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
Examples
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.
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.
Crossrefs
Programs
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Maple
A090739 := proc(n) padic[ordp](9^n-1,2) ; end proc: seq(A090739(n),n=1..80) ; # R. J. Mathar, Jun 28 2025
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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 *)
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PARI
a(n)=valuation(n,2)+3 \\ Charles R Greathouse IV, Mar 14 2014
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Python
def A090739(n): return (~n&n-1).bit_length()+3 # Chai Wah Wu, Jul 11 2022
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
Extensions
More terms from T. D. Noe, Mar 14 2014
Comments