login
This site is supported by donations to The OEIS Foundation.

 

Logo

Annual Appeal: Today, Nov 11 2014, is the 4th anniversary of the launch of the new OEIS web site. 70,000 sequences have been added in these four years, all edited by volunteers. Please make a donation (tax deductible in the US) to help keep the OEIS running.

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A048898 Successive approximations up to 5^n for the 5-adic integer sqrt(-1). 13

%I

%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

%N Successive approximations up to 5^n for the 5-adic integer sqrt(-1).

%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.

%D J. H. Conway, The Sensual Quadratic Form, p. 118.

%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>

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

%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)

%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 *)

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

%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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .

Last modified December 19 17:37 EST 2014. Contains 252238 sequences.