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 A166749 Numbers that are the sum or product of two numbers, such that the sum and product have reversed digits. 1
 0, 4, 18, 27, 49, 72, 81, 94, 499, 994, 4999, 9994, 49999, 99994, 499999, 999994, 4999999, 9999994, 49999999, 99999994, 499999999, 999999994, 4999999999, 9999999994, 49999999999, 99999999994, 499999999999, 999999999994, 4999999999999, 9999999999994, 49999999999999 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. These are in fact all the solutions, shown by a case-by-case analysis. - Wang Pok Lo, Dec 24 2018 LINKS W. P. Lo and Y. Paz, On finding all positive integers a,b such that b±a and ab are palindromic, arXiv:1812.08807 [math.HO] (2018). FORMULA 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. EXAMPLE For instance, 9*9=81 and 9+9=18 are terms; 3*24=72 and 3+24=27 are terms too. MATHEMATICA 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] CROSSREFS Sequence in context: A099565 A063563 A323848 * A103067 A080519 A120407 Adjacent sequences:  A166746 A166747 A166748 * A166750 A166751 A166752 KEYWORD nonn,base,easy AUTHOR Mark Nandor, Oct 21 2009 STATUS approved

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Last modified April 17 11:50 EDT 2021. Contains 343064 sequences. (Running on oeis4.)