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

0, 3, 18, 68, 443, 1068, 1068, 32318, 110443, 1672943, 3626068, 23157318, 120813568, 1097376068, 1097376068, 19407922943, 49925501068, 355101282318, 355101282318, 15613890344818, 15613890344818, 110981321985443, 110981321985443, 9647724486047943, 9647724486047943

Offset: 0

Author

Michael Somos, Jul 26 1999

Keywords

• nonn
• easy

Comments

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

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

From Jianing Song, Sep 06 2022: (Start)

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.

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)

Examples

			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).
		

References

• J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.
• K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1431
• Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.

Crossrefs

Cf. A048898, A066601, A114525, A210851.

Programs

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

Mathematica

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

PARI

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

PARI

a(n) = lift(-sqrt(-1 + O(5^n))); \\ Kevin Ryde, Dec 22 2020

Formula

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

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

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

From Wolfdieter Lang, Apr 28 2012: (Start)

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.

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

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

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

(End)

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

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

Extensions

Example corrected by Wolfdieter Lang, Apr 28 2012

Name clarified by Wolfdieter Lang, Feb 19 2016