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

Author

Simon Plouffe, Mar 23 2009

Keywords

• cons
• nonn
• nice
• base

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

Examples

			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)
		

Links

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

Crossrefs

Cf. A158624, A004094, A023415, A145679.

Programs

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

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

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