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

1,2

Comments

Conjecture: every term after a(4) is a multiple of 10. - Harvey P. Dale, May 28 2015

Links

Harvey P. Dale, Table of n, a(n) for n = 1..150

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

Adjacent sequences: A068776 A068777 A068778 * A068780 A068781 A068782

Keyword

base,easy,nonn

Author

Robert G. Wilson v, Mar 02 2002

Status

approved