cp's OEIS Frontend

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

(m^(2^n)-1)/2^(n+2) is an integer for any odd value of m and n>0.

The next term, a(7), has 106 decimal digits.

The number of digits of a(n) is 1, 1, 5, 17, 54, 167, 506, 1525, ... .

Integer 2*a(n) implies that 5^delta(3^n) == -1 (mod 3^n), n>=1, with the degree delta(3^n) = phi(2*3^n)/2 = 3^(n-1) of the minimal polynomial C(3^n,x) of the algebraic number 2*cos(pi/3^n). For delta and the coefficient array of C see A055034 and A187360, respectively. That 2*a(n) is indeed an even integer can be shown by analyzing the terms of the binomial expansion of (6-1)^(3^(n-1)) + 1.

This congruence implies that floor(5^(3^(n-1))/3^n) = 2*a(n) - 1, i.e., it is odd. Hence 5^delta(3^n) == +1 (Modd 3^n), n>=2. For Modd n (not to be confused with mod n) see a comment on A203571. One can show that 5 is the smallest positive primitive root Modd 3^n for n>=2 (for n=1 one has 1^1 == +1 (Modd 3)). See A206550. The proof uses the fact that the order of 5 using multiplication Modd 3^n has to be a divisor of delta(3^n)=3^(n-1), i.e., a power of 3. This is because the multiplicative group Modd 3^n has order delta(3^n) and the subgroup formed by the cycle has the order of 5 considered Modd 3^n. Then apply Lagrange's theorem. That for n>=2 no number 5^(3^(n-1-j)), with j=1, 2..., n-1, is congruent +1 (Modd 3^n) follows from the above established congruence and an analysis of the relevant expansion for a given smaller power.

The above statements show that for n>=1 the multiplicative group Modd 3^n is cyclic (for n=1 the cycle is [1], and for n>=2 the cycle is generated by 5). For the cyclic moduli see A206551.

The number of digits of a(n) is 1, 4, 22, 117, 593, 2978, 14905, ..., for n >= 1.

Integer a(n) implies that 3^delta(5^n) == -1 (mod 5^n), n>=1, with the degree delta(5^n) = phi(2*5^n)/2 = 2*5^(n-1) of the minimal polynomial C(5^n,x) of the algebraic number 2*cos(Pi/5^n). For delta and the coefficient array of C see A055034 and A187360, respectively. That 2*a(n) is indeed an even integer can be shown by analyzing the terms of the binomial expansion of (10-1)^(5^(n-1)) + 1.

This congruence implies that floor(3^(2*5^(n-1))/5^n) = 2*a(n) - 1, i.e., it is odd. Hence 3^delta(5^n) == +1 (Modd 5^n), n>=1. For Modd n (not to be confused with mod n) see a comment on A203571. One can show that 3 is the smallest positive primitive root Modd 5^n for n>=1. See A206550. The proof uses the fact that the order of 3 using multiplication Modd 5^n has to be a divisor of delta(5^n)=2*5^(n-1), i.e., either 5^e or 2*5^e with certain e>=0. This is because the multiplicative group Modd 5^n has order delta(5^n) and the order of the subgroup formed by the cycle generated by 3 coincides with the order of 3 considered Modd 5^n. Then Lagrange's theorem is applied. That for n>=1 no power of 3 with exponent 2*5^(n-1-j) with j=1,2,..., n-1 is congruent +1 (Modd 5^n) follows by considering the two cases +1 (mod 5^n) and -1 (mod 5^n) separately. The first case is excluded from the above established congruence by an indirect proof. The case -1 (Modd 5^n) can be excluded by an analysis of the relevant expansion for a given smaller power. The other cases 3^k with k = 5^(n-1-j), where j = 0, 1, ..., n-1, are neither -1 (mod 5^n) nor +1 (mod 5^n) because 3^(5^(n-1)) (mod 5^n) is congruent 3 (mod 5) (see A048899), hence neither +1 nor -1 (mod 5^n), respectively. The lower exponents are then excluded in both cases iteratively by an indirect proof taking fifth powers.

The above statements show that for n>=1 the multiplicative group Modd 5^n is cyclic, and for each n the cycle of length 2*5^(n-1) can be generated starting with 3. For the cyclic moduli see A206551.

(m^(2^n) - 1)/2^(n + 2) is an integer for any odd value of m and n > 0.

In particular, for m = 11, a(n) is a multiple of 15.