The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A309605

(Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

A309605

Digits of the 10-adic integer (61/9)^(1/3).
(history; published version)

#19 by Giovanni Resta at Mon Aug 12 02:40:04 EDT 2019

STATUS

reviewed

approved

#18 by Joerg Arndt at Mon Aug 12 01:26:40 EDT 2019

STATUS

proposed

reviewed

#17 by Seiichi Manyama at Sun Aug 11 14:21:04 EDT 2019

STATUS

editing

proposed

#16 by Seiichi Manyama at Sun Aug 11 14:20:57 EDT 2019

CROSSREFS

Cf. A309600, A309643.

STATUS

proposed

editing

#15 by Seiichi Manyama at Sun Aug 11 14:19:45 EDT 2019

STATUS

editing

proposed

#14 by Seiichi Manyama at Sun Aug 11 14:19:38 EDT 2019

LINKS

Seiichi Manyama, <a href="/A309605/b309605.txt">Table of n, a(n) for n = 0..10000</a>

#13 by Seiichi Manyama at Sun Aug 11 14:14:37 EDT 2019

PROG

(Ruby)

def A309605(n)

ary = [9]

a = 9

n.times{|i|

b = (a + 7 * (9 * a ** 3 - 61)) % (10 ** (i + 2))

ary << (b - a) / (10 ** (i + 1))

a = b

}

ary

end

p A309605(100)

#12 by Seiichi Manyama at Sun Aug 11 11:24:03 EDT 2019

FORMULA

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 - 61) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.

b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 - 61) mod 10^n for n > 1.

a(n) = (b(n+1) - b(n))/10^n.

#11 by Seiichi Manyama at Sun Aug 11 09:39:04 EDT 2019

NAME

Digits of the 10-adic integers integer (61/9)^(1/3).

#10 by Seiichi Manyama at Sun Aug 11 09:34:27 EDT 2019

FORMULA

b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 3 7 * (9 * b(n-1)^3 - 61) mod 10^n for n > 1.