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A225405 10-adic integer x such that x^3 = 7. 9
3, 4, 5, 1, 5, 0, 7, 1, 2, 2, 2, 4, 4, 2, 9, 6, 0, 7, 3, 5, 4, 5, 8, 8, 8, 0, 4, 1, 8, 5, 1, 4, 0, 0, 6, 1, 3, 5, 4, 4, 8, 1, 3, 7, 4, 0, 7, 4, 8, 5, 5, 1, 6, 7, 4, 5, 5, 0, 0, 4, 9, 0, 4, 7, 0, 8, 6, 8, 4, 7, 4, 4, 2, 2, 0, 3, 2, 2, 0, 1, 6, 5, 5, 4, 3, 0, 3, 4, 9, 7, 1, 5, 1, 2, 3, 0, 2, 5, 6, 8 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
LINKS
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 - 7) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019
EXAMPLE
3^3 == 7 (mod 10).
43^3 == 7 (mod 10^2).
543^3 == 7 (mod 10^3).
1543^3 == 7 (mod 10^4).
51543^3 == 7 (mod 10^5).
51543^3 == 7 (mod 10^6).
PROG
(PARI) n=0; for(i=1, 100, m=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 05 2019
(Ruby)
def A225405(n)
ary = [3]
a = 3
n.times{|i|
b = (a + 7 * (a ** 3 - 7)) % (10 ** (i + 2))
ary << (b - a) / (10 ** (i + 1))
a = b
}
ary
end
p A225405(100) # Seiichi Manyama, Aug 13 2019
CROSSREFS
Sequence in context: A113138 A010263 A011303 * A051993 A174530 A280490
KEYWORD
nonn,base
AUTHOR
Aswini Vaidyanathan, May 07 2013
STATUS
approved

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Last modified March 29 02:23 EDT 2024. Contains 371264 sequences. (Running on oeis4.)