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

Author

R. J. Cano, Dec 01 2016

Keywords

• nonn
• easy
• base

Comments

The numbers whose binary representation is a palindrome are excluded by definition because 0 is not a power of a positive number.

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.

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).

Crossrefs

Inspired by: A278328.

Cf. A283050.

Programs

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]);

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