OFFSET
1,1
COMMENTS
Digits are all in {0,1,2,3,4} after the first term.
The upper limit is A158624, 0.5265679578796997657885576975995789586775656...
The sequence is not eventually periodic. Assuming any period results in a condition a(1)=0 mod 10 which contradicts a(1)=5. - Cezary Glowacz, Jul 22 2024
LINKS
Jon E. Schoenfield, Table of n, a(n) for n = 1..3000
FORMULA
a(n) >= 0 and is the minimum satisfying (Sum_{i=1..n} a(i)*10^(i-1)) == 0 (mod 5^n), for n >= 2. - Cezary Glowacz, Jul 24 2024
EXAMPLE
5^3 = 125 so the backward value is 0.521, 5^10 = 9765625, so backward value is 0.5265679. The lower limit of all values is a constant, which appears to be 0.521302330431131124210313300023141021034302...
From N. J. A. Sloane, May 11 2018: (Start)
To describe this another way: write down the decimal expansion of powers of 5:
1
5
25
125
625
3125
...
keep going forever.
Write them backwards:
1
5
52
526
5213
...
After a while the beginning digits are all the same.
That's the sequence. (End)
PROG
(Python)
# lower limit of backward sequence of 5^n
a, i=5, 0; x=a
while i < 100:
i+=1; print(x, end=', ')
x=(-a//pow(5, i)*pow(3, i))%5; a+=x*pow(10, i)
# Cezary Glowacz, Jul 29 2024
(Magma) D:=87; e:=6; for d in [2..D-1] do t:=Modexp(5, e, 10^(d+1)); if t div 10^d ge 5 then e+:=2^(d-2); end if; end for; t:=Modexp(5, e, 10^D); IntegerToSequence(t, 10); // Jon E. Schoenfield, Feb 05 2018
CROSSREFS
KEYWORD
AUTHOR
Simon Plouffe, Mar 23 2009
STATUS
approved