Search: id:A048898
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%I A048898
%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
%N A048898 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
%D A048898 J. H. Conway, The Sensual Quadratic Form, p. 118.
%D A048898 K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge,
1973, p. 35.
%H A048898 G. P. Michon,
On the witnesses of a composite integer
%H A048898 G. P. Michon,
Introduction to p-adic integers
%F A048898 If n>0, a(n) = 5^n-A048899(n)
%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.
%o A048898 (PARI) a(n)=if(n<2,2,a(n-1)^5)%5^n
%Y A048898 Cf. A000351 (powers of 5), A048899 (the other pentadic number whose square
is -1), A034939(n)=Min(a(n), A048899(n)).
%Y A048898 Cf. A034935. Different from A034935.
%Y A048898 Sequence in context: A079410 A002658 A034939 this_sequence A034935 A121079
A105183
%Y A048898 Adjacent sequences: A048895 A048896 A048897 this_sequence A048899 A048900
A048901
%K A048898 nonn,easy,nice
%O A048898 0,2
%A A048898 Michael Somos
%E A048898 Additional comments from Gerard P. Michon (g.michon(AT)att.net), Jul
15 2009
%E A048898 Edited by N. J. A. Sloane, Jul 25 2009
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