%I #12 Dec 29 2018 03:55:55
%S 0,4,18,27,49,72,81,94,499,994,4999,9994,49999,99994,499999,999994,
%T 4999999,9999994,49999999,99999994,499999999,999999994,4999999999,
%U 9999999994,49999999999,99999999994,499999999999,999999999994,4999999999999,9999999999994,49999999999999
%N Numbers that are the sum or product of two numbers, such that the sum and product have reversed digits.
%C Note that 0 and 4 are their own reversed-digit sums and products, since 0+0=0*0=0 and 2+2=2*2=4. The pattern of some number of nines and then a four, and a four and some number of nines, continues indefinitely.
%C These are in fact all the solutions, shown by a case-by-case analysis. - _Wang Pok Lo_, Dec 24 2018
%H W. P. Lo and Y. Paz, <a href="https://arxiv.org/abs/1812.08807">On finding all positive integers a,b such that b±a and ab are palindromic</a>, arXiv:1812.08807 [math.HO] (2018).
%F For n>8, a(n)=5*10^((n+1)/2 - 3) - 1 if n odd; a(n)=10^(n/2 - 2) - 6 if n even.
%e For instance, 9*9=81 and 9+9=18 are terms; 3*24=72 and 3+24=27 are terms too.
%t Do[If[IntegerDigits[x y] == Reverse[IntegerDigits[x + y]], Print[{x, y, x + y, x y}]], {x, 0, 20}, {y, x, 100000}] or a[1]=0;a[2]=4;a[3]=18;a[4]=27;a[5]=49;a[6]=72;a[7]=81;a[8]=94 a[n_] := a[n] = If[OddQ[n], 5*10^((n + 1)/2 - 3) - 1, 10^(n/2 - 2) - 6]
%K nonn,base,easy
%O 1,2
%A _Mark Nandor_, Oct 21 2009