

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



2, 36, 100, 2081, 8257, 32897, 131329, 524801, 2098177, 8390657, 33558529, 134225921, 536887297, 2147516417, 8590000129, 34359869441, 137439215617, 549756338177, 2199024304129, 8796095119361, 35184376283137, 140737496743937, 562949970198529, 2251799847239681
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OFFSET

1,1


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(12)=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.


LINKS

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,14,8).


FORMULA

For n>3, a(n) = 1+2*(2^n+4^(n+1)).
From Colin Barker, Dec 02 2016: (Start)
a(n) = 7*a(n1)  14*a(n2) + 8*a(n3) 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)


EXAMPLE

2 in binary is 10, its binary reverse 01 or simply 1 is the decimal number 1, subtracting them gives abs(21)=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(369)=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(10019)=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(20812113) = abs(32) = 2^5.


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 *)


PROG

(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


CROSSREFS

Inspired by: A278328.
Cf. A283050.
Sequence in context: A081310 A187298 A069067 * A258356 A145450 A196558
Adjacent sequences: A278927 A278928 A278929 * A278931 A278932 A278933


KEYWORD

nonn,easy,base


AUTHOR

R. J. Cano, Dec 01 2016


EXTENSIONS

More terms from Colin Barker, Dec 02 2016


STATUS

approved



