A309600 -id:A309600 - 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

A225401 10-adic integer x such that x^3 = -7/9.

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

			       3^3 == 7      (mod 10).
      53^3 == 77     (mod 10^2).
     753^3 == 777    (mod 10^3).
     753^3 == 7777   (mod 10^4).
   60753^3 == 77777  (mod 10^5).
  660753^3 == 777777 (mod 10^6).

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

Cf. A225410, A309600.

b:= proc(n) option remember; `if`(n<2, 3*n,
      irem(b(n-1)+3*(9*b(n-1)^3+7), 10^n))
    end:
a:= n-> (b(n+1)-b(n))/10^n:
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 14 2022

n=0;for(i=1,100,m=7*(10^i-1)/9;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((-7/9+O(2^N))^(1/3), 2^N), Mod((-7/9+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 05 2019

def A225401(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 + 7)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225401(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) + 3 * (9 * 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

A225405 10-adic integer x such that x^3 = 7.

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

			      3^3 == 7 (mod 10).
     43^3 == 7 (mod 10^2).
    543^3 == 7 (mod 10^3).
   1543^3 == 7 (mod 10^4).
  51543^3 == 7 (mod 10^5).
  51543^3 == 7 (mod 10^6).

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

Cf. A225408, A309600.

n=0; for(i=1, 100, m=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 05 2019

def A225405(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 7 * (a ** 3 - 7)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225405(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 - 7) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 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

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

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

A225410 10-adic integer x such that x^3 = 7/9.

Original entry on oeis.org

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

Offset: 0

Aswini Vaidyanathan, May 07 2013

This is the 10's complement of A225401.

			       7^3 == 3      (mod 10).
      47^3 == 23     (mod 10^2).
     247^3 == 223    (mod 10^3).
    9247^3 == 2223   (mod 10^4).
   39247^3 == 22223  (mod 10^5).
  339247^3 == 222223 (mod 10^6).

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

Cf. A225401, A309600.

n=0; for(i=1, 100, m=(2*(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)))

N=100; Vecrev(digits(lift(chinese(Mod((7/9+O(2^N))^(1/3), 2^N), Mod((7/9+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 05 2019

def A225410(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 - 7)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225410(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) + 3 * (9 * 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

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

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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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A225401 10-adic integer x such that x^3 = -7/9.

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A225405 10-adic integer x such that x^3 = 7.

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

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