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A309601
Digits of the 10-adic integer (53/9)^(1/3).
3
3, 7, 3, 8, 6, 7, 0, 5, 3, 0, 8, 5, 3, 4, 8, 1, 3, 0, 9, 0, 3, 2, 9, 2, 3, 6, 3, 2, 4, 3, 5, 1, 5, 2, 9, 8, 0, 7, 6, 0, 3, 9, 9, 4, 2, 5, 3, 0, 3, 2, 0, 3, 2, 8, 2, 8, 7, 8, 3, 1, 0, 0, 4, 6, 4, 1, 8, 9, 4, 8, 5, 3, 5, 3, 7, 3, 1, 6, 7, 9, 1, 1, 8, 5, 0, 2, 5, 7, 6, 3, 8, 9, 4, 2, 7, 3, 0, 3, 6, 6
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) + 3 * (9 * b(n-1)^3 - 53) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.
EXAMPLE
3^3 == 7 (mod 10).
73^3 == 17 (mod 10^2).
373^3 == 117 (mod 10^3).
8373^3 == 1117 (mod 10^4).
68373^3 == 11117 (mod 10^5).
768373^3 == 111117 (mod 10^6).
PROG
(PARI) N=100; Vecrev(digits(lift(chinese(Mod((53/9+O(2^N))^(1/3), 2^N), Mod((53/9+O(5^N))^(1/3), 5^N)))), N)
(Ruby)
def A309601(n)
ary = [3]
a = 3
n.times{|i|
b = (a + 3 * (9 * a ** 3 - 53)) % (10 ** (i + 2))
ary << (b - a) / (10 ** (i + 1))
a = b
}
ary
end
p A309601(100)
CROSSREFS
Cf. A309600.
Sequence in context: A118452 A286090 A225447 * A201903 A133056 A131917
KEYWORD
nonn,base
AUTHOR
Seiichi Manyama, Aug 09 2019
STATUS
approved