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 A158625 Lower limit of backward value of 5^n. 4
 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 (list; constant; graph; refs; listen; history; text; internal format)
 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

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Last modified December 7 15:01 EST 2022. Contains 358667 sequences. (Running on oeis4.)