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Exponent of 2 in 9^n - 1.
14

%I #45 Nov 28 2022 08:08:39

%S 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,

%T 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,

%U 3,4,3,6,3,4,3,5,3,4,3,7,3,4,3,5,3,4,3

%N Exponent of 2 in 9^n - 1.

%C The exponent of 2 in the factorization of Fibonacci(6n). - _T. D. Noe_, Mar 14 2014

%H T. D. Noe, <a href="/A090739/b090739.txt">Table of n, a(n) for n = 1..1000</a>

%H T. Lengyel, <a href="https://www.fq.math.ca/Scanned/33-3/lengyel.pdf">The order of the Fibonacci and Lucas numbers</a>, Fib. Quart. 33 (1995), 234-239.

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

%F a((2*n-1)*2^p) = p + 3, p >= 0. - _Johannes W. Meijer_, Feb 08 2013

%F a(n) = log_2(A006519(9^n - 1)). - _Alonso del Arte_, Feb 08 2013

%F a(n) = 2*tau(4*n)/(tau(4*n) - tau(n)), where tau(n) = A000005(n). - _Peter Bala_, Jan 06 2021

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

%e For n = 2, we see that -1 + 3^4 = 80 = 2^4 * 5 so a(2) = 4.

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

%p 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

%t Table[Part[Flatten[FactorInteger[ -1+3^(2*n)]], 2], {n, 1, 70}]

%t Table[IntegerExponent[Fibonacci[n], 2], {n, 6, 600, 6}] (* _T. D. Noe_, Mar 14 2014 *)

%o (PARI) a(n)=valuation(n,2)+3 \\ _Charles R Greathouse IV_, Mar 14 2014

%o (Python)

%o def A090739(n): return (~n&n-1).bit_length()+3 # _Chai Wah Wu_, Jul 11 2022

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

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

%Y Appears in A161737.

%K nonn,easy

%O 1,1

%A _Labos Elemer_ and _Ralf Stephan_, Jan 19 2004

%E More terms from _T. D. Noe_, Mar 14 2014