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a(n) is an integer (nonnegative) because b(n):=A048899(n) satisfies b(n)^2 + 1 == 0 (mod 5^n), n>=0. The solution of this congruence for n>=1, which satisfies also b(n) == 3 (mod 5), is b(n) = 3^(5^(n-1)) (mod 5^n), but this is inconvenient for computing b(n) for large n. Instead one can use the b(n) recurrence which follows immediately, and this is given in the formula field below. To prove that the given b(n) formula solves the first congruence one can analyze the binomial expansion of (10 - 1)^(5^(n-1)) + 1 and show that it is 0 (mod 5^n) term by term. The second congruence reduces to b(n) == 3^(5^(n-1)) (mod 5) which follows for n>=1 by induction. Because b(n) = 5^n - A048898(n) one could also use the result A048898(n) == 2 (mod 5) once this is proved.

See also the comment on A210848 on two relevant theorems.

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

The digits of u and -u are given in A210850 and A210851, respectively. - Wolfdieter Lang, May 02 2012

For approximations for p-adic square roots see also the W. Lang link under A268922. - Wolfdieter Lang, Apr 03 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 1 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 3 modulo 5. (End)

See A048898 for the successive approximations to this 5-adic integer, called there u.

The digits of -u, the other 5-adic integer sqrt(-1), are given in A210851.

a(n) is the unique solution of the linear congruence 2*A048898(n)*a(n) + A210848(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A210848, eq. (6) on p. 86 adapted to this case. a(0)=2 follows from the formula given below.

If n>0, a(n) == A210848(n) (mod 5), since A048898(n) == 2 (mod 5). - Álvar Ibeas, Feb 21 2017

If a(n)=0 then A048899(n+1) and A048899(n) coincide.

a(n) + A210851(n) = 4 for n >= 1. - Robert Israel, Mar 04 2016

From Jianing Song, Sep 06 2022: (Start)

With a(0) = 1, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 1 modulo 5.

With a(0) = 3, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 3 modulo 5. (End)

This square root of -1 in the 5-adic integers is equal to the 5-adic limit of the sequence {L(5^n,2)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 02 2022

See A048899 for the successive approximations to this 5-adic integer, called -u in a comment on A048898.

The digits of u, the other 5-adic integer sqrt(-1), are given in A210850.

a(n) is the (unique) solution of the linear congruence 2*A048899(n)*a(n) + A210849(n) == 0 (mod 5), n >= 1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A210848, eq. (6) on p. 86 adapted to this case. a(0)=3 follows from the formula given below.

If n>0, a(n) == -(A210849(n)) (mod 5), since A048899(n) == 3 (mod 5). - Álvar Ibeas, Feb 21 2017

If a(n)=0 then A048899(n+1) and A048899(n) coincide.

From Jianing Song, Sep 06 2022: (Start)

With a(0) = 2, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 2 modulo 5.

With a(0) = 4, this is the digits of one of the four 4th root of -4 in the ring of 5-adic integers, the one that is congruent to 4 modulo 5. (End)

This square root of -1 in the 5-adic integers is equal to the 5-adic limit of the sequence {L(5^n,3)}, where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 02 2022

The other approximation for the 5-adic integer sqrt(-4) with numbers 4 (mod 5) is given in A269590.

For the two approximations of the 5-adic integer sqrt(-1) see A048899 and A048898. For comments and some proofs see A210848.

For the digits of this 5-adic integer sqrt(-4), that is the scaled first differences, see A269591. This 5-adic number has the digits read from the right to the left ...32234111132111101130213043113443324032021 = u.

The companion 5-adic number -u has digits ....12210333312333343314231401331001120412424. See A269592.

The recurrence given below (for n >= 1) has been derived from the following facts (i) x^2 + 4 == 0 (mod 5) has the two distinct solutions x(1) = 1 and x(2) = 5 - x(1) = 4. This guarantees the existence of a unique solution x = x1(n) of x^2 + 4 == 0 (mod 5^n) , for n >= 2 , which satisfies also x1(n) == 1 (mod 5). See e.g., Theorem 50, p. 87 of the Nagell reference. The same is true for the solution x = x2(n) with x2(n) == 4 (mod 5^n). (ii) As a consequence of Hensel's lemma (see e.g., the Wikipedia reference under Hensel lifting) one knows that x1(n) (which is treated here) satisfies the congruence x1(n) == x1(n-1) (mod 5^(n-1)) with x1(1) = x1 = 1. (A similar statement holds for x2(n) with input x2(1) = x(2) = 4. This is used in A269590). These two facts allow one to derive a recurrence for x1(n) (and for x2(n)).

