%I #14 Sep 08 2019 10:05:41

%S 1,2,6,8,9,10,14,18,19,20,22,26,27,28,30,31,32,34,40,41,42,52,53,55,

%T 59,61,64,66,71,73,74,77,79,85,87,89,92,93,94,95,97,99,101,107,109,

%U 113,115,116,117,120,121,123,125,127,128,129,130,133,135,138,143,146,147,149

%N n such that tau(Fibonacci(n)) is a perfect square.

%C tau = A000005 is the number of divisors of n.

%H Amiram Eldar, <a href="/A174331/b174331.txt">Table of n, a(n) for n = 1..423</a>

%F {n: A063375(n) in A000290}. - _R. J. Mathar_, Jul 09 2012

%e 40 is in the sequence because tau(Fibonacci(40)) = tau(102334155) = 64 is square.

%t Select[Range[150], IntegerQ[Sqrt[DivisorSigma[0,Fibonacci[#]]]] &]

%o (PARI) is(n) = issquare(numdiv(fibonacci(n))) \\ _Felix Fröhlich_, Sep 08 2019

%Y Cf. A000005, A000045, A000290, A036436, A063375.

%K nonn

%O 1,2

%A _Michel Lagneau_, Mar 15 2010