A174279 Smallest k such that tau(Fibonacci(k)) = 2^n.

1, 3, 6, 15, 18, 44, 30, 54, 128, 80, 138, 90, 162, 198, 308, 294, 210, 460, 288, 270, 378, 510, 680, 594, 920, 570, 690, 1280, 1190, 630, 1040, 1386, 810

Offset: 0

Author

Michel Lagneau, Mar 15 2010

Keywords

• nonn
• more

Comments

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

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

Examples

			a(0) =  1 because tau(Fibonacci(1))  = tau(1)   = 2^0 = 1.
a(1) =  3 because tau(Fibonacci(3))  = tau(2)   = 2^1 = 2.
a(2) =  6 because tau(Fibonacci(6))  = tau(8)   = 2^2 = 4.
a(3) = 15 because tau(Fibonacci(15)) = tau(610) = 2^3 = 8.
		

References

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

Crossrefs

Cf. A085077, A063375.

Programs

Maple

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:

Extensions

a(27)-a(32) from Amiram Eldar, Oct 14 2019