A306554 Expansion of the 10-adic cube root of 1/13, that is, the 10-adic integer solution to x^3 = 1/13.

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

Author

Jianing Song, Feb 23 2019

Keywords

• nonn
• base

Comments

10's complement of A306555.

Examples

			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.
		

Links

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

Crossrefs

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).

Programs

Maple

op([1,3],padic:-rootp(13*x^3-1,10,100)); # Robert Israel, Mar 24 2019

PARI

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.

Formula

a(n) = 9 - A306555(n) for n >= 2.