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A217101
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Minimal number such that the number of contiguous palindromic bit patterns in its binary representation is n, or 0, if there is no such number.
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4
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1, 2, 3, 4, 0, 7, 8, 18, 17, 15, 16, 42, 33, 68, 31, 32, 133, 65, 267, 130, 63, 64, 260, 129, 341, 258, 447, 127, 128, 682, 257, 1040, 514, 895, 1029, 255, 256, 1919, 513, 2056, 1026, 1791, 2053, 2052, 511, 512, 5376, 1025, 5461, 2050, 3583, 4101, 4100, 8203, 1023, 1024, 8200
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OFFSET
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1,2
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COMMENTS
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The set of numbers which have n contiguous palindromic bit patterns (in their binary representation) is not empty, provided n<>5. Proof: For even n we have A206925(A206927(n/2)) = 2*(n/2) = n. For n=1,3,7,9 we get A206925(k)=n if we set k=1,3,8,17. For odd n>10 we define b(n) := 14*2^((n-9)/2)+A206927((n-9)/2). The b(n) have the binary expansion 11110, 111100, 1111001, 11110010, 111100101, 1111001011, 11110010110, 111100101100, 1111001011001, 11110010110010, 111100101100101, ..., for n=11, 13, 15, 17, ... . Evidently, b(n) is constructed by the concatenation of 111 with repeated bit patterns of 100101 (=37) truncated to 4+(n-9)/2 digits. As a result, the number of contiguous palindromic bit patterns of b(n) is A206925(111_2) + 3 + A206925(A206927((n-9)/2)) = 6 + 3 + n - 9 = n. This proves that there is always a number with n contiguous palindromic bit patterns.
a(5)=0, and this is the only zero term. Proof: The inequality A206925(n) >= 2*floor(log_2(n)) (cf. A206925) implies A206925(n) > 5 for n >= 8. By direct search we find A206925(n)<>5 for n=1..7. Thus, there is no k with A206925(k)=5, which implies a(5)=0.
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LINKS
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FORMULA
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a(n) = min(k | A206925(k) = n), for n<>5.
a(n) <= A217100(n), equality holds for n = 1, 2, 3 and 5, only.
a(A000217(n)+6) = 2^(n+3) - 2^n - 1, n>5.
a(A000217(n)+10) = 2^(n+4) - 2^n - 1, n>9.
a(A000217(n)+12) = 2^(n+4)+8, n>11.
a(A000217(n)+13) = 2^(n+5)+18, n>12.
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EXAMPLE
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a(3) = 3, since 3=11_2 has 3 contiguous palindromic bit patterns, and this is the least such number.
a(6) = 7. Since 7=111_2 has 6 contiguous palindromic bit patterns, and this is the least such number.
a(8) = 18. Since 18=10010_2 has 8 contiguous palindromic bit patterns (1, 0, 0, 1, 0, 00, 010 and 1001), and this is the least such number.
a(9) = 17. Since 17=10001_2 has 9 contiguous palindromic bit patterns (1, 0, 0, 0, 1, 00, 00, 000, and 10001), and this is the least such number.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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