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A278930
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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.
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2
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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
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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For n>3, a(n) = 1+2*(2^n+4^(n+1)).
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)
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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