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A048898
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Successive approximations up to 5^n for the 5-adic integer sqrt(-1).
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4
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0, 2, 7, 57, 182, 2057, 14557, 45807, 280182, 280182, 6139557, 25670807, 123327057, 123327057, 5006139557, 11109655182, 102662389557, 407838170807, 3459595983307, 3459595983307, 79753541295807, 365855836217682
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| This is the root congruent to 2 mod 5.
Or, residues modulo 5^n of a 5-adic solution of x^2+1=0.
The radix-5 expansion of a(n) is obtained from the n rightmost digits in the expansion of the following pentadic integer:
...422331102414131141421404340423140223032431212 = u
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:
...022113342030313303023040104021304221412013233 = -u
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REFERENCES
| J. H. Conway, The Sensual Quadratic Form, p. 118.
K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.
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LINKS
| G. P. Michon, On the witnesses of a composite integer
G. P. Michon, Introduction to p-adic integers
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FORMULA
| If n>0, a(n) = 5^n-A048899(n)
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EXAMPLE
| 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.
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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}] (* From Jean-François Alcover, Nov 24 2011, after Pari *)
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PROG
| (PARI) a(n)=if(n<2, 2, a(n-1)^5)%5^n
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CROSSREFS
| Cf. A000351 (powers of 5), A048899 (the other pentadic number whose square is -1), A034939(n)=Min(a(n), A048899(n)).
Cf. A034935. Different from A034935.
Sequence in context: A002658 A175818 A034939 * A034935 A178769 A121079
Adjacent sequences: A048895 A048896 A048897 * A048899 A048900 A048901
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KEYWORD
| nonn,easy,nice
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AUTHOR
| Michael Somos
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EXTENSIONS
| Additional comments from Gerard P. Michon (g.michon(AT)att.net), Jul 15 2009
Edited by N. J. A. Sloane, Jul 25 2009
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