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 A210847 a(n) = (3^(2*5^(n-1)) + 1)/(2*5^n), n >= 1. 0
 1, 1181, 2871591950767410355081 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. LINKS FORMULA a(n) = (3^(2*5^(n-1)) + 1)/(2*5^n), n >= 1. EXAMPLE n = 1: (9 + 1)/(10) = 1; n = 2: (3^10 + 1)/50 = 59050/50 = 1181; n = 3: (3^50 + 1)/250 = 717897987691852588770250/250 = 2871591950767410355081. CROSSREFS Cf. A068531(n-2), n>=3, (the case p=2), A210846 (the case p=3). Sequence in context: A258912 A237094 A321867 * A320716 A269017 A328064 Adjacent sequences:  A210844 A210845 A210846 * A210848 A210849 A210850 KEYWORD nonn,bref AUTHOR Wolfdieter Lang, May 03 2012 STATUS approved

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Last modified December 8 10:20 EST 2021. Contains 349594 sequences. (Running on oeis4.)