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A056169 Number of unitary prime divisors of n. 51
0, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 1, 0, 2, 0, 1, 1, 3, 1, 0, 2, 2, 2, 0, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 1, 0, 1, 2, 1, 1, 2, 3, 1, 1, 0, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 0, 1, 3, 1, 1, 3 (list; graph; refs; listen; history; text; internal format)
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

LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Catalan's Conjecture

Index entries for sequences computed from exponents in factorization of n

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.

a(n) = #{k: A124010(n,k) = 1, k = 1..A001221}. - Reinhard Zumkeller, Sep 10 2013

From Antti Karttunen, Nov 28 2017: (Start)

a(1) = 0; for n > 1, a(n) = A063524(A067029(n)) + a(A028234(n)).

a(n) = A001221(A055231(n)) = A001222(A055231(n)).

a(n) = A001221(n) - A056170(n) = A001221(n) - A001221(A000188(n)).

a(n) = A001222(n) - A275812(n).

a(n) = A162642(n) - A295662(n).

a(n) <= A162642(n) <= a(n) + A295659(n).

a(n) <= A295664(n).

(End)

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(n)

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 xrange(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

Cf. A001221, A001694, A002110, A034444, A056170, A055231, A076445, A162642, A275812, A295659, A295662, A295664, A001694.

See also A060355.

Sequence in context: A263251 A318370 A050326 * A286852 A125070 A125071

Adjacent sequences:  A056166 A056167 A056168 * A056170 A056171 A056172

KEYWORD

nice,nonn

AUTHOR

Labos Elemer, Jul 27 2000

STATUS

approved

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Last modified September 17 19:02 EDT 2019. Contains 327137 sequences. (Running on oeis4.)