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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 start from starting at the rightmost digit for odd primes p, while for p = 2 they differ start from starting at the second 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.

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Jianing Song, <a href="/A318961/b318961_1.txt">Table of n, a(n) for n = 2..999</a> (offset corrected by Jianing Song)

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One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). Here This is the 3 (mod 4) case.

The 2-adic integers are very different from p-adic ones where p is any 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 start from the rightmost digit for odd primes p, while for p = 2 they differ start from the second rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

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Jianing Song, <a href="/A318961/b318961_1.txt">Table of n, a(n) for n = 32..1000999</a> (offset corrected by Jianing Song)

a(32) = 3; for n >= 4, 3, a(n) = a(n-1) if a(n-1)^2 + 7 is divisible by 2^(n, +1), otherwise a(n-1) + 2^(n-21).

a(n) = 2^(n-1) - A318960(n).

a(n) = Sum_{i=0..n-21} A318963(i)*2^i.

The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^2 + 7 is divisible by 8 is 3, so a(32) = 3.

a(32)^2 + 7 = 16 which is divisible by 16, so a(43) = a(32) = 3.

a(43)^2 + 7 = 16 which is not divisible by 32, so a(54) = a(43) + 2^3 = 11.

a(54)^2 + 7 = 128 which is divisible by 64, so a(65) = a(54) = 11.

a(75)^2 + 7 = 128 which is divisible by 128, so a(76) = a(65) = 11.

(PARI) a(n) = if(n==3, 2, 3, truncate(sqrt(-7+O(2^(n+1)))))

Offset corrected by Jianing Song, Aug 28 2019

3,2,1

a(n) is the unique number k in [1, 2^(n-1)] 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 any 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). The two successive approximation of sqrt(a) up to p^n are in range [1, p^n] for odd primes p and [1, 2^(n-1)] for p = 2. If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ start from the rightmost digit for odd primes p, while for p = 2 they differ start from the second rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

approved

editing