A225404 -id:A225404 - cp's OEIS frontend

A309600 Digits of the 10-adic integer (17/9)^(1/3).

7, 1, 6, 8, 7, 0, 3, 3, 3, 6, 5, 2, 7, 8, 7, 2, 6, 7, 1, 1, 0, 3, 3, 2, 4, 5, 6, 5, 3, 6, 5, 3, 3, 3, 7, 5, 2, 4, 7, 5, 0, 2, 9, 0, 6, 7, 0, 8, 8, 6, 6, 7, 0, 1, 2, 4, 5, 3, 2, 8, 6, 9, 7, 3, 1, 6, 6, 9, 5, 0, 1, 6, 4, 6, 8, 0, 3, 8, 5, 9, 6, 1, 3, 5, 3, 7, 9, 7, 2, 3, 6, 6, 9, 0, 0, 0, 5, 3, 7, 7, 2

Offset: 0

Seiichi Manyama, Aug 09 2019

			      7^3 == 3      (mod 10).
     17^3 == 13     (mod 10^2).
    617^3 == 113    (mod 10^3).
   8617^3 == 1113   (mod 10^4).
  78617^3 == 11113  (mod 10^5).
  78617^3 == 111113 (mod 10^6).

Seiichi Manyama, Table of n, a(n) for n = 0..10000

10-adic integer x.
A225404       (x^3 = ...000003).
A225405       (x^3 = ...000007).
A225406       (x^3 = ...000009).
A153042       (x^3 = ...111111).
this sequence (x^3 = ...111113).
A309601       (x^3 = ...111117).
A309602       (x^3 = ...111119).
A309603       (x^3 = ...222221).
A225410       (x^3 = ...222223).
A309604       (x^3 = ...222227).
A309605       (x^3 = ...222229).
A309606       (x^3 = ...333331).
A225402       (x^3 = ...333333).
A309569       (x^3 = ...333337).
A309570       (x^3 = ...333339).
A309595       (x^3 = ...444441).
A309608       (x^3 = ...444443).
A309609       (x^3 = ...444447).
A309610       (x^3 = ...444449).
A309611       (x^3 = ...555551).
A309612       (x^3 = ...555553).
A309613       (x^3 = ...555557).
A309614       (x^3 = ...555559).
A309640       (x^3 = ...666661).
A309641       (x^3 = ...666663).
A225411       (x^3 = ...666667).
A309642       (x^3 = ...666669).
A309643       (x^3 = ...777771).
A309644       (x^3 = ...777773).
A225401       (x^3 = ...777777).
A309645       (x^3 = ...777779).
A309646       (x^3 = ...888881).
A309647       (x^3 = ...888883).
A309648       (x^3 = ...888887).
A225412       (x^3 = ...888889).
A225409       (x^3 = ...999991).
A225408       (x^3 = ...999993).
A225407       (x^3 = ...999997).

N=100; Vecrev(digits(lift(chinese(Mod((17/9+O(2^N))^(1/3), 2^N), Mod((17/9+O(5^N))^(1/3), 5^N)))), N)

def A309600(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 - 17)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309600(100)

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 17) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

A225407 10-adic integer x such that x^3 = -3.

Original entry on oeis.org

3, 1, 4, 5, 6, 8, 4, 0, 1, 2, 1, 9, 9, 4, 1, 6, 7, 5, 9, 3, 8, 4, 3, 3, 7, 0, 5, 1, 8, 6, 9, 6, 3, 8, 1, 5, 1, 5, 3, 8, 0, 7, 7, 8, 6, 6, 1, 3, 1, 2, 1, 1, 0, 9, 3, 4, 0, 3, 6, 4, 6, 9, 7, 5, 7, 6, 9, 0, 9, 4, 7, 8, 3, 9, 2, 9, 8, 0, 0, 5, 2, 4, 2, 0, 6, 5, 1, 3, 3, 6, 6, 8, 8, 5, 6, 6, 3, 3, 7, 5

Offset: 0

Aswini Vaidyanathan, May 07 2013

This is the 10's complement of A225404.

			       3^3 == -7 (mod 10).
      13^3 == -7 (mod 10^2).
     413^3 == -7 (mod 10^3).
    5413^3 == -7 (mod 10^4).
   65413^3 == -7 (mod 10^5).
  865413^3 == -7 (mod 10^6).

