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Numbers n whose reversal divides n+1.
2

%I #13 Nov 26 2014 22:40:02

%S 1,10,41,73,100,793,1000,7993,9100,9320,10000,73000,79993,100000,

%T 340000,735221,799993,1000000,3070000,7999993,9382711,9910000,

%U 10000000,73000000,79000000,79999993,100000000,361300000,799999993,1000000000,1281010000

%N Numbers n whose reversal divides n+1.

%C Similar to A160946.

%C Sequence is infinite because any number of the form 10^k, for k integer, belongs to the sequence. In fact Rev(10^k)=1 and (10^k + 1) / 1 = 10^k + 1. Also concat(7,9…9,3) = 73, 793, 7993, 79993, etc. is part of the sequence: Rev(concat(7,9…9,3)) = concat(3,9…9,7) and concat(7,9…9,4) / concat(3,9…9,7) = 2.

%e Rev(1) = 1 and 2 / 1 = 2.

%e Rev(10) = 1 and 11 /1 = 11.

%e Rev(41) = 14 and 42 / 14 = 3.

%e Rev(73) = 37 and 74 / 37 = 2. Etc.

%p with(numtheory): T:=proc(w) local x, y, z; x:=0; y:=w;

%p for z from 1 to ilog10(w)+1 do x:=10*x+(y mod 10); y:=trunc(y/10); od; x; end;

%p P:=proc(q) local n; for n from 1 to q do if type((n+1)/T(n),integer)

%p then print(n); fi; od; end: P(10^9);

%o (PARI) rev(n)=s="";for(i=1,#(d=digits(n)),s=concat(d[i],s));eval(s)

%o for(n=1,10^5,if(!((n+1)%rev(n)),print1(n,", "))) \\ _Derek Orr_, Nov 26 2014

%Y Cf. A160946.

%K nonn,base

%O 1,2

%A _Paolo P. Lava_, Nov 25 2014