cp's OEIS Frontend

For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^4 - 6 is divisible by 5^n.

For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.

a(n) is the unique number k in [1, 2^n] and congruent to 1 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

The 2-adic integers are very different from p-adic ones where p is an odd prime. For example, provided that there is at least one solution, the number of solutions to x^n = a over p-adic integers is gcd(n, p-1) for odd primes p and gcd(n, 2) for p = 2. For odd primes p, x^2 = a is solvable iff a is a quadratic residue modulo p, while for p = 2 it's solvable iff a == 1 (mod 8). If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ starting at the rightmost digit for odd primes p, while for p = 2 they differ starting at the next-to-rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

a(n) is the unique number k in [1, 2^n] and congruent to 3 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

The 2-adic integers are very different from p-adic ones where p is an odd prime. For example, provided that there is at least one solution, the number of solutions to x^n = a over p-adic integers is gcd(n, p-1) for odd primes p and gcd(n, 2) for p = 2. For odd primes p, x^2 = a is solvable iff a is a quadratic residue modulo p, while for p = 2 it's solvable iff a == 1 (mod 8). If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ starting at the rightmost digit for odd primes p, while for p = 2 they differ starting at the next-to-rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

For n > 0, a(n) is the unique solution to x^2 == -6 (mod 5^n) in the range [0, 5^n - 1] and congruent to 2 modulo 5.

A324028 is the approximation (congruent to 3 mod 5) of another square root of 6 over the 5-adic field.

For n > 0, a(n) is the unique solution to x^2 == -6 (mod 5^n) in the range [0, 5^n - 1] and congruent to 3 modulo 5.

A324027 is the approximation (congruent to 3 mod 5) of another square root of -6 over the 5-adic field.

This is the root congruent to 2 mod 5.

a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^2 + 9 is divisible by 5^n.

a(n) is the unique number k in [1, 5^n] and congruent to 1 mod 5 such that k^2 + 9 is divisible by 5^n.

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.

For any positive integer n, if a number of the form m^2+1 is divisible by 5^n, then m mod 5^n must take one of two values--one even, the other odd. This sequence gives the even residue. (The odd residues are in A259266.)