A158625 Lower limit of backward value of 5^n.

5, 2, 1, 3, 0, 2, 3, 3, 0, 4, 3, 1, 1, 3, 1, 1, 2, 4, 2, 1, 0, 3, 1, 3, 3, 0, 0, 0, 2, 3, 1, 4, 1, 0, 2, 1, 0, 3, 4, 3, 0, 2, 1, 2, 2, 1, 1, 4, 4, 3, 4, 0, 2, 0, 4, 0, 2, 2, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 3, 3, 2, 0, 1, 1, 4, 4, 2, 0, 1, 4, 2, 4, 1, 2, 0, 4

Offset

1,1

Comments

Digits are all in {0,1,2,3,4} after the first term.

The upper limit is A158624, 0.5265679578796997657885576975995789586775656...

Links

Jon E. Schoenfield, Table of n, a(n) for n = 1..3000

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=', ')
     for j in range(5):
         if (a+j*10**i)%(5**(i+1))==0: x=j; a+=j*10**i
# Cezary Glowacz, Mar 11 2017
(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

Cf. A158624, A004094, A023415, A145679.

Sequence in context: A243258 A275704 A038631 * A133615 A136161 A350688

Adjacent sequences: A158622 A158623 A158624 * A158626 A158627 A158628

Keyword

cons,nonn,nice,base

Author

Simon Plouffe, Mar 23 2009

Status

approved