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A064560 Numbers n such that reciprocal of n terminates with an infinite repetition of digit 1. Multiples of 10 are omitted. 8
9, 576, 36864, 140625, 2359296, 150994944, 2197265625, 9663676416, 618475290624, 34332275390625, 39582418599936, 2533274790395904, 162129586585337856, 536441802978515625, 10376293541461622784, 664082786653543858176 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 73-82.
LINKS
FORMULA
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
EXAMPLE
1/36864 = 0.0000271267361111111111...
PROG
(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;
CROSSREFS
Sequence in context: A347844 A291910 A074731 * A264121 A095968 A233067
KEYWORD
nonn,base,easy
AUTHOR
Patrick De Geest, Sep 20 2001
STATUS
approved

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Last modified May 25 16:17 EDT 2024. Contains 372801 sequences. (Running on oeis4.)