cp's OEIS Frontend

A006889 Exponent of least power of 2 having n consecutive 0's in its decimal representation.

%I A006889 M4710 #69 Nov 07 2019 13:23:29
%S A006889 0,10,53,242,377,1491,1492,6801,14007,100823,559940,1148303,4036338,
%T A006889 4036339,53619497,119476156,146226201,918583174,4627233991,11089076233
%N A006889 Exponent of least power of 2 having n consecutive 0's in its decimal representation.
%C A006889 A224782(a(n)) = n and A224782(m) <> n for m < a(n). - _Reinhard Zumkeller_, Apr 30 2013
%C A006889 The name of this sequence previously began "Least power of 2 having exactly n consecutive 0's...". The word "exactly" was unnecessary because the least power of 2 having at least n consecutive 0's in its decimal representation will always have exactly n consecutive 0's. The previous power of two will have had n-1 consecutive 0's with a "5" immediately to the left. - _Clive Tooth_, Jan 22 2016
%C A006889 a(20) is greater than 12*10^9. - _Benjamin Chaffin_, Jan 18 2017
%D A006889 Julian Havil, Impossible?: Surprising Solutions to Counterintuitive Conundrums, Princeton University Press 2008, chapter 15, p. 176ff
%D A006889 Popular Computing (Calabasas, CA), Zeros in Powers of 2, Vol. 3 (No. 25, Apr 1975), page PC25-16 [Gives a(1)-a(8)]
%D A006889 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A006889 E. Karst and U. Karst, <a href="http://dx.doi.org/10.1090/S0025-5718-64-99163-X">The first power of 2 with 8 consecutive zeros</a>, Math. Comp. 18 (1964), 508-508.
%H A006889 Popular Computing (Calabasas, CA), <a href="/A094776/a094776.jpg">Two Tables</a>, Vol. 1, (No. 9, Dec 1973), page PC9-16.
%e A006889 2^53619497 is the smallest power of 2 to contain a run of 14 consecutive zeros in its decimal form.
%e A006889 2^119476156 (a 35965907-digit number) contains the sequence ...40030000000000000008341... about one third of the way through.
%e A006889 2^4627233991 (a 1392936229-digit number) contains the sequence "813000000000000000000538" about 99.5% of the way through. The computation took about six months.
%p A006889 A[0]:= 0:
%p A006889 m:= 1:
%p A006889 for n from 1 while m <= 9 do
%p A006889   S:= convert(2^n,string);
%p A006889   if StringTools:-Search(StringTools:-Fill("0",m),S) <> 0 then
%p A006889     A[m]:= n;
%p A006889     m:= m+1;
%p A006889   fi
%p A006889 od:
%p A006889 seq(A[i],i=0..9); # _Robert Israel_, Jan 22 2016
%t A006889 a = ""; Do[ a = StringJoin[a, "0"]; k = 1; While[ StringPosition[ ToString[2^k], a] == {}, k++ ]; Print[k], {n, 1, 10} ] (* _Robert G. Wilson v_, edited by _Clive Tooth_, Jan 25 2016 *)
%o A006889 (Haskell)
%o A006889 import Data.List (elemIndex)
%o A006889 import Data.Maybe (fromJust)
%o A006889 a006889 = fromJust . (`elemIndex` a224782_list)
%o A006889 -- _Reinhard Zumkeller_, Apr 30 2013
%o A006889 (PARI) conseczerorec(n) = my(d=digits(n), i=0, r=0, x=#Str(n)); while(x > 0, while(d[x]==0, i++; x--); if(i > r, r=i); i=0; x--); r
%o A006889 a(n) = my(k=0); while(conseczerorec(2^k) < n, k++); k \\ _Felix Fröhlich_, Sep 27 2016
%Y A006889 Cf. A031146.
%Y A006889 Cf. also A131535, A131536, A259089, A063565, A259091, A259092.
%K A006889 nonn,hard,base,more
%O A006889 0,2
%A A006889 P. D. Mitchelmore (dh115(AT)city.ac.uk)
%E A006889 3 more terms from _Clive Tooth_, Jan 24 2001
%E A006889 One more term from _Clive Tooth_, Nov 28 2001
%E A006889 One more term from Sacha Roscoe (scarletmanuka(AT)iprimus.com.au), Dec 16 2002
%E A006889 a(17) from Sacha Roscoe (scarletmanuka(AT)iprimus.com.au), Feb 06 2007
%E A006889 a(18) from _Clive Tooth_, Sep 30 2012
%E A006889 Name clarified by _Clive Tooth_, Jan 22 2016
%E A006889 Definition clarified by _Felix Fröhlich_, Sep 27 2016
%E A006889 a(19) from _Benjamin Chaffin_, Jan 18 2017