A212153 -id:A212153 - cp's OEIS frontend

A309454 The successive approximations up to 7^n for 7-adic integer 6^(1/5).

0, 6, 20, 265, 1980, 11584, 11584, 246882, 1070425, 29894430, 29894430, 1159795426, 9069102398, 9069102398, 202847123212, 2237516341759, 2237516341759, 201635099759365, 1132157155708193, 6017397949439540, 17416293134812683, 496169890920484689, 1613261619087052703

Offset: 0

Seiichi Manyama, Aug 03 2019

nonn

			a(1) = (   6)_7 = 6,
a(2) = (  26)_7 = 20,
a(3) = ( 526)_7 = 265,
a(4) = (5526)_7 = 1980.

Cf. A309449.
Expansions of p-adic integers:
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A309450 (7-adic, 2^(1/5));
A309451 (7-adic, 3^(1/5));
A309452 (7-adic, 4^(1/5));
A309453 (7-adic, 5^(1/5)).

PARI
```
{a(n) = truncate((6+O(7^n))^(1/5))}
```

a(0) = 0 and a(1) = 6, a(n) = a(n-1) + 4 * (a(n-1)^5 - 6) mod 7^n for n > 1.

A212155 Digits of one of the three 7-adic integers (-1)^(1/3).

Original entry on oeis.org

5, 2, 0, 3, 6, 4, 0, 4, 2, 3, 2, 2, 1, 4, 5, 4, 5, 2, 0, 5, 5, 3, 1, 6, 4, 3, 2, 5, 3, 2, 3, 1, 0, 0, 4, 4, 4, 6, 4, 2, 6, 0, 0, 5, 1, 2, 5, 4, 3, 2, 5, 3, 2, 6, 3, 3, 4, 2, 2, 2, 1, 5, 6, 2, 6, 4, 6, 3, 5, 6, 4, 0, 5, 1, 4, 1, 1, 0, 6, 0, 4, 2, 2, 4, 5, 0, 3, 2, 1, 1, 5, 6, 2, 4, 2, 2, 1, 1, 5, 3

Offset: 0

Wolfdieter Lang, May 02 2012

See A210853 for comments and an approximation to this 7-adic number, called there v.  See also A048898 for references on p-adic numbers.
a(n), n>=1, is the (unique) solution of the linear congruence 3 * b(n)^2 * a(n) + c(n) == 0 (mod 7), with  b(n):=A212153(n) and c(n):=A212154(n). a(0) = 5, one of the three solutions of X^3+1 == 0 (mod 7).
Since b(n) == 5 (mod 7), a(n) == 4 * c(n) (mod 7) for n>0. - Álvar Ibeas, Feb 20 2017
With a(0) = 4, this is the digits of one of the three cube root of 1, the one that is congruent to 4 modulo 7. - Jianing Song, Aug 26 2022

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A212153 (approximations of (-1)^(1/3)), A212152 (digits of another cube root of -1), 6*A000012 (digits of -1).

Cf. A210850, A210851 (digits of the 5-adic integers sqrt(-1)); A319297, A319305, A319555 (digits of the 7-adic integers 6^(1/3)).

Join[{5}, MapIndexed[#/7^#2[[1]] &, Differences[FoldList[PowerMod[#, 7, 7^#2] &, 5, Range[2, 100]]]]] (* Paolo Xausa, Jan 14 2025 *)

a(n) = (b(n+1) - b(n))/7^n, n>=1, with b(n):=A212153(n), defined by a recurrence given there. One also finds there a Maple program for b(n). a(0)=5.

a(n) = 6 - A212152(n), for n>0. - Álvar Ibeas, Feb 21 2017

A318960 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

Original entry on oeis.org

1, 5, 5, 21, 53, 53, 181, 181, 181, 181, 181, 181, 181, 16565, 49333, 49333, 49333, 49333, 573621, 1622197, 1622197, 1622197, 10010805, 10010805, 10010805, 77119669, 211337397, 479772853, 479772853, 479772853, 2627256501, 6922223797, 15512158389, 15512158389

Offset: 2

Jianing Song, Sep 06 2018

nonn

a(n) is the unique number k in [1, 2^n] and congruent to 1 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

The 2-adic integers are very different from p-adic ones where p is an odd prime. For example, provided that there is at least one solution, the number of solutions to x^n = a over p-adic integers is gcd(n, p-1) for odd primes p and gcd(n, 2) for p = 2. For odd primes p, x^2 = a is solvable iff a is a quadratic residue modulo p, while for p = 2 it's solvable iff a == 1 (mod 8). If gcd(n, p-1) > 1 and gcd(a, p) = 1, then the solutions to x^n = a differ starting at the rightmost digit for odd primes p, while for p = 2 they differ starting at the next-to-rightmost digit. As a result, the formulas and the program here are different from those in other entries related to p-adic integers.