The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 3^(7^(n-1)) (mod 7^n), n >= 1, which satisfies a(n)^3 + 1 == 0 (mod 7^n), n >= 1. a(0) = 0 satisfies this congruence as well. The proof can be done by showing that each term in the binomial expansion of (3^(7^(n-1)))^3 +1 = (28 -1)^(7^(n-1)) + 1 has a factor 7^n.

a(n) == 3 (mod 7), n >= 1. This follows from the formula given above, and 3^(7^(n-1)) == 3 (mod 7), n >= 1 (proof by induction).

The digit t(n), n >= 0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to this sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n >= 1, with b(n):= (a(n)^3+1)/7^n = A210853(n). t(0):=3, one of the three solutions of X^3 + 1 == 0 (mod 7). For these digits see A212152. The 7-adic number is, read from right to left, ...3143214516604202226653431432053116412125443426203643 =: u.

a(n) is obtained from reading u in base 7, and adding the first n terms.

One can show directly that a(n) = 7^n + 1 - y(n), n >= 1, with y(n) = A212153(n) and z(n) = 7^n - 1 = 6*A023000(n), n >= 0.

Iff a(n+1) = a(n) then t(n) = A212152(n) = 0.

See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 + 1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 + 1 == 0 (mod 7^n) exactly three solutions for each n >= 1, which can be chosen as a(n) == 3 (mod 7), y(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The y- and z- sequences are given in A212153 and 6*A023000, respectively.

For n > 0, a(n) - 1 (== a(n)^2 (mod 7^n)) and 7^n - a(n) (== a(n)^4 (mod 7^n)) are the two primitive cubic roots of unity in Z/(7^n Z). - Álvar Ibeas, Feb 20 2017

From Jianing Song, Aug 26 2022: (Start)

a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 3 modulo 7 (if n>0).

A212153(n) is the multiplicative inverse of a(n) modulo 7^n. (End)

See A210852 for comments and the approximation of one of the other three 7-adic integers (-1)^(1/3), called there u.

The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 5^(7^(n-1)) (mod 7^n), n>= 1, which satisfies a(n)^3 + 1 == 0 (mod 7^n), n>=1. a(0) = 0 satisfies this congruence also. The proof can be done by showing that each term in the binomial expansion of (5^(7^(n-1)))^3 + 1 = (2*3^2*7 - 1)^(7^(n-1)) + 1 has a factor 7^n.

a(n) == 5 (mod 7), n>=1. This follows from the formula given above, and 5^(7^(n-1)) == 5 (mod 7), n>=1 (proof by induction).

The digit t(n), n>=0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to the present sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n>=1, with b(n):= (a(n)^3 + 1)/7^n = A212154(n). t(0):=5. For these digits see A212155. The 7-adic number is, read from right to left,

...3452150062464440013235234613550254541223240463025 =: v.

a(n) is obtained from reading v in base 7, and adding the first n terms.

One can show directly that a(n) = 7^n + 1 - x(n), n>=1, with x(n) = A210852(n), and z(n) = 7^n - 1 = 6*A023000(n), n>=0.

Iff a(n+1) = a(n) then t(n) = A212155(n) = 0.

See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 +1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 +1 == 0 (mod 7^n) exactly three solutions for each n>=1, which can be chosen as x(n) == 3 (mod 7), a(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The x- and z- sequences are given in A210852 and 6*A023000, respectively.

For n>0, a(n) - 1 (== a(n)^2 mod 7^n) and 7^n - a(n) (== a(n)^4 mod 7^n) are the two primitive cubic roots of unity in Z/(7^n Z). - Álvar Ibeas, Feb 20 2017

From Jianing Song, Aug 26 2022: (Start)

a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 5 modulo 7 (if n>0).

A210852(n) is the multiplicative inverse of a(n) modulo 7^n. (End)