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One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.
7

%I #45 Dec 17 2021 08:26:07

%S 1,5,5,21,53,53,181,181,181,181,181,181,181,16565,49333,49333,49333,

%T 49333,573621,1622197,1622197,1622197,10010805,10010805,10010805,

%U 77119669,211337397,479772853,479772853,479772853,2627256501,6922223797,15512158389,15512158389

%N One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

%C 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).

%C 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.

%H Jianing Song, <a href="/A318960/b318960.txt">Table of n, a(n) for n = 2..999</a> (offset corrected by Jianing Song)

%H G. P. Michon, <a href="http://www.numericana.com/answer/p-adic.htm#integers">Introduction to p-adic integers</a>, Numericana.

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

%F a(n) = 2^n - A318961(n).

%F a(n) = Sum_{i=0..n-1} A318962(i)*2^i.

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

%e a(2)^2 + 7 = 8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 5.

%e a(3)^2 + 7 = 32 which is divisible by 32, so a(4) = a(3) = 5.

%e a(4)^2 + 7 = 32 which is divisible by 64, so a(5) = a(4) + 2^4 = 21.

%e a(5)^2 + 7 = 448 which is divisible by 128, so a(6) = a(5) + 2^5 = 53.

%e ...

%o (PARI) a(n) = truncate(-sqrt(-7+O(2^(n+1))))

%Y Cf. A318962.

%Y Expansions of p-adic integers:

%Y this sequence, A318961 (2-adic, sqrt(-7));

%Y A268924, A271222 (3-adic, sqrt(-2));

%Y A268922, A269590 (5-adic, sqrt(-4));

%Y A048898, A048899 (5-adic, sqrt(-1));

%Y A290567 (5-adic, 2^(1/3));

%Y A290568 (5-adic, 3^(1/3));

%Y A290800, A290802 (7-adic, sqrt(-6));

%Y A290806, A290809 (7-adic, sqrt(-5));

%Y A290803, A290804 (7-adic, sqrt(-3));

%Y A210852, A212153 (7-adic, (1+sqrt(-3))/2);

%Y A290557, A290559 (7-adic, sqrt(2));

%Y A286840, A286841 (13-adic, sqrt(-1));

%Y A286877, A286878 (17-adic, sqrt(-1)).

%Y Also expansions of 10-adic integers:

%Y A007185, A010690 (nontrivial roots to x^2-x);

%Y A216092, A216093, A224473, A224474 (nontrivial roots to x^3-x).

%K nonn

%O 2,2

%A _Jianing Song_, Sep 06 2018

%E Offset corrected by _Jianing Song_, Aug 28 2019