A225405 -id:A225405 - 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.

A225408 10-adic integer x=.....8457 satisfying x^3 = -7.

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

This is the 10's complement of A225405.

			       7^3 == -7 (mod 10).
      57^3 == -7 (mod 10^2).
     457^3 == -7 (mod 10^3).
    8457^3 == -7 (mod 10^4).
   48457^3 == -7 (mod 10^5).
  948457^3 == -7 (mod 10^6).

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

Cf. A091663, A225405, A309600.

n=0; for(i=1, 100, m=(10^i-7); 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((-7+O(2^N))^(1/3), 2^N), Mod((-7+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 04 2019

def A225408(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 7 * (a ** 3 + 7)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225408(100) # Seiichi Manyama, Aug 13 2019

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

A319739 The 10-adic integer cube root of one seventh (1/7), that is, satisfying 7 * x^3 == 1 (mod 10^(n+1)), for all n.

Original entry on oeis.org

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

Offset: 0

Patrick A. Thomas, Sep 26 2018

			25380616954407^3 * 7 == 1 (mod 10^14).

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

Digits of 10-adic integers:
A225405       (     7^(1/3));
A225411       ( (1/3)^(1/3));
A225412       ( (1/9)^(1/3));
A225451       ( (1/3)^(1/7));
this sequence ( (1/7)^(1/3));
A319740       ((1/11)^(1/3)).

seq(n)={my(v=vector(n), t=0, b=1); for(i=1, #v, for(q=0, 9, if(lift(7*Mod(t, 10*b)^3)==1, v[i]=q; break); t+=b); b*=10); v} \\ Andrew Howroyd, Nov 26 2018

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)} \\ Andrew Howroyd, Nov 26 2018

a(55)-a(89) from Andrew Howroyd, Nov 26 2018
a(90)-a(199) from Patrick A. Thomas, Jan 13 2019
Offset changed to 0 by Seiichi Manyama, Aug 17 2019

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.

A322933 Digits of the 8-adic integer 7^(1/3).

Original entry on oeis.org

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

Offset: 0

Patrick A. Thomas, Dec 31 2018

The octal version of A225405.

			56027726627^3 == 7 (mod 8^11) in octal.

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

Cf. A225405 (decimal version), A322931, A322932.

N=100; Vecrev(digits(lift((7+O(2^(3*N)))^(1/3)), 8), N) \\ Seiichi Manyama, Aug 14 2019

def A322933(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 5 * (a ** 3 - 7)) % (8 ** (i + 2))
    ary << (b - a) / (8 ** (i + 1))
    a = b
  }
  ary
end
p A322933(100) # Seiichi Manyama, Aug 14 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, b(n) = b(n-1) + 5 * (b(n-1)^3 - 7) 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).

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A225408 10-adic integer x=.....8457 satisfying x^3 = -7.

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A319739 The 10-adic integer cube root of one seventh (1/7), that is, satisfying 7 * x^3 == 1 (mod 10^(n+1)), for all n.

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

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

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

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

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