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%I #15 Dec 27 2018 00:32:42
%S 9,576,36864,140625,2359296,150994944,2197265625,9663676416,
%T 618475290624,34332275390625,39582418599936,2533274790395904,
%U 162129586585337856,536441802978515625,10376293541461622784,664082786653543858176
%N Numbers n such that reciprocal of n terminates with an infinite repetition of digit 1. Multiples of 10 are omitted.
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 73-82.
%H <a href="/index/1#1overn">Index entries for sequences related to decimal expansion of 1/n</a>
%F These are the numbers 9*2^(6i) and 9*5^(6i). Proof: We have 1/n = (k+ (1/9))/ 10^m so n = 9*10^m/(9k+1) so 9 divides n, say n=9x and 10 does not divide x. Then x = 10^m/(9k+1), so 9k+1 = 2^r 5^s (say) and m = max(r, s). Also 9 | 2^r 5^s - 1 (*). But phi(9) = 6, 2^6 == 1 mod 9, 5^6 == 1 mod 9 so only solution to (*) is r == s mod 6. If r >= s, say r = s + 6i then n = 9*5^(6i). Similarly if s >= r, n = 9*2^(6i). - _N. J. A. Sloane_, Sep 21 2001
%e 1/36864 = 0.0000271267361111111111...
%o (ARIBAS): a064560(1,200000000,1,36). Definition of a064560: function a064560(minarg,maxarg,d,len: integer); var n,r,decpos,sign: integer; s,sd: string; begin set_floatprec(round(len*3.3)); r := len div 3; sd := alloc(string,r,ftoa(d)[0]); n := minarg - 1; while n < maxarg do inc(n); if n mod 10 <> 0 then s := float_ecvt(1/n,len,decpos,sign); if s[len-r-1..len-2] = sd then write(n," "); end; end; end; end;
%Y Cf. A064561-A064567.
%K nonn,base,easy
%O 1,1
%A _Patrick De Geest_, Sep 20 2001
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