cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A090129 Smallest exponent such that -1 + 3^a(n) is divisible by 2^n.

Original entry on oeis.org

1, 2, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592, 17179869184
Offset: 1

Views

Author

Labos Elemer and Ralf Stephan, Jan 19 2004

Keywords

Comments

A131577 and A011782 are companions, A131577(n) + A011782(n) = 2^n, (and differences each other). - Paul Curtz, Jan 18 2009
A090127 with offset 0: (1, 2, 2, 4, 8, ...) = A(x) / A(x^2), when A(x) = (1 + 2x + 4x^2 + 8x^3 + ...). - Gary W. Adamson, Feb 20 2010
From Wolfdieter Lang, Apr 18 2012: (Start)
a(n) is the order of 3 modulo 2^n. For n=1 and 2 this is obviously 1 and 2, respectively, and for n >= 3 it is 2^(n-2).
For a proof see, e.g., the Graeme McRae link under A068531, the section 'A Different Approach', proposed by Alexander Monnas, the first part, where the result from the expansion of (4-1)^(2^(k-2)) holds only for k >= 3. See also the Charles R Greathouse IV program below where this result has been used.
This means that the cycle generated by 3, taken modulo 2^n, has length a(n), and that 3 is not a primitive root modulo 2^n, if n >= 3 (because Euler's phi(2^n) = 2^(n-1), n >= 1, see A000010).
(End)
Let r(x) = (1 + 2x + 2x^2 + 4x^3 + ...). Then (1 + 2x + 4x^2 + 8x^3 + ...) = (r(x) * r(x^2) * r(x)^4 * r(x^8) * ...). - Gary W. Adamson, Sep 13 2016

Examples

			a(1) = 1 since -1 + 3 = 2 is divisible by 2^1;
a(2) = a(3) = 2 since -1 + 9 = 8 is divisible by 4 = 2^2 and also by 8 = 2^3;
a(5) = 8 since -1 + 6561 = 6560 = 32*205 is divisible by 2^5.
From _Wolfdieter Lang_, Apr 18 2012: (Start)
n=3: the order of 3 (mod 8) is a(3)=2 because the cycle generated by 3 is [3, 3^2==1 (mod 8)].
n=5: a(5) = 2^3 = 8 because the cycle generated by 3 is [3^1=3, 3^2=9, 3^3=27, 17, 19, 25, 11, 1] (mod 32).
  The multiplicative group mod 32 is non-cyclic (see A033949(10)) with the additional four cycles  [5, 25, 29, 17, 21, 9, 13, 1], [7, 17, 23, 1], [15, 1], and [31, 1]. This is the cycle structure of the (Abelian) group Z_8 x Z_2 (see one of the cycle graphs shown in the Wikipedia link 'List of small groups' for the order phi(32)=16, given under A192005).
(End)
		

Crossrefs

Essentially the same as A000079.

Programs

  • Mathematica
    t=Table[Part[Flatten[FactorInteger[ -1+3^(n)]], 2], {n, 1, 130}] Table[Min[Flatten[Position[t, j]]], {j, 1, 10}]
    Join[{1,2},2^Range[30]] (* or *) Join[{1,2},NestList[2#&,2,30]] (* Harvey P. Dale, Nov 08 2012 *)
  • PARI
    a(n)=2^(n+(n<3)-2) \\ Charles R Greathouse IV, Apr 09 2012
    
  • Python
    def A090129(n): return n if n<3 else 1<Chai Wah Wu, Jul 11 2022

Formula

a(n) = 2^(n-2) if n >= 3, 1 for n=1 and 2 for n=2 (see the order comment above).
a(n+2) = A152046(n) + A152046(n+1) = 2*A011782(n). - Paul Curtz, Jan 18 2009

Extensions

a(11) through a(20) from R. J. Mathar, Aug 08 2008
More terms (powers of 2, see a comment above) from Wolfdieter Lang, Apr 18 2012