A153042 -id:A153042 - 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.

A225412 Digits of the 10-adic integer (1/9)^(1/3).

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

This is the 10's complement of A153042.

Equivalently, the 10-adic cube root of 1/9, i.e., x such that 9*x^3 = 1 (mod 10^n) for all n. - M. F. Hasler, Jan 02 2019

			       9^3 == -1      (mod 10).
      29^3 == -11     (mod 10^2).
     529^3 == -111    (mod 10^3).
    1529^3 == -1111   (mod 10^4).
   11529^3 == -11111  (mod 10^5).
  711529^3 == -111111 (mod 10^6).

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

Cf. A309600, A319740 (10-adic cube root of 1/11).
Digits of 10-adic integers:
A153042 ((-1/9)^(1/3));
A225406 (     9^(1/3));
A225409 (  (-9)^(1/3)).

op([1,3],padic:-rootp(9*x^3  -1,  10, 101)); # Robert Israel, Aug 04 2019

n=0; for(i=1, 100, m=(8*(10^i-1)/9)+1; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))

upto(N=100, m=1/3)=Vecrev(digits(lift(chinese(Mod((1/9+O(5^N))^m, 5^N), Mod((1/9+O(2^N))^m, 2^N)))), N) \\ Following Andrew Howroyd's code for A319740. - M. F. Hasler, Jan 02 2019

Vecrev(digits(truncate(-(-1/9+O(10^100))^(1/3)))) \\ Seiichi Manyama, Aug 04 2019

def A225412(n)
  ary = [9]
  a = 9
  n.times{|i|
    b = (a + 7 * (9 * a ** 3 - 1)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225412(100) # Seiichi Manyama, Aug 13 2019

p = ...711529, q = A153042 = ...288471, p + q = 0. - Seiichi Manyama, Aug 04 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019

Name edited by Seiichi Manyama, Aug 04 2019

A225406 Digits of the 10-adic integer 9^(1/3).

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

			       9^3 == 9 (mod 10).
      69^3 == 9 (mod 10^2).
     569^3 == 9 (mod 10^3).
     569^3 == 9 (mod 10^4).
   60569^3 == 9 (mod 10^5).
  660569^3 == 9 (mod 10^6).

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A309600.
Digits of 10-adic integers:
A153042 ((-1/9)^(1/3));
A225409 (  (-9)^(1/3));
A225412 ( (1/9)^(1/3)).

op([1,3],padic:-rootp((x)^3  -9,  10, 101)); # Robert Israel, Aug 04 2019

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

Vecrev(digits(truncate(-(-9+O(10^100))^(1/3)))) \\ Seiichi Manyama, Aug 04 2019

N=100; Vecrev(digits(lift(chinese(Mod((9+O(2^N))^(1/3), 2^N), Mod((9+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 04 2019

def A225406(n)
  ary = [9]
  a = 9
  n.times{|i|
    b = (a + 3 * (a ** 3 - 9)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225406(100) # Seiichi Manyama, Aug 13 2019

p = ...660569, q = A225409 = ...339431, p + q = 0. - Seiichi Manyama, Aug 04 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 3 * (b(n-1)^3 - 9) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019

A225409 Digits of the 10-adic integers (-9)^(1/3).

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

This is the 10's complement of A225406.

			       1^3 == -9 (mod 10).
      31^3 == -9 (mod 10^2).
     431^3 == -9 (mod 10^3).
    9431^3 == -9 (mod 10^4).
   39431^3 == -9 (mod 10^5).
  339431^3 == -9 (mod 10^6).

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

Cf. A309600,
Digits of 10-adic integers:
A153042 ((-1/9)^(1/3));
A225406 (     9^(1/3));
A225412 ( (1/9)^(1/3)).

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

Vecrev(digits(truncate((-9+O(10^100))^(1/3)))) \\ Seiichi Manyama, Aug 04 2019

N=100; Vecrev(digits(lift(chinese(Mod((-9+O(2^N))^(1/3), 2^N), Mod((-9+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 04 2019

def A225409(n)
  ary = [1]
  a = 1
  n.times{|i|
    b = (a + 3 * (a ** 3 + 9)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225409(100) # Seiichi Manyama, Aug 13 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 1, b(n) = b(n-1) + 3 * (b(n-1)^3 + 9) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 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.

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

A225412 Digits of the 10-adic integer (1/9)^(1/3).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

PARI

PARI

Ruby

Formula

Extensions

A225406 Digits of the 10-adic integer 9^(1/3).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

PARI

PARI

PARI

Ruby

Formula

A225409 Digits of the 10-adic integers (-9)^(1/3).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

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