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 A158625 Lower limit of backward value of 5^n. 4

%I

%S 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,

%T 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,

%U 1,3,3,2,0,1,1,4,4,2,0,1,4,2,4,1,2,0,4

%N Lower limit of backward value of 5^n.

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

%C The upper limit is A158624, 0.5265679578796997657885576975995789586775656...

%H Jon E. Schoenfield, <a href="/A158625/b158625.txt">Table of n, a(n) for n = 1..3000</a>

%e 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...

%e From _N. J. A. Sloane_, May 11 2018 (Start)

%e To describe this another way: write down the decimal expansion of powers of 5:

%e 1

%e 5

%e 25

%e 125

%e 625

%e 3125

%e ...

%e keep going forever.

%e Write them backwards:

%e 1

%e 5

%e 52

%e 526

%e 5213

%e ...

%e After a while the beginning digits are all the same.

%e That's the sequence. (End)

%o (Python)

%o # lower limit of backward sequence of 5^n

%o a,i=5,0; x=a

%o while 1:

%o i+=1; print x,',',

%o for j in range(5):

%o if (a+j*10**i)%(5**(i+1))==0: x=j; a+=j*10**i

%o # _Cezary Glowacz_, Mar 11 2017

%o (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

%Y Cf. A158624, A004094, A023415, A145679.

%K cons,nonn,nice,base

%O 1,1

%A _Simon Plouffe_, Mar 23 2009

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Last modified January 20 21:36 EST 2019. Contains 319336 sequences. (Running on oeis4.)