

A309645


Digits of the 10adic integer (11/9)^(1/3).


3



9, 5, 8, 6, 6, 5, 0, 7, 6, 0, 5, 1, 0, 8, 9, 4, 8, 5, 8, 5, 6, 2, 2, 5, 2, 1, 8, 7, 6, 9, 9, 8, 2, 3, 8, 1, 0, 5, 8, 5, 7, 0, 0, 9, 6, 7, 4, 2, 0, 6, 6, 7, 7, 1, 2, 4, 1, 7, 9, 1, 2, 5, 4, 8, 7, 7, 3, 4, 4, 9, 1, 7, 6, 9, 6, 7, 0, 7, 4, 1, 3, 3, 6, 9, 7, 4, 6, 9, 8, 8, 5, 9, 0, 0, 1, 5, 4, 0, 5, 4
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OFFSET

0,1


LINKS

Seiichi Manyama, Table of n, a(n) for n = 0..10000


FORMULA

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n1) + 7 * (9 * b(n1)^3  11) mod 10^n for n > 1, then a(n) = (b(n+1)  b(n))/10^n.


EXAMPLE

9^3 == 9 (mod 10).
59^3 == 79 (mod 10^2).
859^3 == 779 (mod 10^3).
6859^3 == 7779 (mod 10^4).
66859^3 == 77779 (mod 10^5).
566859^3 == 777779 (mod 10^6).


PROG

(PARI) N=100; Vecrev(digits(lift(chinese(Mod((11/9+O(2^N))^(1/3), 2^N), Mod((11/9+O(5^N))^(1/3), 5^N)))), N)
(Ruby)
def A309645(n)
ary = [9]
a = 9
n.times{i
b = (a + 7 * (9 * a ** 3  11)) % (10 ** (i + 2))
ary << (b  a) / (10 ** (i + 1))
a = b
}
ary
end
p A309645(100)


CROSSREFS

Cf. A309600, A309603.
Sequence in context: A245292 A203081 A329715 * A146483 A090463 A272795
Adjacent sequences: A309642 A309643 A309644 * A309646 A309647 A309648


KEYWORD

nonn,base


AUTHOR

Seiichi Manyama, Aug 11 2019


STATUS

approved



