A169830 - OEIS

%I #32 Aug 13 2020 14:02:28

%S 1,73,793,7993,79993,799993,7999993,79999993,799999993,7999999993,

%T 79999999993,799999999993,7999999999993,79999999999993,

%U 799999999999993,7999999999999993,79999999999999993,799999999999999993,7999999999999999993,79999999999999999993

%N Numbers n such that 2*reverse(n) - n = 1.

%C The sequence is infinite since it contains all numbers of the form 799...9993. (Cf. A101155, A101849.) [Ulrich Krug (leuchtfeuer37(AT)gmx.de), Jun 02 2010]

%C All numbers of the form 8*10^k-7 are members, but are there any others? - _Robert G. Wilson v_, Jun 01 2010

%C All solutions are of the form 8*10^k-7. - _David Radcliffe_, Jul 25 2015

%H Matthew House, <a href="/A169830/b169830.txt">Table of n, a(n) for n = 1..996</a>

%H Erich Friedman, <a href="https://erich-friedman.github.io/numbers.html">What's Special About This Number?</a> (See entry 73.)

%H David Radcliffe, <a href="/A169830/a169830.pdf">Numbers that are nearly doubled when reversed</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10).

%F a(n) = 8*10^(n-1) - 7. - _David Radcliffe_, Jul 25 2015

%F From _Matthew House_, Feb 12 2017: (Start)

%F G.f.: x*(1+62*x)/((1-x)*(1-10*x)).

%F a(n) = 11*a(n-1) - 10*a(n-2). (End)

%t k = 1; lst = {}; fQ[n_] := 2 FromDigits@ Reverse@ IntegerDigits@n == 1 + n; While[k < 10^8, If[fQ@k, Print@k; AppendTo[lst, k]]; k++ ]; lst (* _Robert G. Wilson v_, Jun 01 2010 *)

%t Rest@ CoefficientList[Series[x (1 + 62 x)/((1 - x) (1 - 10 x)), {x, 0, 20}], x] (* or *)

%t Table[If[n == 1, 1, FromDigits@ Join[{7}, ConstantArray[9, n - 2], {3}]], {n, 20}] (* or *)

%t LinearRecurrence[{11, -10}, {1, 73}, 20] (* _Michael De Vlieger_, Feb 12 2017 *)

%o (PARI) isok(n) = 2*fromdigits(Vecrev(digits(n))) - n == 1; \\ _Michel Marcus_, Feb 12 2017

%Y Same sequence as A100412. Digit reversals of A083818.

%K nonn,base

%O 1,2

%A _N. J. A. Sloane_, May 31 2010

%E a(6)-a(8) from _Robert G. Wilson v_, Jun 01 2010

%E More terms from _David Radcliffe_, Jul 25 2015