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A225402, A225404, A309600.

n=0; for(i=1, 100, m=(10^i-3); for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))

N=100; Vecrev(digits(lift(chinese(Mod((-3+O(2^N))^(1/3), 2^N), Mod((-3+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 06 2019

def A225407(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 7 * (a ** 3 + 3)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225407(100) # Seiichi Manyama, Aug 13 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 7 * (b(n-1)^3 + 3) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019

A309698 Digits of the 4-adic integer 3^(1/3).

Original entry on oeis.org

3, 2, 3, 1, 1, 0, 3, 3, 1, 0, 2, 0, 3, 3, 0, 3, 1, 3, 0, 1, 1, 3, 0, 3, 3, 3, 3, 3, 1, 0, 3, 2, 0, 2, 0, 0, 1, 2, 3, 2, 0, 3, 1, 0, 1, 1, 1, 2, 1, 2, 0, 1, 0, 1, 3, 2, 2, 1, 1, 1, 3, 2, 2, 0, 3, 3, 3, 0, 3, 0, 0, 0, 3, 0, 2, 3, 3, 0, 3, 2, 1, 2, 1, 2, 2, 1, 0, 0, 0, 2, 0, 1, 3, 0

Offset: 0

Seiichi Manyama, Aug 13 2019

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, Hensel's Lemma.

Digits of the k-adic integer (k-1)^(1/(k-1)): this sequence (k=4), A309699 (k=6), A309700 (k=8), A225458 (k=10).

Cf. A225404, A290563, A322931.

N=100; Vecrev(digits(lift((3+O(2^(2*N)))^(1/3)), 4), N)

def A309698(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + a ** 3 - 3) % (4 ** (i + 2))
    ary << (b - a) / (4 ** (i + 1))
    a = b
  }
  ary
end
p A309698(100)

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + b(n-1)^3 - 3 mod 4^n for n > 1, then a(n) = (b(n+1) - b(n))/4^n.

A306552 Expansion of the 10-adic cube root of -1/7, that is, the 10-adic integer solution to x^3 = -1/7.

Original entry on oeis.org

3, 9, 5, 5, 4, 0, 3, 8, 3, 9, 1, 6, 4, 7, 2, 6, 5, 2, 9, 6, 2, 4, 5, 7, 0, 0, 9, 0, 6, 1, 9, 3, 8, 2, 5, 1, 4, 1, 8, 4, 1, 0, 2, 4, 4, 7, 8, 5, 0, 6, 2, 4, 3, 8, 4, 2, 0, 2, 4, 7, 3, 3, 4, 7, 1, 9, 9, 3, 5, 3, 9, 7, 0, 4, 4, 6, 3, 7, 7, 1, 7, 6, 3, 5, 5, 9, 6

Offset: 1

Jianing Song, Feb 23 2019

10's complement of A319739.

			3^3 == 7 == -1/7 (mod 10).
93^3 == 57 == -1/7 (mod 100).
593^3 == 857 == -1/7 (mod 1000).
5593^3 == 2857 == -1/7 (mod 10000).
...
...619383045593^3 = ...142857142857 = ...999999999999/7 = -1/7.

Robert Israel, Table of n, a(n) for n = 1..10000

10-adic cube root of p/q:
q=1: A225409 (p=-9), A225408 (p=-7), A225407 (p=-3), A225404 (p=3), A225405 (p=7), A225406 (p=9);
q=3: A225402 (p=-1), A225411 (p=1);
q=7: this sequence (p=-1), A319739 (p=1);
q=9: A225401 (p=-7), A153042 (p=-1), A225412 (p=1), A225410 (p=7);
q=11: A306553 (p=-1), A319740 (p=1);
q=13: A306555 (p=-1), A306554 (p=1).

op([1, 3], padic:-rootp(7*x^3+1, 10, 100)); # Robert Israel, Mar 31 2019

seq(n)={Vecrev(digits(lift(chinese( Mod((-1/7 + O(5^n))^(1/3), 5^n), Mod((-1/7 + O(2^n))^(1/3), 2^n)))), n)} \\ Following Andrew Howroyd's code for A319740.

a(n) = 9 - A319739(n) for n >= 2.

A306553 Expansion of the 10-adic cube root of -1/11, that is, the 10-adic integer solution to x^3 = -1/11.

