

A174279


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


0



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

Smallest k such that A000005(A000045(k)) = 2^n.
The multiplicative property of the taufunction 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


REFERENCES

Mohammad K. Azarian, The Generating Function for the Fibonacci Sequence, Missouri Journal of Mathematical Sciences, Vol. 2, No. 2, Spring 1990, pp. 7879. Zentralblatt MATH, Zbl 1097.11516.
Majorie Bicknell and Verner E Hoggatt, Fibonacci's Problem Book, Fibonacci Association, San Jose, Calif., 1974.


LINKS

Table of n, a(n) for n=0..26.
N. D. Cahill and D. A. Narayan, Fibonacci and Lucas Numbers as Tridiagonal Matrix Determinants, Fibonacci Quarterly, 42(3):216221, 2004.
S. Happersett, Mathematical meditations , J. Math. Arts 1 (1) (2007) 2933.


EXAMPLE

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.


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:


CROSSREFS

Cf. A085077, A063375.
Sequence in context: A212060 A249246 A248969 * A233554 A276546 A265486
Adjacent sequences: A174276 A174277 A174278 * A174280 A174281 A174282


KEYWORD

nonn,more


AUTHOR

Michel Lagneau, Mar 15 2010


STATUS

approved



