|
%I #48 Feb 07 2022 01:40:51
%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 i < 100:
%o i+=1; print(x, end=',')
%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
|
