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A048899 One of the two successive approximations up to 5^n for 5-adic integer sqrt(-1). 33

%I #60 Dec 03 2022 07:51:41

%S 0,3,18,68,443,1068,1068,32318,110443,1672943,3626068,23157318,

%T 120813568,1097376068,1097376068,19407922943,49925501068,355101282318,

%U 355101282318,15613890344818,15613890344818,110981321985443

%N One of the two successive approximations up to 5^n for 5-adic integer sqrt(-1).

%C This is the root congruent to 3 (mod 5) for n>0.

%C The other case with the 2 (mod 5) numbers (except for n=0) is given in A048898. - _Wolfdieter Lang_, Feb 19 2016

%C From _Jianing Song_, Sep 06 2022: (Start)

%C 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 2 modulo 5.

%C 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 4 modulo 5. (End)

%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 Seiichi Manyama, <a href="/A048899/b048899.txt">Table of n, a(n) for n = 0..1431</a>

%H 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.

%F a(n) = 5^n - A048898(n), n>=1.

%F a(n) = A066601(5^n), n>=0.

%F 0 <= a(n) < 5^n. 5^n divides a(n)^2 + 1.

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

%F Recurrence: a(n) = a(n-1)^5 (mod 5^n), a(1) = 3, n>=2. See the Pari program below, and the J.- F. Alcover Mathematica program for A048898.

%F a(n) = 3^(5^(n-1)) (mod 5^n), n>=1. Compare with the above given formula involving A066601.

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

%F (a(n)^2 + 1)/5^n = A210849(n), n>=0.

%F (End)

%F Another recurrence: a(n) = modp(a(n-1) + 4*(a(n-1)^2 + 1), 5^n), n >= 2, a(1) = 3. 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 a(n) == L(5^n,3) (mod 5^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - _Peter Bala_, Nov 20 2022

%e a(2) = 18 because the two roots of x^2 + 1 == 0 (mod 5^2) are 7 and 18 and 18 == 3 (mod 5). For 7 see A048898(2).

%t Join[{0}, RecurrenceTable[{a[1] == 3, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* _Vincenzo Librandi_, Feb 29 2016 * )

%o (PARI) {a(n) = if( n<2, 3, a(n - 1)^5) % 5^n}

%o (PARI) a(n) = lift(-sqrt(-1 + O(5^n))); \\ _Kevin Ryde_, Dec 22 2020

%o (Magma) [n le 2 select 3*(n-1) else Self(n-1)^5 mod 5^(n-1): n in [1..30]]; // _Vincenzo Librandi_, Feb 29 2016

%Y Cf. A048898, A066601, A114525, A210851.

%K nonn,easy

%O 0,2

%A _Michael Somos_, Jul 26 1999

%E Example corrected by _Wolfdieter Lang_, Apr 28 2012

%E Name clarified by _Wolfdieter Lang_, Feb 19 2016

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