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
Offset: 0
Examples
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).
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-7); 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((-7+O(2^N))^(1/3), 2^N), Mod((-7+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 04 2019
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Ruby
def A225408(n) ary = [7] a = 7 n.times{|i| b = (a + 7 * (a ** 3 + 7)) % (10 ** (i + 2)) ary << (b - a) / (10 ** (i + 1)) a = b } ary end p A225408(100) # Seiichi Manyama, Aug 13 2019
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
Comments