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%I #12 Sep 07 2020 19:09:45
%S 2,3,5,7,11,12,13,17,18,19,20,23,28,29,31,32,37,41,43,44,45,47,48,50,
%T 52,53,59,61,63,67,68,71,73,75,76,79,80,83,89,92,97,98,99,101,103,107,
%U 109,112,113,116,117,124,127,131,137,139,147,148,149,151,153
%N Numbers n for which the number of divisors of n is congruent to 2 mod 4.
%C The product of any 2 terms a(i)*a(j) is not a member of the sequence.
%C tau(n) is congruent to 2 modulo 4 iff only one prime in the prime factorization of n has exponent of the form 4*m + 1, and no prime in the prime factorization of n has exponent of the form 4*k + 3.
%H Charles R Greathouse IV, <a href="/A182358/b182358.txt">Table of n, a(n) for n = 1..10000</a>
%F Terms are of the form p * m^2 * n^4 for any prime p, m coprime to p, and n. - _Charles R Greathouse IV_, Apr 26 2012
%e The divisors of 12 are: 1, 2, 3, 4, 6, 12 [6 divisors]. 6 is congruent to 2 modulo 4. Thus 12 is a member of this sequence.
%t Select[Range[200],Mod[DivisorSigma[0,#],4]==2&] (* _Harvey P. Dale_, Sep 07 2020 *)
%o (PARI) {plnt=1 ; for(k=1, 10^7,
%o if(numdiv(k) % 4 == 2, print1(k, ", "); plnt++ ; if(100 < plnt, break() )))}
%o (PARI) is(n)=my(p=core(n));isprime(p)&&valuation(n,p)%4==1 \\ _Charles R Greathouse IV_, Apr 26 2012
%o (PARI) list(lim)=my(v=List(),t);forprime(p=2,lim,for(m=1,sqrtint(lim\p), if(m%p==0, next); t=p*m^2; for(n=1,sqrtint(sqrtint(lim\t)), listput(v, t*n^4)))); vecsort(Vec(v),,8) \\ _Charles R Greathouse IV_, Apr 26 2012
%Y This is an extension of A000040 (the prime numbers, which each have 2 divisors). The definition of this sequence uses A000005 (the number of divisors of n).
%K nonn,easy
%O 1,1
%A _Douglas Latimer_, Apr 26 2012
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