A268922 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-4). These are the 1 mod 5 numbers, except for n = 0.

0, 1, 11, 11, 261, 2136, 2136, 64636, 220886, 1392761, 7252136, 46314636, 241627136, 974049011, 2194752136, 8298267761, 99851002136, 710202564636, 710202564636, 12154294361511, 31227780689636

Offset: 0

Author

Wolfdieter Lang, Mar 02 2016

Keywords

• nonn
• easy

Comments

The other approximation for the 5-adic integer sqrt(-4) with numbers 4 (mod 5) is given in A269590.

For the two approximations of the 5-adic integer sqrt(-1) see A048899 and A048898. For comments and some proofs see A210848.

For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269591. This 5-adic number has the digits read from the right to the left ...32234111132111101130213043113443324032021 = u.

The companion 5-adic number -u has digits ....12210333312333343314231401331001120412424. See A269592.

The recurrence given below (for n >= 1) has been derived from the following facts (i) x^2 + 4 == 0 (mod 5) has the two distinct solutions x(1) = 1 and x(2) = 5 - x(1) = 4. This guarantees the existence of a unique solution x = x1(n) of x^2 + 4 == 0 (mod 5^n) , for n >= 2 , which satisfies also x1(n) == 1 (mod 5). See e.g., Theorem 50, p. 87 of the Nagell reference. The same is true for the solution x = x2(n) with x2(n) == 4 (mod 5^n). (ii) As a consequence of Hensel's lemma (see e.g., the Wikipedia reference under Hensel lifting) one knows that x1(n) (which is treated here) satisfies the congruence x1(n) == x1(n-1) (mod 5^(n-1)) with x1(1) = x1 = 1. (A similar statement holds for x2(n) with input x2(1) = x(2) = 4. This is used in A269590). These two facts allow one to derive a recurrence for x1(n) (and for x2(n)).

Examples

			n=2:  11^2 + 4 = 125  == 0 (mod 5^2), and 125 is the only solution from {0, 1, ..., 24} which is congruent to 1 modulo 5.
n=3:  the only solution of x1^2 + 4 == 0 (mod 5^3) with x1 from {0, ..., 124} and x1 ==  1 (mod 4) is also 11. The number 114 satisfies also the first congruence but not the second one: 114 == 2 (mod 4).
		

References

• Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, p. 87.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1430
• Wolfdieter Lang, Note on a Recurrence for Approximation Sequences of p-adic Square Roots
• Wikipedia, Hensel's lemma.

Crossrefs

Cf. A000032, A048899, A048898, A144837, A210848, A269590, A269591, A269592.

Programs

Maple

with(padic):D1:=op(3,op([evalp(RootOf(x^2+4),5,20)][1])): 0,seq(sum('D1[k]*5^(k-1)','k'=1..n), n=1..20);
# alternative program - see A144837
a := proc (n) option remember; if n = 1 then 1 else irem( a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1), 5^n) end if; end: seq(a(n), n = 1..20); # Peter Bala, Nov 14 2022

PARI

a(n) = truncate(sqrt(-4+O(5^(n)))); \\ Michel Marcus, Mar 04 2016

Formula

Recurrence for n >= 1: a(n) = modp( a(n-1) + 2*(a(n-1)^2 + 4), 5^n), n >= 2, with a(1) = 1. Here modp(a, m) is used to pick the representative of the residue class a modulo m from the smallest nonnegative complete residue system {0, 1, ... , m-1}.

a(n) = 5^n - A269590(n), n >= 1.

a(n) == Lucas(5^n) (mod 5^n). - Peter Bala, Nov 10 2022