A090739 Exponent of 2 in 9^n - 1.

3, 4, 3, 5, 3, 4, 3, 6, 3, 4, 3, 5, 3, 4, 3, 7, 3, 4, 3, 5, 3, 4, 3, 6, 3, 4, 3, 5, 3, 4, 3, 8, 3, 4, 3, 5, 3, 4, 3, 6, 3, 4, 3, 5, 3, 4, 3, 7, 3, 4, 3, 5, 3, 4, 3, 6, 3, 4, 3, 5, 3, 4, 3, 9, 3, 4, 3, 5, 3, 4, 3, 6, 3, 4, 3, 5, 3, 4, 3, 7, 3, 4, 3, 5, 3, 4, 3

Offset

1,1

Comments

The exponent of 2 in the factorization of Fibonacci(6n). - T. D. Noe, Mar 14 2014

Links

T. D. Noe, Table of n, a(n) for n = 1..1000

T. Lengyel, The order of the Fibonacci and Lucas numbers, Fib. Quart. 33 (1995), 234-239.

Formula

a(n) = A007814(n) + 3.

a((2*n-1)*2^p) = p + 3, p >= 0. - Johannes W. Meijer, Feb 08 2013

a(n) = log_2(A006519(9^n - 1)). - Alonso del Arte, Feb 08 2013

a(n) = 2*tau(4*n)/(tau(4*n) - tau(n)), where tau(n) = A000005(n). - Peter Bala, Jan 06 2021

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 4. - Amiram Eldar, Nov 28 2022

Example

For n = 2, we see that -1 + 3^4 = 80 = 2^4 * 5 so a(2) = 4.
For n = 3, we see that -1 + 3^6 = 728 = 2^3 * 7 * 13, so a(3) = 3.

Maple

nmax:=70: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := p+3: od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 08 2013

Mathematica

Table[Part[Flatten[FactorInteger[ -1+3^(2*n)]], 2], {n, 1, 70}]
Table[IntegerExponent[Fibonacci[n], 2], {n, 6, 600, 6}] (* T. D. Noe, Mar 14 2014 *)

Prog

(PARI) a(n)=valuation(n, 2)+3 \\ Charles R Greathouse IV, Mar 14 2014
(Python)
def A090739(n): return (~n&n-1).bit_length()+3 # Chai Wah Wu, Jul 11 2022

Crossrefs

Cf. A024101, A069895, A091512, A088660, A090740, A220466.

Cf. A000005, A006519, A120738 (partial sums).

Appears in A161737.

Sequence in context: A223169 A201420 A244055 * A076400 A363194 A121889

Adjacent sequences: A090736 A090737 A090738 * A090740 A090741 A090742

Keyword

nonn,easy

Author

Labos Elemer and Ralf Stephan, Jan 19 2004

Extensions

More terms from T. D. Noe, Mar 14 2014

Status

approved