A211338 Numbers k for which the number of divisors, tau(k), is congruent to 2 modulo 3.
2, 3, 5, 7, 11, 13, 16, 17, 19, 23, 24, 29, 30, 31, 37, 40, 41, 42, 43, 47, 53, 54, 56, 59, 61, 66, 67, 70, 71, 73, 78, 79, 81, 83, 88, 89, 97, 101, 102, 103, 104, 105, 107, 109, 110, 113, 114, 127, 128, 130, 131, 135, 136, 137, 138, 139, 149, 151, 152, 154
Offset: 1
Keywords
Examples
The divisors of 16 are: 1, 2, 4, 8, 16 (5 divisors). 5 is congruent to 2 modulo 3. Thus 16 is a member of this sequence.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Douglas Latimer)
Crossrefs
Programs
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Mathematica
Select[Range[154], Mod[DivisorSigma[0, #], 3] == 2 &] (* T. D. Noe, Apr 21 2012 *)
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PARI
{plnt=1 ; mxind=100 ;for(k=1, 10^6, if(numdiv(k) % 3 == 2, print(k); plnt++; if(mxind+1 == plnt, break() )))}
Formula
Conjecture: a(n) ~ k*n where k = 2/prod(1 - (p-1)/(p^(3*k))) = 2.7290077... where p ranges over the primes and k ranges over the positive integers. - Charles R Greathouse IV, Apr 13 2012
Comments