A278930 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.
2, 36, 100, 2081, 8257, 32897, 131329, 524801, 2098177, 8390657, 33558529, 134225921, 536887297, 2147516417, 8590000129, 34359869441, 137439215617, 549756338177, 2199024304129, 8796095119361, 35184376283137, 140737496743937, 562949970198529, 2251799847239681
Offset: 1
Examples
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. 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. 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. 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: 1.....1....1 1....1.....1, 1......1.....1 1.....1......1, 1.......1......1 1......1.......1, 1........1.......1 1.......1........1, 1.........1........1 1........1.........1, 1..........1.........1 1.........1..........1, 1...........1..........1 1..........1...........1, 1............1...........1 1...........1............1, 1.............1............1 1............1.............1; 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.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (7,-14,8).
Programs
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Mathematica
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 *) LinearRecurrence[{7,-14,8},{2,36,100,2081,8257,32897},30] -
PARI
a(n)=if(n>3,1+2*(2^n+4^(n+1)),[2,36,100][n]);
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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
Formula
For n>3, a(n) = 1+2*(2^n+4^(n+1)).
From Colin Barker, Dec 02 2016: (Start)
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3) for n>6.
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)).
(End)
Extensions
More terms from Colin Barker, Dec 02 2016
Comments