly, there can be no more than 23 consecutive values with none divisible by both 2 and 3 but neither 4 nor 9 (so a(n) >= 2), etc. In general, this gives an upper bound that is a (relatively) small multiple of the k-th primorial number (prime(k)#). One suspects that the actual upper bounds for such subsequences are quite a bit lower; e.g., Erdős conjectured that there are no three consecutive powerful numbers. - Franklin T. Adams-Watters, Aug 08 2006
In particular, for every A048670(k)*A002110(k) consecutive terms, at least one is greater than or equal to k. - Charlie Neder, Jan 03 2019
Following Catalan's conjecture (which became Mihăilescu's theorem in 2002), the first case of two consecutive zeros in this sequence is for a(8) and a(9), because 8 = 2^3 and 9 = 3^2, and there are no other consecutive zeros for consecutive powers. However, there are other pairs of consecutive zeros at powerful numbers (A001694, A060355). The next example is a(288) = a(289) = 0, because 288 = 2^5 * 3^2 and 289 = 17^2, then also a(675) and a(676). - Bernard Schott, Jan 06 2019
a(2k-1) is the number of primes p such that p || x + y and p^2 || x^(2k-1) + y^(2k-1) for some positive integers x and y. For any positive integers x, y and k > 1, there is no prime p such that p || x + y and p^2 || x^(2k) + y^(2k). - Jinyuan Wang, Apr 08 2020
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Catalan's Conjecture.
FORMULA
A prime factor of n is unitary iff its exponent is 1 in prime factorization of n. In general, gcd(p, n/p) = 1 or = p.
Additive with a(p^e) = 1 if e = 1, 0 otherwise.
From Antti Karttunen, Nov 28 2017: (Start)
a(n) <= A295664(n).
(End)
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B - C), where B is Mertens's constant (A077761) and C = Sum_{p prime} (1/p^2) = 0.452247... (A085548). - Amiram Eldar, Sep 28 2023
EXAMPLE
9 = 3^2 so a(9) = 0; 10 = 2 * 5 so a(10) = 2; 11 = 11^1 so a(11) = 1.
MAPLE
a:= n-> nops(select(i-> i[2]=1, ifactors(n)[2])):
seq(a(n), n=1..120); # Alois P. Heinz, Mar 27 2017
MATHEMATICA
Join[{0}, Table[Count[Transpose[FactorInteger[n]][[2]], 1], {n, 2, 110}]] (* Harvey P. Dale, Mar 15 2012 *)
Table[DivisorSum[n, 1 &, And[PrimeQ@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Nov 28 2017 *)
PROG
(Haskell)
a056169 = length . filter (== 1) . a124010_row
-- Reinhard Zumkeller, Sep 10 2013
(PARI) a(n)=my(f=factor(n)[, 2]); sum(i=1, #f, f[i]==1) \\ Charles R Greathouse IV, Apr 29 2015
(Python)
from sympy import factorint
def a(n):
f=factorint(n)
return 0 if n==1 else sum(1 for i in f if f[i]==1)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 19 2017
(Scheme, with memoization-macro definec) (definec (A056169 n) (if (= 1 n) 0 (+ (if (= 1 (A067029 n)) 1 0) (A056169 (A028234 n))))) ;; Antti Karttunen, Nov 28 2017
CROSSREFS
KEYWORD
nonn,nice,easy
AUTHOR
Labos Elemer, Jul 27 2000
STATUS
approved