%I #33 Mar 26 2021 06:31:33
%S 1089,2178,8712,9801,10989,21978,87912,98901,109989,219978,879912,
%T 989901,1099989,2199978,8799912,9899901,10891089,10999989,21782178,
%U 21999978,87128712,87999912,98019801,98999901,108901089,109999989
%N Non-palindromic numbers n, not divisible by 10, such that either n divides R(n) or R(n) divides n, where R(n) is the digit-reversal of n.
%C The quotient R(n)/n or n/R(n) is always 4 or 9.
%C This is the union of the four sequence A001232, A222814, A008918, A222815. Equivalently, the union of A008919 and A031877.
%C There are 4*Fibonacci(floor((n-2)/2)) terms with n digits (this is 2*A214927 or essentially 4*A103609). - _Ray Chandler_, Oct 12 2017
%C Conjecture: every term mod 100 is equal to 1, 12, 78, or 89. - _Harvey P. Dale_, Dec 13 2017
%H Ray Chandler, <a href="/A071685/b071685.txt">Table of n, a(n) for n = 1..10000</a>
%H N. J. A. Sloane, <a href="http://arxiv.org/abs/1307.0453">2178 And All That</a>, Fib. Quart., 52 (2014), 99-120.
%F x = q*R(x), q is integer q<>1, q<>10^j and neither of x or R(x) is divisible by 10.
%e 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.
%t 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}]
%t npnQ[n_]:=Module[{r=IntegerReverse[n]},!PalindromeQ[n]&&!Divisible[ n,10] &&(Mod[n,r]==0||Mod[r,n]==0)]; Select[Range[11*10^7],npnQ] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Dec 13 2017 *)
%Y Cf. A001232, A222814, A008918, A222815, A008919, A031877.
%Y Cf. A004086, A055483, A069554, A103609, A214927.
%K base,easy,nonn
%O 1,1
%A _Labos Elemer_, Jun 03 2002
%E Corrected and extended by _Harvey P. Dale_, Jul 01 2013
%E Edited by _N. J. A. Sloane_, Jul 02 2013
%E Missing terms inserted by _Ray Chandler_, Oct 09 2017
%E Incorrect comment removed by _Ray Chandler_, Oct 12 2017
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