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A083971 Reverse of k concatenated with k, divided by k, where k = A083970(n).

%I A083971 #32 Aug 06 2015 03:30:03
%S A083971 11,11,11,11,11,11,11,11,11,11,101,176,341,451,11,101,176,209,11,101,
%T A083971 176,11,101,121,176,11,101,11,101,11,77,101,11,101,11,101,11,1001,101,
%U A083971 1001,176,1001,4169,1001,1751,1001,341,1001,1001,3401,5126,1001,451,1001,1001,4501,11,1001,1001
%N A083971 Reverse of k concatenated with k, divided by k, where k = A083970(n).
%H A083971 Robert Israel, <a href="/A083971/b083971.txt">Table of n, a(n) for n = 1..10000</a>
%F A083971 Note that when n is a palindrome, R(n)=n, so R(n) concat n = (10^d)n + n, where d is the number of digits of n, and R(n) is the reverse of n. Dividing by n, we obtain (R(n) concat n)/n = 10^d + 1. - _Sam Alexander_, Oct 21 2003; edited by _Kevin Southwick_, Jul 26 2015
%p A083971 f:= proc(n) local L,m,i,v;
%p A083971     L:= convert(n,base,10);
%p A083971      m:= nops(L);
%p A083971     v:= add(10^(2*m-i)*L[i],i=1..m)/n+1;
%p A083971     if v::integer then v else NULL fi
%p A083971 end proc:
%p A083971 map(f, [$1..1000]); # _Robert Israel_, Jul 26 2015
%t A083971 A083970 =
%t A083971   Select[ Range[ 250 ],
%t A083971    Divisible[
%t A083971      FromDigits[
%t A083971       Flatten[ { Reverse[ IntegerDigits[ # ] ], IntegerDigits[ # ] } ] ], # ] & ];
%t A083971 Table[ FromDigits[
%t A083971    Flatten[ { Reverse[ IntegerDigits[ tmp ] ], IntegerDigits[ tmp ] } ] ] /
%t A083971   tmp, {tmp, A083970} ]
%t A083971 (* _Kevin Southwick_, Jul 26 2015 *)
%Y A083971 Cf. A071273, A083970.
%K A083971 base,nonn
%O A083971 1,1
%A A083971 _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 21 2003
%E A083971 Corrected and extended by _Sam Alexander_, Oct 21 2003
%E A083971 Name edited by _Charles R Greathouse IV_, Aug 05 2015