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A068779 Least number k such that floor( k / digit reversal of k ) = n. 1

%I #6 May 28 2015 09:47:04

%S 1,31,51,71,120,350,230,450,890,10,650,430,320,850,530,950,210,2110,

%T 730,520,830,1800,310,3310,4910,720,3110,4610,410,4410,920,4310,4210,

%U 510,5510,4110,2700,610,5310,7910,710,29600,6410,5110,810,6310,910

%N Least number k such that floor( k / digit reversal of k ) = n.

%C Conjecture: every term after a(4) is a multiple of 10. - _Harvey P. Dale_, May 28 2015

%H Harvey P. Dale, <a href="/A068779/b068779.txt">Table of n, a(n) for n = 1..150</a>

%t Reversal = ToExpression @ StringReverse @ ToString[ # ] &; a = Table[0, {70}]; Do[b = Floor[n/Reversal[n]]; If[b < 71 && a[[b]] == 0, a[[b]] = n], {n, 1, 10^6}]; a

%t lnk[n_]:=Module[{k=1},While[Floor[k/FromDigits[Reverse[ IntegerDigits[k]]]] !=n, k++]; k]; Array[lnk,50] (* _Harvey P. Dale_, May 28 2015 *)

%Y Cf. A068159.

%K base,easy,nonn

%O 1,2

%A _Robert G. Wilson v_, Mar 02 2002

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Last modified April 20 02:14 EDT 2024. Contains 371798 sequences. (Running on oeis4.)