A225409 Digits of the 10-adic integers (-9)^(1/3).
1, 3, 4, 9, 3, 3, 6, 5, 1, 3, 3, 9, 5, 1, 4, 5, 4, 0, 5, 4, 8, 8, 0, 5, 9, 3, 9, 1, 8, 6, 2, 9, 3, 3, 0, 5, 1, 6, 0, 0, 6, 9, 7, 5, 7, 9, 6, 4, 0, 1, 3, 4, 4, 9, 0, 3, 2, 2, 5, 1, 9, 2, 5, 3, 8, 9, 6, 7, 0, 1, 4, 1, 7, 8, 4, 2, 9, 0, 9, 0, 1, 1, 8, 3, 9, 3, 1, 3, 9, 6, 0, 4, 9, 0, 0, 4, 3, 4, 6, 2
Offset: 0
Examples
1^3 == -9 (mod 10).
31^3 == -9 (mod 10^2).
431^3 == -9 (mod 10^3).
9431^3 == -9 (mod 10^4).
39431^3 == -9 (mod 10^5).
339431^3 == -9 (mod 10^6).
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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PARI
n=0; for(i=1, 100, m=(10^i-9); 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
Vecrev(digits(truncate((-9+O(10^100))^(1/3)))) \\ Seiichi Manyama, Aug 04 2019
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PARI
N=100; Vecrev(digits(lift(chinese(Mod((-9+O(2^N))^(1/3), 2^N), Mod((-9+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 04 2019
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Ruby
def A225409(n) ary = [1] a = 1 n.times{|i| b = (a + 3 * (a ** 3 + 9)) % (10 ** (i + 2)) ary << (b - a) / (10 ** (i + 1)) a = b } ary end p A225409(100) # Seiichi Manyama, Aug 13 2019
Formula
Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 1, b(n) = b(n-1) + 3 * (b(n-1)^3 + 9) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019
Comments