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%I #10 Aug 30 2019 04:16:55
%S 1,4,9,14,15,21,25,49,51,55,57,63,95,98,99,115,116,121,147,161,169,
%T 172,175,188,195,203,236,244,245,247,265,284,287,289,297,299,322,328,
%U 329,351,356,361,363,370,371,374,387,406,412,413,418,423,425,437,465,488
%N 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.
%C 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.
%H Amiram Eldar, <a href="/A071864/b071864.txt">Table of n, a(n) for n = 1..10000</a>
%t aQ[n_] := ! PrimeQ[n] && Length@ContinuedFraction[DivisorSigma[1, n]/n] == DivisorSigma[0, n]; Select[Range[488], aQ] (* _Amiram Eldar_, Aug 30 2019 *)
%o (PARI) for(n=1,1000,if(length(contfrac(sumdiv(n,d,1/d)))==numdiv(n)*(1-isprime(n)),print1(n,",")))
%Y Cf. A000005, A017665, A017666, A071862.
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, Jun 09 2002