			The unique number k in [1, 4] and congruent to 1 modulo 4 such that k^2 + 7 is divisible by 8 is 1, so a(2) = 1.
a(2)^2 + 7 = 8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 5.
a(3)^2 + 7 = 32 which is divisible by 32, so a(4) = a(3) = 5.
a(4)^2 + 7 = 32 which is divisible by 64, so a(5) = a(4) + 2^4 = 21.
a(5)^2 + 7 = 448 which is divisible by 128, so a(6) = a(5) + 2^5 = 53.
...

Jianing Song, Table of n, a(n) for n = 2..999 (offset corrected by Jianing Song)
G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A318962.
Expansions of p-adic integers:
this sequence, A318961 (2-adic, sqrt(-7));
A268924, A271222 (3-adic, sqrt(-2));
A268922, A269590 (5-adic, sqrt(-4));
A048898, A048899 (5-adic, sqrt(-1));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A286840, A286841 (13-adic, sqrt(-1));
A286877, A286878 (17-adic, sqrt(-1)).
Also expansions of 10-adic integers:
A007185, A010690 (nontrivial roots to x^2-x);
A216092, A216093, A224473, A224474 (nontrivial roots to x^3-x).

PARI
```
a(n) = truncate(-sqrt(-7+O(2^(n+1))))
```

a(2) = 1; for n >= 3, a(n) = a(n-1) if a(n-1)^2 + 7 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A318961(n).
a(n) = Sum_{i=0..n-1} A318962(i)*2^i.

Offset corrected by Jianing Song, Aug 28 2019

A318961 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 3 (mod 4) case.

Original entry on oeis.org

3, 3, 11, 11, 11, 75, 75, 331, 843, 1867, 3915, 8011, 16203, 16203, 16203, 81739, 212811, 474955, 474955, 474955, 2572107, 6766411, 6766411, 23543627, 57098059, 57098059, 57098059, 57098059, 593968971, 1667710795, 1667710795, 1667710795, 1667710795, 18847579979

Offset: 2

Jianing Song, Sep 06 2018

nonn

a(n) is the unique number k in [1, 2^n] and congruent to 3 (mod 4) such that k^2 + 7 is divisible by 2^(n+1).

			The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^2 + 7 is divisible by 8 is 3, so a(2) = 3.
a(2)^2 + 7 = 16 which is divisible by 16, so a(3) = a(2) = 3.
a(3)^2 + 7 = 16 which is not divisible by 32, so a(4) = a(3) + 2^3 = 11.
a(4)^2 + 7 = 128 which is divisible by 64, so a(5) = a(4) = 11.
a(5)^2 + 7 = 128 which is divisible by 128, so a(6) = a(5) = 11.
...

Jianing Song, Table of n, a(n) for n = 2..999 (offset corrected by Jianing Song)
G. P. Michon, Introduction to p-adic integers, Numericana.

Cf. A318963.
Expansions of p-adic integers:
A318960, this sequence (2-adic, sqrt(-7));
A268924, A271222 (3-adic, sqrt(-2));
A268922, A269590 (5-adic, sqrt(-4));
A048898, A048899 (5-adic, sqrt(-1));
A290567 (5-adic, 2^(1/3));
A290568 (5-adic, 3^(1/3));
A290800, A290802 (7-adic, sqrt(-6));
A290806, A290809 (7-adic, sqrt(-5));
A290803, A290804 (7-adic, sqrt(-3));
A210852, A212153 (7-adic, (1+sqrt(-3))/2);
A290557, A290559 (7-adic, sqrt(2));
A286840, A286841 (13-adic, sqrt(-1));
A286877, A286878 (17-adic, sqrt(-1)).
Also expansions of 10-adic integers:
A007185, A010690 (nontrivial roots to x^2-x);
A216092, A216093, A224473, A224474 (nontrivial roots to x^3-x).

a(n) = if(n==2, 3, truncate(sqrt(-7+O(2^(n+1)))))

a(2) = 3; for n >= 3, a(n) = a(n-1) if a(n-1)^2 + 7 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).
a(n) = 2^n - A318960(n).
a(n) = Sum_{i=0..n-1} A318963(i)*2^i.

Offset corrected by Jianing Song, Aug 28 2019

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A309454 The successive approximations up to 7^n for 7-adic integer 6^(1/5).

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A212155 Digits of one of the three 7-adic integers (-1)^(1/3).

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A318960 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 1 (mod 4) case.

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A318961 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-7). This is the 3 (mod 4) case.

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