A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1).

0, 2, 7, 57, 182, 2057, 14557, 45807, 280182, 280182, 6139557, 25670807, 123327057, 123327057, 5006139557, 11109655182, 102662389557, 407838170807, 3459595983307, 3459595983307, 79753541295807, 365855836217682, 2273204469030182, 2273204469030182, 49956920289342682

Offset: 0

Author

Michael Somos, Jul 26 1999

Keywords

• nonn
• easy
• nice

Comments

This is the root congruent to 2 mod 5.

Or, residues modulo 5^n of a 5-adic solution of x^2+1=0.

The radix-5 expansion of a(n) is obtained from the n rightmost digits in the expansion of the following pentadic integer:

...422331102414131141421404340423140223032431212 = u

The residues modulo 5^n of the other 5-adic solution of x^2+1=0 are given by A048899 which corresponds to the pentadic integer -u:

...022113342030313303023040104021304221412013233 = -u

The digits of u and -u are given in A210850 and A210851, respectively. - Wolfdieter Lang, May 02 2012

For approximations for p-adic square roots see also the W. Lang link under A268922. - Wolfdieter Lang, Apr 03 2016

From Jianing Song, Sep 06 2022: (Start)

For n > 0, a(n)-1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 1 modulo 5.

For n > 0, a(n)+1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 3 modulo 5. (End)

Examples

			a(0)=0 because 0 satisfies any equation in integers modulo 1.
a(1)=2 because 2 is one solution of x^2+1=0 modulo 5. (The other solution is 3, which gives rise to A048899.)
a(2)=7 because the equation (5y+a(1))^2+1=0 modulo 25 means that y is 1 modulo 5.
		

References

• J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.
• K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1431 (terms 0..200 from Vincenzo Librandi)
• Peter Bala, Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3, Dec 2022.
• G. P. Michon, On the witnesses of a composite integer, Numericana.
• G. P. Michon, Introduction to p-adic integers, Numericana.

Crossrefs

The two successive approximations up to p^n for p-adic integer sqrt(-1): this sequence and A048899 (p=5), A286840 and A286841 (p=13), A286877 and A286878 (p=17).

Cf. A000351 (powers of 5), A034939(n) = Min(a(n), A048899(n)).

Different from A034935.

Programs

Magma

[n le 2 select 2*(n-1) else Self(n-1)^5 mod 5^(n-1): n in [1..30]]; // Vincenzo Librandi, Feb 29 2016

Mathematica

a[0] = 0; a[1] = 2; a[n_] := a[n] = Mod[a[n-1]^5, 5^n]; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Nov 24 2011, after PARI *)
Join[{0}, RecurrenceTable[{a[1] == 2, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* Vincenzo Librandi, Feb 29 2016 *)

PARI

{a(n) = if( n<2, 2, a(n-1)^5) % 5^n}

PARI

a(n) = lift(sqrt(-1 + O(5^n))); \\ Kevin Ryde, Dec 22 2020

Formula

If n>0, a(n) = 5^n - A048899(n).

From Wolfdieter Lang, Apr 28 2012: (Start)

Recurrence: a(n) = a(n-1)^5 (mod 5^n), a(1) = 2, n>=2. See the J.-F. Alcover Mathematica program and the PARI program below.

a(n) == 2^(5^(n-1)) (mod 5^n), n>=1.

a(n)*a(n-1) + 1 == 0 (mod 5^(n-1)), n>=1.

(a(n)^2 + 1)/5^n = A210848(n), n>=0.

(End)

Another recurrence: a(n) = modp(a(n-1) + a(n-1)^2 + 1, 5^n), n >= 2, a(1) = 2. Here modp(a, m) is the representative from {0, 1, ..., |m|-1} of the residue class a modulo m. Note that a(n) is in the residue class of a(n-1) modulo 5^(n-1) (see Hensel lifting). - Wolfdieter Lang, Feb 28 2016

a(n) == L(5^n,2) (mod 5^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

Extensions

Additional comments from Gerard P. Michon, Jul 15 2009

Edited by N. J. A. Sloane, Jul 25 2009

Name clarified by Wolfdieter Lang, Feb 19 2016