A225464 10-adic integer x such that x^9 = 1/9.
9, 0, 6, 5, 1, 4, 8, 8, 0, 6, 4, 1, 3, 1, 2, 9, 1, 6, 1, 5, 4, 7, 3, 1, 2, 7, 1, 7, 4, 1, 0, 1, 7, 9, 4, 9, 1, 2, 5, 6, 3, 3, 0, 5, 5, 6, 3, 7, 7, 1, 9, 7, 7, 6, 2, 4, 7, 9, 4, 4, 3, 0, 7, 1, 7, 2, 8, 2, 8, 9, 1, 9, 3, 6, 9, 1, 1, 1, 3, 2, 3, 2, 4, 2, 5, 0, 1, 8, 4, 4, 7, 8, 2, 9, 6, 8, 3, 9, 3, 0
Offset: 0
Examples
9^9 == -1 (mod 10). 9^9 == -11 (mod 10^2). 609^9 == -111 (mod 10^3). 5609^9 == -1111 (mod 10^4). 15609^9 == -11111 (mod 10^5). 415609^9 == -111111 (mod 10^6).
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000
Crossrefs
Cf. A225455.
Programs
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PARI
n=0;for(i=1,100,m=(8*(10^i-1)/9)+1;for(x=0,9,if(((n+(x*10^(i-1)))^9)%(10^i)==m,n=n+(x*10^(i-1));print1(x", ");break)))
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PARI
N=100; Vecrev(digits(lift(chinese(Mod((1/9+O(2^N))^(1/9), 2^N), Mod((1/9+O(5^N))^(1/9), 5^N)))), N) \\ Seiichi Manyama, Aug 07 2019
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Ruby
def A225464(n) ary = [9] a = 9 n.times{|i| b = (a + 9 * (9 * a ** 9 - 1)) % (10 ** (i + 2)) ary << (b - a) / (10 ** (i + 1)) a = b } ary end p A225464(100) # Seiichi Manyama, Aug 14 2019
Formula
Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 9 * (9 * b(n-1)^9 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 14 2019
Comments