

A268922


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


24



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

0,3


COMMENTS

The other approximation for the 5adic integer sqrt(4) with numbers 4 (mod 5) is given in A269590.
For the two approximations of the 5adic integer sqrt(1) see A048899 and A048898. For comments and some proofs see A210848.
For the digits of this 5adic integer sqrt(4), that is the scaled first differences, see A269591. This 5adic number has the digits read from the right to the left ...32234111132111101130213043113443324032021 = u.
The companion 5adic 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(n1) (mod 5^(n1)) 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)).


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 padic Square Roots
Wikipedia, Hensel's lemma.


FORMULA

Recurrence for n >= 1: a(n) = modp( a(n1) + 2*(a(n1)^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, ... , m1}.
a(n) = 5^n  A269590(n), n >= 1.


EXAMPLE

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).


MAPLE

with(padic):D1:=op(3, op([evalp(RootOf(x^2+4), 5, 20)][1])): 0, seq(sum('D1[k]*5^(k1)', 'k'=1..n), n=1..20);


PROG

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


CROSSREFS

Cf. A048899, A048898, A210848, A269590, A269591, A269592.
Sequence in context: A110381 A165832 A038325 * A328918 A062129 A290298
Adjacent sequences: A268919 A268920 A268921 * A268923 A268924 A268925


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Mar 02 2016


STATUS

approved