Original entry on oeis.org

9, 6, 8, 2, 3, 8, 1, 4, 2, 0, 2, 0, 6, 9, 8, 3, 8, 9, 4, 5, 4, 0, 6, 0, 0, 9, 6, 8, 6, 1, 3, 4, 7, 8, 0, 6, 6, 7, 1, 6, 5, 5, 3, 6, 4, 9, 9, 0, 2, 7, 1, 7, 4, 2, 6, 5, 1, 4, 0, 6, 9, 0, 7, 0, 8, 7, 8, 1, 4, 1, 2, 6, 6, 9, 4, 2, 5, 3, 5, 7, 4, 9, 6, 4, 4, 0, 5

Offset: 1

Jianing Song, Feb 23 2019

10's complement of A319740.

			9^3 == 9 == -1/11 (mod 10).
69^3 == 9 == -1/11 (mod 100).
869^3 == 909 == -1/11 (mod 1000).
2869^3 == 909 == -1/11 (mod 10000).
...
...020241832869^3 = ...090909090909 = ...999999999999/11 = -1/11.

Robert Israel, Table of n, a(n) for n = 1..10000

10-adic cube root of p/q:
q=1: A225409 (p=-9), A225408 (p=-7), A225407 (p=-3), A225404 (p=3), A225405 (p=7), A225406 (p=9);
q=3: A225402 (p=-1), A225411 (p=1);
q=7: A306552 (p=-1), A319739 (p=1);
q=9: A225401 (p=-7), A153042 (p=-1), A225412 (p=1), A225410 (p=7);
q=11: this sequence (p=-1), A319740 (p=1);
q=13: A306555 (p=-1), A306554 (p=1).

op([1,3],padic:-rootp(11*x^3+1,10,100)); # Robert Israel, Mar 24 2019

seq(n)={Vecrev(digits(lift(chinese( Mod((-1/11 + O(5^n))^(1/3), 5^n), Mod((-1/11 + O(2^n))^(1/3), 2^n)))), n)} \\ Following Andrew Howroyd's code for A319740.

a(n) = 9 - A319740(n) for n >= 2.

A306554 Expansion of the 10-adic cube root of 1/13, that is, the 10-adic integer solution to x^3 = 1/13.

Original entry on oeis.org

3, 5, 6, 4, 1, 9, 3, 2, 8, 7, 4, 0, 8, 3, 6, 5, 7, 7, 0, 9, 8, 2, 7, 5, 1, 4, 8, 0, 9, 5, 1, 6, 0, 6, 2, 1, 3, 2, 2, 6, 4, 2, 7, 0, 6, 8, 6, 1, 3, 3, 2, 2, 0, 0, 1, 5, 6, 7, 9, 6, 2, 7, 8, 4, 2, 6, 3, 6, 3, 0, 1, 0, 4, 5, 5, 6, 6, 1, 3, 5, 4, 3, 3, 3, 1, 7, 0

Offset: 1

Jianing Song, Feb 23 2019

10's complement of A306555.

			3^3 == 7 == 1/13 (mod 10).
53^3 == 77 == 1/13 (mod 100).
653^3 == 77 == 1/13 (mod 1000).
4653^3 == 3077 == 1/13 (mod 10000).
...
...047823914653^3 = ...923076923077 = 1 + (...999999999999)*(12/13) = 1 - 12/13 = 1/13.

Robert Israel, Table of n, a(n) for n = 1..10000

10-adic cube root of p/q:
q=1: A225409 (p=-9), A225408 (p=-7), A225407 (p=-3), A225404 (p=3), A225405 (p=7), A225406 (p=9);
q=3: A225402 (p=-1), A225411 (p=1);
q=7: A306552 (p=-1), A319739 (p=1);
q=9: A225401 (p=-7), A153042 (p=-1), A225412 (p=1), A225410 (p=7);
q=11: A306553 (p=-1), A319740 (p=1);
q=13: A306555 (p=-1), this sequence (p=1).

op([1,3],padic:-rootp(13*x^3-1,10,100)); # Robert Israel, Mar 24 2019

seq(n)={Vecrev(digits(lift(chinese( Mod((1/13 + O(5^n))^(1/3), 5^n), Mod((1/13 + O(2^n))^(1/3), 2^n)))), n)} \\ Following Andrew Howroyd's code for A319740.

a(n) = 9 - A306555(n) for n >= 2.

A306555 Expansion of the 10-adic cube root of -1/13, that is, the 10-adic integer solution to x^3 = -1/13.

