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A068779
Least number k such that floor( k / digit reversal of k ) = n.
1
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
1,2
COMMENTS
Conjecture: every term after a(4) is a multiple of 10. - Harvey P. Dale, May 28 2015
LINKS
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 *)
CROSSREFS
Cf. A068159.
Sequence in context: A194129 A288291 A248904 * A068473 A275628 A176507
KEYWORD
base,easy,nonn
AUTHOR
Robert G. Wilson v, Mar 02 2002
STATUS
approved