A319740 -id:A319740 - cp's OEIS frontend

A225402 10-adic cube root of -1/3.

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

Previous name was: 10-adic integer x such that x^3 == (10^(n+1)-1)/3 mod 10^(n+1) for n >= 0.

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

			       7^3 == 3      (mod 10).
      77^3 == 33     (mod 10^2).
     477^3 == 333    (mod 10^3).
    6477^3 == 3333   (mod 10^4).
   46477^3 == 33333  (mod 10^5).
  446477^3 == 333333 (mod 10^6).

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

Cf. A225411, A309600, A319740 (10-adic cube root of 1/11).

b:= proc(n) option remember; `if`(n<2, 7*n,
      irem(b(n-1)+9*(3*b(n-1)^3+1), 10^n))
    end:
a:= n-> (b(n+1)-b(n))/10^n:
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 15 2022

n=0;for(i=1,100,m=(10^i-1)/3;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((-m+O(5^N))^m, 5^N), Mod((-m+O(2^N))^m, 2^N)))), N) \\ Following Andrew Howroyd's code for A319740. - M. F. Hasler, Jan 02 2019

def A225402(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 9 * (3 * a ** 3 + 1)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225402(100) # Seiichi Manyama, Aug 12 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, b(n) = b(n-1) + 9 * (3 * b(n-1)^3 + 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 12 2019

New name (using M. F. Hasler's comment) from Joerg Arndt, Aug 08 2019

A225411 10-adic integer x such that x^3 == (6*(10^(n+1)-1)/9)+1 mod 10^(n+1) for all n.

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

This is the 10's complement of A225402.

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

			       3^3 ==      7 (mod 10).
      23^3 ==     67 (mod 10^2).
     523^3 ==    667 (mod 10^3).
    3523^3 ==   6667 (mod 10^4).
   53523^3 ==  66667 (mod 10^5).
  553523^3 == 666667 (mod 10^6).

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

Cf. A225402, A309600, A319740 (10-adic cube root of 1/11).

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

n=0; for(i=1, 100, m=(2*(10^i-1)/3)+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((m+O(5^N))^m, 5^N), Mod((m+O(2^N))^m, 2^N)))),N) \\ Following Andrew Howroyd's code for A319740. - M. F. Hasler, Jan 02 2019

def A225411(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 9 * (3 * a ** 3 - 1)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225411(100) # Seiichi Manyama, Aug 12 2019

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 9 * (3 * b(n-1)^3 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 12 2019

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

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.

cp's OEIS Frontend

A225402 10-adic cube root of -1/3.

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A225411 10-adic integer x such that x^3 == (6*(10^(n+1)-1)/9)+1 mod 10^(n+1) for all n.

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

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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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