A071864 Nonprime n such that the number of elements in the continued fraction for Sum_{d|n} 1/d equals tau(n), the number of divisors of n.

1, 4, 9, 14, 15, 21, 25, 49, 51, 55, 57, 63, 95, 98, 99, 115, 116, 121, 147, 161, 169, 172, 175, 188, 195, 203, 236, 244, 245, 247, 265, 284, 287, 289, 297, 299, 322, 328, 329, 351, 356, 361, 363, 370, 371, 374, 387, 406, 412, 413, 418, 423, 425, 437, 465, 488

Offset: 1

Author

Benoit Cloitre, Jun 09 2002

Keywords

• easy
• nonn

Comments

If p is prime p^2 is in the sequence since the continued fraction for Sum_{d|n} 1/d is [1, p-1, p+1] and there are 3 divisors for p^2.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A017665, A017666, A071862.

Programs

Mathematica

aQ[n_] := ! PrimeQ[n] && Length@ContinuedFraction[DivisorSigma[1, n]/n] ==  DivisorSigma[0, n]; Select[Range[488], aQ] (* Amiram Eldar, Aug 30 2019 *)

PARI

for(n=1,1000,if(length(contfrac(sumdiv(n,d,1/d)))==numdiv(n)*(1-isprime(n)),print1(n,",")))