A225402 -id:A225402 - cp's OEIS frontend

A352282 Indices k where A225402(k) = 0.

17, 56, 68, 71, 75, 77, 79, 93, 94, 103, 104, 110, 116, 118, 119, 124, 151, 167, 186, 197, 222, 238, 239, 249, 253, 262, 263, 277, 283, 298, 303, 317, 325, 341, 343, 345, 348, 351, 362, 363, 367, 370, 373, 384, 385, 391, 398, 404, 405, 411, 420, 425, 428, 430, 445, 451, 475, 489

Offset: 1

Michel Marcus, Apr 25 2022

These are also the indices k such that A352995(k+1) < A352995(k).

Cf. A225402, A352995.

lista(nn) = {my(n=0, list = List()); for(i=1, nn, m=(10^i-1)/9; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); if (x==0, listput(list, i-1)); break; ); ); ); Vec(list); }

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

Original entry on oeis.org

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.

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

A319740 The 10-adic integer cube root of one eleventh (1/11), that is, satisfying 11 * x^3 == 1 (mod 10^n), for all n.

Original entry on oeis.org

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

Offset: 1

Patrick A. Thomas, Sep 26 2018

			45016103979758167131^3 * 11 == 1 (mod 10^20).

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

Cf. A225402, A225411, A225412 (10-adic cube root of -1/3, 1/3, 1/9).

op([1,3], padic:-rootp(11*x^3-1,10,100)); # Robert Israel, Jan 03 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)} \\ Andrew Howroyd, Nov 26 2018

Terms a(56) and beyond from Andrew Howroyd, Nov 26 2018

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

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.

A352995 Smallest positive integer whose cube ends with exactly n 3's.

Original entry on oeis.org

1, 7, 77, 477, 6477, 46477, 446477, 5446477, 85446477, 385446477, 4385446477, 44385446477, 644385446477, 8644385446477, 38644385446477, 138644385446477, 5138644385446477, 115138644385446477, 15138644385446477, 5015138644385446477

Offset: 0

Bernard Schott, Apr 24 2022

nonn

When A225402(k) = 0, i.e., k is a term of A352282, then a(k) > a(k+1); 1st example is for k = 17 with a(17) = 115138644385446477 > a(18) = 15138644385446477; otherwise, a(n) < a(n+1).
When n <> k, a(n) coincides with the 'backward concatenation' of A225402(n-1) up to A225402(0), where A225402 is the 10-adic integer x such that x^3 = -1/3 (see table in Example section); when n = k, a(k) must be calculated directly with the definition.
Without "exactly" in the name, terms a'(n) should be also: 1, 7, 77, 477, 6477, 46477, 446477, ..., first difference arrives for n = 17.
There are similar sequences when cubes end with 1, 7, 8 or 9.

			a(0) = 1 because 1^3 = 1;
a(1) = 7 because 7^3 = 343;
a(2) = 77 because 77^3 = 456533;
a(3) = 477 because 477^3 = 108531333;
  ------------------------------------------------------------------------------
  |    |     a(n)          |      a'(n)        | A225402(n-1) | concatenation  |
  | n  | with "exactly"    | without "exactly" |  = b(n-1)    | b(n-1)...b(0)  |
  ------------------------------------------------------------------------------
    1                    7                    7     7                      ...7
    2                   77                   77     7                     ...77
    3                  477                  477     4                    ...477
   ............................................................................
   15      138644385446477      138644385446477     1        ...138644385446477
   16     5138644385446477     5138644385446477     5       ...5138644385446477
   17   115138644385446477    15138644385446477     1      ...15138644385446477
   18    15138644385446477    15138644385446477     0     ...015138644385446477
   19  5015138644385446477  5015138644385446477     5    ...5015138644385446477
  ------------------------------------------------------------------------------

Robert Israel, Table of n, a(n) for n = 0..996

Cf. A225402, A352282, A352992 (similar, with 7).

f:= proc(n) local t,x;
       t:= 3/9*(10^n-1);
       x:= rhs(op(msolve(x^3=t,10^n)));
       while x^3 mod 10^(n+1) = 10*t+3 do x:= x + 10^n od;
       x
end proc:
f(0):= 1:
map(f, [$0..20]); # Robert Israel, Jul 29 2025

When n is not in A352282, a(n) = Sum_{k=0..n-1} A225402(k) * 10^k.

cp's OEIS Frontend

A352282 Indices k where A225402(k) = 0.

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A309600 Digits of the 10-adic integer (17/9)^(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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A319740 The 10-adic integer cube root of one eleventh (1/11), that is, satisfying 11 * x^3 == 1 (mod 10^n), for all n.

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

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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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