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A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1). Here the 2 (mod 5) numbers (except for n=0). 17


%S 0,2,7,57,182,2057,14557,45807,280182,280182,6139557,25670807,

%T 123327057,123327057,5006139557,11109655182,102662389557,407838170807,

%U 3459595983307,3459595983307,79753541295807,365855836217682,2273204469030182,2273204469030182,49956920289342682

%N One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1). Here the 2 (mod 5) numbers (except for n=0).

%C This is the root congruent to 2 mod 5.

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

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

%C ...422331102414131141421404340423140223032431212 = u

%C 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:

%C ...022113342030313303023040104021304221412013233 = -u

%C The digits of u and -u are given in A210850 and A210851, respectively. - From _Wolfdieter Lang_, May 02 2012.

%C For approximations for p-adic square roots see also the W. Lang link under A268922. - From _Wolfdieter Lang_, Apr 03 2016.

%D J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.

%D K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

%H Vincenzo Librandi, <a href="/A048898/b048898.txt">Table of n, a(n) for n = 0..200</a>

%H G. P. Michon, <a href="http://www.numericana.com/answer/pseudo.htm#witness">On the witnesses of a composite integer</a>, Numericana.

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

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

%F From _Wolfdieter Lang_, Apr 28 2012: (Start)

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

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

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

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

%F (End)

%F 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

%e a(0)=0 because 0 satisfies any equation in integers modulo 1.

%e 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.)

%e a(2)=7 because the equation (5y+a(1))^2+1=0 modulo 25 means that y is 1 modulo 5.

%t a[0] = 0; a[1] = 2; a[n_] := a[n] = Mod[a[n-1]^5, 5^n]; Table[a[n], {n, 0, 21}] (* _Jean-Fran├žois Alcover_, Nov 24 2011, after Pari *)

%t Join[{0}, RecurrenceTable[{a[1] == 2, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* _Vincenzo Librandi_, Feb 29 2016 *)

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

%o (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

%Y Cf. A000351 (powers of 5), A048899 (the other pentadic number whose square is -1), A034939(n)=Min(a(n), A048899(n)).

%Y Cf. A034935. Different from A034935.

%K nonn,easy,nice

%O 0,2

%A _Michael Somos_, Jul 26 1999

%E Additional comments from _Gerard P. Michon_, Jul 15 2009

%E Edited by _N. J. A. Sloane_, Jul 25 2009

%E Name clarified by _Wolfdieter Lang_, Feb 19 2016

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Last modified April 23 04:14 EDT 2017. Contains 285313 sequences.