A225407 10-adic integer x such that x^3 = -3.
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
Examples
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).
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000
Programs
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PARI
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)))
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PARI
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
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Ruby
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
Formula
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
Comments