

A071685


Nonpalindromic numbers n, not divisible by 10, such that either n divides R(n) or R(n) divides n, where R(n) is the digitreversal of n.


2



1089, 2178, 8712, 9801, 10989, 21978, 87912, 98901, 109989, 219978, 879912, 989901, 1099989, 2199978, 8799912, 9899901, 10891089, 10999989, 21782178, 21999978, 87128712, 87999912, 98019801, 98999901, 108901089, 109999989
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OFFSET

1,1


COMMENTS

The quotient R(n)/n or n/R(n) is always 4 or 9.
There are 4*Fibonacci(floor((n2)/2)) terms with n digits (this is 2*A214927 or essentially 4*A103609).  Ray Chandler, Oct 12 2017
Conjecture: every term mod 100 is equal to 1, 12, 78, or 89.  Harvey P. Dale, Dec 13 2017


LINKS



FORMULA

x = q*R(x), q is integer q<>1, q<>10^j and neither of x or R(x) is divisible by 10.


EXAMPLE

Palindromic solutions like 12021 or also solutions divisible by 10 were filtered out like {8380,838; q=10} or {8400,48; q=175}. In case of m>R(m), q=m/R(m)=4 or 9.


MATHEMATICA

nd[x_, y_] := 10*x+y tn[x_] := Fold[nd, 0, x] ed[x_] := IntegerDigits[x] red[x_] := Reverse[IntegerDigits[x]] Do[s=Mod[Max[{n, tn[red[n]]}], Min[{n, r=tn[red[n]]}]]; If[Equal[s, 0]&&!Equal[Mod[n, 10], 0] &&!Equal[n, r], Print[{n, r/n}]], {n, 1, 1000000}]
npnQ[n_]:=Module[{r=IntegerReverse[n]}, !PalindromeQ[n]&&!Divisible[ n, 10] &&(Mod[n, r]==0Mod[r, n]==0)]; Select[Range[11*10^7], npnQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 13 2017 *)


CROSSREFS



KEYWORD

base,easy,nonn


AUTHOR



EXTENSIONS



STATUS

approved



