A210846 (5^(3^(n-1)) + 1)/(2*3^n).

1, 7, 36169, 45991238252616223, 851008860651263039991161205833295116837255258128476241

Offset: 1

Author

Wolfdieter Lang, Apr 24 2012

Keywords

• nonn
• easy

Comments

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.

Examples

			n=1: (5^1+1)/6  = 1; n=2: (5^3 + 1)/18 = 126/18 = 7;
n=3: (5^9 +1)/(2*27) = 1953126/54 = 36169.
		

Crossrefs

Cf. A000244 (powers of 3), A068531, A090129 (the case Modd 2^n).

Formula

a(n) = (5^(3^(n-1)) + 1)/(2*3^n).