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%I #52 Mar 04 2017 16:00:07
%S 2,36,100,2081,8257,32897,131329,524801,2098177,8390657,33558529,
%T 134225921,536887297,2147516417,8590000129,34359869441,137439215617,
%U 549756338177,2199024304129,8796095119361,35184376283137,140737496743937,562949970198529,2251799847239681
%N a(n) is the least positive integer that differs (in absolute value) by an (n+1)-st power from the reverse of its binary representation.
%C The numbers whose binary representation is a palindrome are excluded by definition because 0 is not a power of a positive number.
%C It might be thought that the first term should be 1 instead of 2, since by prepending its binary representation (itself) with a zero we get 01 with reverse 10 (decimal 2), and their difference in absolute value is abs(1-2)=1, which is itself its 1st power 1^1. However, leading zeros are ignored. Another alternative interpretation is to consider 1 as a palindrome, which also excludes it from this sequence.
%H Colin Barker, <a href="/A278930/b278930.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-14,8).
%F For n>3, a(n) = 1+2*(2^n+4^(n+1)).
%F From _Colin Barker_, Dec 02 2016: (Start)
%F a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3) for n>6.
%F G.f.: x*(2 + 22*x - 124*x^2 + 1869*x^3 - 5198*x^4 + 3432*x^5) / ((1 - x)*(1 - 2*x)*(1 - 4*x)).
%F (End)
%e 2 in binary is 10, its binary reverse 01 or simply 1 is the decimal number 1, subtracting them gives abs(2-1)=1, and since 1 is its own square, a(1)=2.
%e 36 in binary is 100100, its binary reverse 1001 is the decimal number 9, subtracting them abs(36-9)=27=3^3, a third power, therefore a(2)=36.
%e 100 in binary is 1100100, its binary reverse 10011 is the decimal number 19, subtracting them abs(100-19)=81=3^4, a fourth power, therefore a(3)=100.
%e For n>3 if we represent zeros with dots and place the binary representation for each term followed by its reverse, up to n=12 we obtain the graph:
%e 1.....1....1
%e 1....1.....1,
%e 1......1.....1
%e 1.....1......1,
%e 1.......1......1
%e 1......1.......1,
%e 1........1.......1
%e 1.......1........1,
%e 1.........1........1
%e 1........1.........1,
%e 1..........1.........1
%e 1.........1..........1,
%e 1...........1..........1
%e 1..........1...........1,
%e 1............1...........1
%e 1...........1............1,
%e 1.............1............1
%e 1............1.............1;
%e which illustrates better why the absolute value should be part of the definition, and how the difference is an (n+1)th power: From the first two rows for a(4) we have abs(2081-2113) = abs(-32) = 2^5.
%t Rest@ CoefficientList[Series[x (2 + 22 x - 124 x^2 + 1869 x^3 - 5198 x^4 + 3432 x^5)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 24}], x] (* _Michael De Vlieger_, Dec 07 2016 *)
%o (PARI) a(n)=if(n>3,1+2*(2^n+4^(n+1)),[2,36,100][n]);
%o (PARI) Vec(x*(2 + 22*x - 124*x^2 + 1869*x^3 - 5198*x^4 + 3432*x^5) / ((1 - x)*(1 - 2*x)*(1 - 4*x)) + O(x^30)) \\ _Colin Barker_, Dec 02 2016
%Y Inspired by: A278328.
%Y Cf. A283050.
%K nonn,easy,base
%O 1,1
%A _R. J. Cano_, Dec 01 2016
%E More terms from _Colin Barker_, Dec 02 2016
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