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A100468
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Numbers n such that n appears in the decimal representation of 1/n.
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1
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1, 3, 6, 7, 10, 14, 17, 28, 58, 59, 83, 86, 87, 89, 97, 100, 118, 167, 197, 228, 281, 313, 316, 339, 367, 379, 383, 456, 458, 469, 529, 541, 543, 569, 577, 587, 593, 607, 618, 626, 629, 647, 669, 673, 677, 678, 683, 687, 701, 709, 719, 722, 727, 729, 767, 771, 772
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OFFSET
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1,2
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COMMENTS
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Terminating decimals are zero-extended, so that 10 appears in 0.100... = 1/10.
Let n >= 2 and let m = 10^floor(log_10(n)+1) be the smallest power of 10 > n. Then n is in this sequence iff ceiling(n^2/m) < ceiling((n^2+n)/m) and there exists k >= 0 with 10^k == ceiling(n^2/m) (mod n). - David W. Wilson, Dec 14 2006
As n increases, the average number of digits in the decimal expansion of 1/n grows fairly steadily. It takes a jump at powers of ten. So the density of the present sequence grows steadily until a new power of ten is reached, when there is an abrupt drop. - Franklin T. Adams-Watters, Dec 17 2006
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LINKS
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EXAMPLE
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a(6)=17 because 1/17 = ".0588235294117647" which contains "17"
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PROG
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(PARI) f1(n, prec) = for(x=1, n, default(realprecision, prec); a=1./x; if(countmatch(a, x), print1(x", "))) countmatch(str, match) = \Count the unique occurrences of string match in string str { local(lnm, lns, x, c); str=Str(str); \This allows leaving quotes off input match=Str(match); c=0; i=0; lns=length(str); lnm=length(match); if(lnm>1, i=1); x=1; while(x<=lns-lnm+1, if(mid(str, x, lnm)== match, c++; x+=lnm, x++); ); return(c) } - Cino Hilliard, Jan 04 2005
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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