

A182358


Numbers n for which the number of divisors of n is congruent to 2 mod 4.


2



2, 3, 5, 7, 11, 12, 13, 17, 18, 19, 20, 23, 28, 29, 31, 32, 37, 41, 43, 44, 45, 47, 48, 50, 52, 53, 59, 61, 63, 67, 68, 71, 73, 75, 76, 79, 80, 83, 89, 92, 97, 98, 99, 101, 103, 107, 109, 112, 113, 116, 117, 124, 127, 131, 137, 139, 147, 148, 149, 151, 153
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OFFSET

1,1


COMMENTS

The product of any 2 terms a(i)*a(j) is not a member of the sequence.
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.


LINKS



FORMULA



EXAMPLE

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.


MATHEMATICA

Select[Range[200], Mod[DivisorSigma[0, #], 4]==2&] (* Harvey P. Dale, Sep 07 2020 *)


PROG

(PARI) {plnt=1 ; for(k=1, 10^7,
if(numdiv(k) % 4 == 2, print1(k, ", "); plnt++ ; if(100 < plnt, break() )))}
(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


CROSSREFS

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).


KEYWORD

nonn,easy


AUTHOR



STATUS

approved



