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A225408

10-adic integer x=.....8457 satisfying x^3 = -7.

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

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OFFSET

0,1

COMMENTS

This is the 10's complement of A225405.

LINKS

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

FORMULA

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

EXAMPLE

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).

PROG

(PARI) 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)))

(PARI) 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

(Ruby)

def A225408(n)