A225401 -id:A225401 - cp's OEIS frontend

A353003 Indices k where A225401(k) = 0.

3, 23, 26, 32, 42, 46, 48, 51, 54, 84, 100, 101, 119, 123, 136, 184, 185, 202, 206, 216, 241, 263, 265, 272, 273, 284, 293, 323, 325, 332, 351, 352, 362, 392, 400, 406, 408, 410, 425, 432, 447, 449, 450, 466, 484, 488, 490, 497, 498, 509, 512, 522, 532, 534, 573, 585, 593, 595

Offset: 1

Michel Marcus, Apr 15 2022

These are also the indices k such that A352992(k+1) < A352992(k).

			3 is a term because A225401(3) = 0; and we have A352992(4)=753 < A352992(3)=1753.

Cf. A225401, A352992.

lista(nn) = {my(n=0, list = List()); for(i=1, nn, 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)); if (x==0, listput(list, i-1)); break;););); Vec(list);} \\ after A225401

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.

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

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.

A352992 Smallest positive integer whose cube ends with exactly n 7's.

Original entry on oeis.org

1, 3, 53, 1753, 753, 60753, 660753, 9660753, 99660753, 899660753, 3899660753, 33899660753, 233899660753, 7233899660753, 97233899660753, 497233899660753, 1497233899660753, 31497233899660753, 631497233899660753, 9631497233899660753, 59631497233899660753, 559631497233899660753

Offset: 0

Bernard Schott, Apr 14 2022

When A225401(k) = 0, i.e. k is a term of A353003, then a(k) > a(k+1); 1st example is for k = 3 with a(3) = 1753 > a(4) = 753; otherwise, a(n) < a(n+1).
When n <> k, a(n) coincides with the 'backward concatenation' of A225401(n-1) up to A225401(0), where A225401 is the 10-adic integer x such that x^3 = -7/9 (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: 1, 3, 53, 753, 753, 60753, 660753, ...
There are similar sequences when cubes end with 1, 3, 8 or 9; but there's no similar sequence for squares, because when a square ends in more than three identical digits, these digits are necessarily 0.

			a(1) = 3 because 3^3 = 27;
a(2) = 53 because 53^2 = 148877;
a(3) = 1753 because 1753^3 = 5386984777;
a(4) = 753 because 753^2 = 426957777;
a(5) = 60753 because 60753^3 = 224234888577777.
Table with a(n) and A225401(n-1)
   ---------------------------------------------------------------------------
   |    |     a(n)       |      a'(n)        | A225401(n-1) |  concatenation |
   | n  | with "exactly" | without "exactly" |  = b(n-1)    |  b(n-1)...b(0) |
   ---------------------------------------------------------------------------
     0           1                 1
     1           3                 3               3                  ...3
     2          53                53               5                 ...53
     3        1753               753               7                ...753
     4         753               753               0               ...0753
     5       60753             60753               6              ...60753
     6      660753            660753               6             ...660753
     7     9660753           9660753               9            ...9660753
  ..........................................................................
Also, as A225401(23) = 0, we have from a(21) up to a(25):
a(21) =     559631497233899660753;
a(22) =    3559631497233899660753;
a(23) =  193559631497233899660753, found by _Marius A. Burtea_;
a(24) =   93559631497233899660753;
a(25) = 2093559631497233899660753.

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

Cf. A000578, A017307, A039685, A225401, A353003.

f:= proc(n) local t,x;
       t:= 7/9*(10^n-1);
       x:= rhs(op(msolve(x^3=t,10^n)));
       while x^3 mod 10^(n+1) = 10*t+7 do x:= x + 10^n od;
       x
end proc:
f(0):= 1:
map(f, [$0..30]); # Robert Israel, Jul 29 2025

def a(n):
    k, s, target = 1, "1", "7"*n
    while s.rstrip("7") + target != s: k += 1; s = str(k**3)
    return k
print([a(n) for n in range(8)]) # Michael S. Branicky, Apr 14 2022

When n is not in A353003, a(n) = Sum_{k=0..n-1} A225401(k) * 10^k.

a(8)-a(9) from Marius A. Burtea, Apr 14 2022

cp's OEIS Frontend

A353003 Indices k where A225401(k) = 0.

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

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

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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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