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%I A048898 #95 Dec 03 2022 07:51:30
%S A048898 0,2,7,57,182,2057,14557,45807,280182,280182,6139557,25670807,
%T A048898 123327057,123327057,5006139557,11109655182,102662389557,407838170807,
%U A048898 3459595983307,3459595983307,79753541295807,365855836217682,2273204469030182,2273204469030182,49956920289342682
%N A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1).
%C A048898 This is the root congruent to 2 mod 5.
%C A048898 Or, residues modulo 5^n of a 5-adic solution of x^2+1=0.
%C A048898 The radix-5 expansion of a(n) is obtained from the n rightmost digits in the expansion of the following pentadic integer:
%C A048898   ...422331102414131141421404340423140223032431212 = u
%C A048898 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 A048898   ...022113342030313303023040104021304221412013233 = -u
%C A048898 The digits of u and -u are given in A210850 and A210851, respectively. - _Wolfdieter Lang_, May 02 2012
%C A048898 For approximations for p-adic square roots see also the W. Lang link under A268922. - _Wolfdieter Lang_, Apr 03 2016
%C A048898 From _Jianing Song_, Sep 06 2022: (Start)
%C A048898 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.
%C A048898 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)
%D A048898 J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.
%D A048898 K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.
%H A048898 Seiichi Manyama, <a href="/A048898/b048898.txt">Table of n, a(n) for n = 0..1431</a> (terms 0..200 from Vincenzo Librandi)
%H A048898 Peter Bala, <a href="/A210850/a210850.pdf">Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3/a>, Dec 2022.
%H A048898 G. P. Michon, <a href="http://www.numericana.com/answer/pseudo.htm#witness">On the witnesses of a composite integer</a>, Numericana.
%H A048898 G. P. Michon, <a href="http://www.numericana.com/answer/p-adic.htm#integers">Introduction to p-adic integers</a>, Numericana.
%F A048898 If n>0, a(n) = 5^n - A048899(n).
%F A048898 From _Wolfdieter Lang_, Apr 28 2012: (Start)
%F A048898 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.
%F A048898 a(n) == 2^(5^(n-1)) (mod 5^n), n>=1.
%F A048898 a(n)*a(n-1) + 1 == 0 (mod 5^(n-1)), n>=1.
%F A048898 (a(n)^2 + 1)/5^n = A210848(n), n>=0.
%F A048898 (End)
%F A048898 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
%F A048898 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
%e A048898 a(0)=0 because 0 satisfies any equation in integers modulo 1.
%e A048898 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 A048898 a(2)=7 because the equation (5y+a(1))^2+1=0 modulo 25 means that y is 1 modulo 5.
%t A048898 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 *)
%t A048898 Join[{0}, RecurrenceTable[{a[1] == 2, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* _Vincenzo Librandi_, Feb 29 2016 *)
%o A048898 (PARI) {a(n) = if( n<2, 2, a(n-1)^5) % 5^n}
%o A048898 (PARI) a(n) = lift(sqrt(-1 + O(5^n))); \\ _Kevin Ryde_, Dec 22 2020
%o A048898 (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 A048898 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).
%Y A048898 Cf. A000351 (powers of 5), A034939(n) = Min(a(n), A048899(n)).
%Y A048898 Different from A034935.
%K A048898 nonn,easy,nice
%O A048898 0,2
%A A048898 _Michael Somos_, Jul 26 1999
%E A048898 Additional comments from _Gerard P. Michon_, Jul 15 2009
%E A048898 Edited by _N. J. A. Sloane_, Jul 25 2009
%E A048898 Name clarified by _Wolfdieter Lang_, Feb 19 2016

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