A068779 Least number k such that floor( k / digit reversal of k ) = n.
1, 31, 51, 71, 120, 350, 230, 450, 890, 10, 650, 430, 320, 850, 530, 950, 210, 2110, 730, 520, 830, 1800, 310, 3310, 4910, 720, 3110, 4610, 410, 4410, 920, 4310, 4210, 510, 5510, 4110, 2700, 610, 5310, 7910, 710, 29600, 6410, 5110, 810, 6310, 910
Offset: 1
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..150
Crossrefs
Cf. A068159.
Programs
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Mathematica
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 lnk[n_]:=Module[{k=1},While[Floor[k/FromDigits[Reverse[ IntegerDigits[k]]]] !=n, k++]; k]; Array[lnk,50] (* Harvey P. Dale, May 28 2015 *)
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