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Numbers k in base 10 such that the digits of 2 + k are the digits of 2k written in reverse order.
1

%I #48 Sep 07 2024 15:44:43

%S 2,47,497,4997,49997,499997,4999997,49999997,499999997,4999999997,

%T 49999999997,499999999997,4999999999997,49999999999997,

%U 499999999999997,4999999999999997,49999999999999997,499999999999999997,4999999999999999997,49999999999999999997,499999999999999999997

%N Numbers k in base 10 such that the digits of 2 + k are the digits of 2k written in reverse order.

%H Xander Faber and Jon Grantham, <a href="https://arxiv.org/abs/2108.13441">On Integers Whose Sum is the Reverse of their Product</a>, arXiv:2108.13441 [math.NT], 2021.

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

%F a(n) = 5 * 10^(n - 1) - 3. - _Peter Bala_, Sep 06 2016

%F G.f.: x*(2 + 25*x)/(1 - 11*x + 10*x^2). - _Michael De Vlieger_, Sep 06 2016

%F E.g.f.: (exp(10*x) - 6*exp(x) + 5)/2. - _Stefano Spezia_, Mar 04 2023

%e 47 is in the sequence because 47 + 2 = 49 and 47*2 = 94.

%e 497 is in the sequence because 497 + 2 = 499 and 497*2 = 994.

%t Select[Range[10^6], IntegerDigits[# + 2] == Reverse@ IntegerDigits[2 #] &] (* or *)

%t Table[5 (10^(n - 1)) - 3, {n, 22}] (* or *)

%t CoefficientList[Series[x(2 + 25 x)/(1 - 11 x + 10 x^2), {x, 0, 21}], x] (* or *)

%t {2}~Join~Table[FromDigits@ Join[{4}, ConstantArray[9, {n - 2}], {7}], {n, 2, 22}] (* _Michael De Vlieger_, Sep 06 2016 *)

%o (PARI) isok(n) = digits(n+2) == Vecrev(digits(2*n)); \\ _Michel Marcus_, Sep 07 2016

%o (Magma) [5*10^(n-1)-3: n in [1..25]]; // _Vincenzo Librandi_, Sep 09 2016

%K nonn,base,easy

%O 1,1

%A _Debapriyay Mukhopadhyay_, Sep 06 2016