cp's OEIS Frontend

A056169 Number of unitary prime divisors of n.

%I A056169 #104 Feb 16 2025 08:32:43
%S A056169 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,
%T A056169 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,
%U A056169 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
%N A056169 Number of unitary prime divisors of n.
%C A056169 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
%C A056169 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
%C A056169 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
%C A056169 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
%H A056169 T. D. Noe, <a href="/A056169/b056169.txt">Table of n, a(n) for n = 1..10000</a>
%H A056169 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CatalansConjecture.html">Catalan's Conjecture</a>.
%H A056169 <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.
%F A056169 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.
%F A056169 Additive with a(p^e) = 1 if e = 1, 0 otherwise.
%F A056169 a(n) = #{k: A124010(n,k) = 1, k = 1..A001221}. - _Reinhard Zumkeller_, Sep 10 2013
%F A056169 From _Antti Karttunen_, Nov 28 2017: (Start)
%F A056169 a(1) = 0; for n > 1, a(n) = A063524(A067029(n)) + a(A028234(n)).
%F A056169 a(n) = A001221(A055231(n)) = A001222(A055231(n)).
%F A056169 a(n) = A001221(n) - A056170(n) = A001221(n) - A001221(A000188(n)).
%F A056169 a(n) = A001222(n) - A275812(n).
%F A056169 a(n) = A162642(n) - A295662(n).
%F A056169 a(n) <= A162642(n) <= a(n) + A295659(n).
%F A056169 a(n) <= A295664(n).
%F A056169 (End)
%F A056169 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
%e A056169 9 = 3^2 so a(9) = 0; 10 = 2 * 5 so a(10) = 2; 11 = 11^1 so a(11) = 1.
%p A056169 a:= n-> nops(select(i-> i[2]=1, ifactors(n)[2])):
%p A056169 seq(a(n), n=1..120);  # _Alois P. Heinz_, Mar 27 2017
%t A056169 Join[{0},Table[Count[Transpose[FactorInteger[n]][[2]],1],{n,2,110}]] (* _Harvey P. Dale_, Mar 15 2012 *)
%t A056169 Table[DivisorSum[n, 1 &, And[PrimeQ@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* _Michael De Vlieger_, Nov 28 2017 *)
%o A056169 (Haskell)
%o A056169 a056169 = length . filter (== 1) . a124010_row
%o A056169 -- _Reinhard Zumkeller_, Sep 10 2013
%o A056169 (PARI) a(n)=my(f=factor(n)[,2]); sum(i=1,#f,f[i]==1) \\ _Charles R Greathouse IV_, Apr 29 2015
%o A056169 (Python)
%o A056169 from sympy import factorint
%o A056169 def a(n):
%o A056169     f=factorint(n)
%o A056169     return 0 if n==1 else sum(1 for i in f if f[i]==1)
%o A056169 print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Jun 19 2017
%o A056169 (Scheme)
%o A056169 ;; With memoization-macro definec.
%o A056169 (definec (A056169 n) (if (= 1 n) 0 (+ (if (= 1 (A067029 n)) 1 0) (A056169 (A028234 n))))) ;; _Antti Karttunen_, Nov 28 2017
%Y A056169 Cf. A001221, A001694, A002110, A034444, A056170, A055231, A076445, A162642, A275812, A295659, A295662, A295664, A001694.
%Y A056169 See also A060355.
%Y A056169 Cf. A077761, A085548.
%K A056169 nonn,nice,easy
%O A056169 1,6
%A A056169 _Labos Elemer_, Jul 27 2000