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A059269 Numbers n for which the number of divisors, tau(n), is divisible by 3. 7


%S 4,9,12,18,20,25,28,32,36,44,45,49,50,52,60,63,68,72,75,76,84,90,92,

%T 96,98,99,100,108,116,117,121,124,126,132,140,144,147,148,150,153,156,

%U 160,164,169,171,172,175,180,188,196,198,200,204,207,212,220,224,225,228

%N Numbers n for which the number of divisors, tau(n), is divisible by 3.

%C tau(n) is divisible by 3 iff at least one prime in the prime factorization of n has exponent of the form 3*m + 2. This sequence is an extension of the sequence A038109 in which the numbers has at least one prime with exponent 2 (the case of m = 0 here ) in their prime factorization.

%C The union of A211337 and A211338 is the complementary sequence to this one. [_Douglas Latimer_, Apr 12 2012]

%H Charles R Greathouse IV, <a href="/A059269/b059269.txt">Table of n, a(n) for n = 1..10000</a>

%F Conjecture: a(n) ~ k*n where k = 1/(1 - prod(1 - (p-1)/(p^(3*k)))) = 3.743455... where p ranges over the primes and k ranges over the positive integers. - _Charles R Greathouse IV_, Apr 13 2012

%e a(7) = 28 because the number of divisors of 28 d(28) = 6

%p with(numtheory): for n from 1 to 1000 do if tau(n) mod 3 = 0 then printf(`%d,`,n) fi: od:

%o (PARI) is(n)=vecmax(factor(n)[,2]%3)==2 \\ _Charles R Greathouse IV_, Apr 10 2012

%o (PARI) is(n)=numdiv(n)%3==0 \\ _Charles R Greathouse IV_, Sep 18 2015

%Y Cf. A038109, A000005, A211337, A211338.

%K nonn,easy

%O 1,1

%A Avi Peretz (njk(AT)netvision.net.il), Jan 24 2001

%E More terms from _James A. Sellers_, Jan 24 2001

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Last modified January 16 23:44 EST 2019. Contains 319206 sequences. (Running on oeis4.)