Original entry on oeis.org

7, 4, 3, 5, 8, 0, 6, 7, 1, 2, 5, 9, 1, 6, 3, 4, 2, 2, 9, 0, 1, 7, 2, 4, 8, 5, 1, 9, 0, 4, 8, 3, 9, 3, 7, 8, 6, 7, 7, 3, 5, 7, 2, 9, 3, 1, 3, 8, 6, 6, 7, 7, 9, 9, 8, 4, 3, 2, 0, 3, 7, 2, 1, 5, 7, 3, 6, 3, 6, 9, 8, 9, 5, 4, 4, 3, 3, 8, 6, 4, 5, 6, 6, 6, 8, 2, 9

Offset: 1

Jianing Song, Feb 23 2019

10's complement of A306554.

			7^3 == 3 == -1/13 (mod 10).
47^3 == 23 == -1/13 (mod 100).
347^3 == 923 == -1/13 (mod 1000).
5347^3 == 6923 == -1/13 (mod 10000).
...
...952176085347^3 = ...076923076923 = ...999999999999/13 = -1/13.

Robert Israel, Table of n, a(n) for n = 1..10000

10-adic cube root of p/q:
q=1: A225409 (p=-9), A225408 (p=-7), A225407 (p=-3), A225404 (p=3), A225405 (p=7), A225406 (p=9);
q=3: A225402 (p=-1), A225411 (p=1);
q=7: A306552 (p=-1), A319739 (p=1);
q=9: A225401 (p=-7), A153042 (p=-1), A225412 (p=1), A225410 (p=7);
q=11: A306553 (p=-1), A319740 (p=1);
q=13: this sequence (p=-1), A306554 (p=1).

op([1,3],padic:-rootp(13*x^3+1,10,100)); # Robert Israel, Mar 24 2019

seq(n)={Vecrev(digits(lift(chinese( Mod((-1/13 + O(5^n))^(1/3), 5^n), Mod((-1/13 + O(2^n))^(1/3), 2^n)))), n)} \\ Following Andrew Howroyd's code for A319740.

a(n) = 9 - A306554(n) for n >= 2.

A322931 Digits of the 8-adic integer 3^(1/3).

Original entry on oeis.org

3, 7, 5, 0, 7, 3, 0, 1, 7, 1, 7, 6, 4, 2, 3, 6, 7, 7, 7, 0, 3, 1, 2, 0, 1, 7, 2, 6, 1, 2, 5, 4, 1, 1, 1, 2, 3, 5, 5, 2, 3, 5, 4, 7, 3, 6, 0, 0, 3, 4, 7, 1, 3, 3, 6, 4, 6, 0, 0, 4, 4, 6, 0, 5, 6, 4, 1, 5, 5, 6, 0, 0, 0, 6, 2, 6, 1, 0, 7, 1, 6, 0, 0, 4, 6, 5, 0, 7, 0, 1, 3, 7, 3, 7, 0, 0, 2, 0, 6, 1

Offset: 0

Patrick A. Thomas, Dec 31 2018

The octal version of A225404.

			4671710370573^3 == 3 (mod 8^13) in octal.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, Hensel's Lemma.

Cf. A225404 (decimal version), A290563, A309698, A322932, A322933.

N=100; Vecrev(digits(lift((3+O(2^(3*N)))^(1/3)), 8), N) \\ Seiichi Manyama, Aug 14 2019

def A322931(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 5 * (a ** 3 - 3)) % (8 ** (i + 2))
    ary << (b - a) / (8 ** (i + 1))
    a = b
  }
  ary
end
p A322931(100) # Seiichi Manyama, Aug 14 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 5 * (b(n-1)^3 - 3) mod 8^n for n > 1, then a(n) = (b(n+1) - b(n))/8^n. - Seiichi Manyama, Aug 14 2019

cp's OEIS Frontend

A309600 Digits of the 10-adic integer (17/9)^(1/3).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

Ruby

Formula

A225407 10-adic integer x such that x^3 = -3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Ruby

Formula

A309698 Digits of the 4-adic integer 3^(1/3).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Ruby

Formula

A306552 Expansion of the 10-adic cube root of -1/7, that is, the 10-adic integer solution to x^3 = -1/7.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

A306553 Expansion of the 10-adic cube root of -1/11, that is, the 10-adic integer solution to x^3 = -1/11.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

A306554 Expansion of the 10-adic cube root of 1/13, that is, the 10-adic integer solution to x^3 = 1/13.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

A306555 Expansion of the 10-adic cube root of -1/13, that is, the 10-adic integer solution to x^3 = -1/13.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs