A174279 - OEIS

%I #30 Oct 14 2019 05:07:45

%S 1,3,6,15,18,44,30,54,128,80,138,90,162,198,308,294,210,460,288,270,

%T 378,510,680,594,920,570,690,1280,1190,630,1040,1386,810

%N Smallest k such that tau(Fibonacci(k)) = 2^n.

%C Smallest k such that A000005(A000045(k)) = 2^n.

%C The multiplicative property of the tau-function implies that the Fibonacci(k) has a prime factor representation p_1^e_1*p_2^e_2*... where (e_1+1)*(e_2+1)*... is a power of 2, that is, the exponents are in {1,3,7,15,...}. This adds for example the squarefree Fibonacci numbers with indices from A037918 to the list of candidates. - _R. J. Mathar_, Oct 11 2011

%D Majorie Bicknell and Verner E Hoggatt, Fibonacci's Problem Book, Fibonacci Association, San Jose, Calif., 1974.

%e a(0) =  1 because tau(Fibonacci(1))  = tau(1)   = 2^0 = 1.

%e a(1) =  3 because tau(Fibonacci(3))  = tau(2)   = 2^1 = 2.

%e a(2) =  6 because tau(Fibonacci(6))  = tau(8)   = 2^2 = 4.

%e a(3) = 15 because tau(Fibonacci(15)) = tau(610) = 2^3 = 8.

%p with(numtheory):for p from 1 to 100 do:indic:=0:u0:=0:u1:=1:for n from 2 to 1000 while(indic=0)do:s:=u0+u1:u0:=u1:u1:=s:if tau(s)= 2^p and indic=0 then print(p): print(n): indic:=1:else fi:od:od:

%Y Cf. A085077, A063375.

%K nonn,more

%O 0,2

%A _Michel Lagneau_, Mar 15 2010

%E a(27)-a(32) from _Amiram Eldar_, Oct 14 2019