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
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
Programs
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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
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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
Comments