A217101 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.

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

Offset

1,2

Comments

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.

Links

Hieronymus Fischer, Table of n, a(n) for n = 1..300

Formula

a(n) = min(k | A206925(k) = n), for n<>5.

A206925(a(n)) = n, n<>5.

a(n) <= A217100(n), equality holds for n = 1, 2, 3 and 5, only.

a(A000217(n)) = 2^n - 1.

a(A000217(n)+1) = 2^n.

a(A000217(n)+3) = 2^(n+1)+1, n>2.

a(A000217(n)+5) = 2^(n+2)+2, n>4.

a(A000217(n)+6) = 2^(n+3) - 2^n - 1, n>5.

a(A000217(n)+7) = 2^(n+3)+5, n>6.

a(A000217(n)+8) = 2^(n+3)+4, n>7.

a(A000217(n)+9) = 2^(n+4)+11, n>8.

a(A000217(n)+10) = 2^(n+4) - 2^n - 1, n>9.

a(A000217(n)+11) = 21*2^n, n>10.

a(A000217(n)+12) = 2^(n+4)+8, n>11.

a(A000217(n)+13) = 2^(n+5)+18, n>12.

Example

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.

Crossrefs

Cf. A006995, A206923, A206924, A206925, A206926, A070939, A217100.

Sequence in context: A011418 A214158 A054425 * A265516 A256295 A179078

Adjacent sequences: A217098 A217099 A217100 * A217102 A217103 A217104

Keyword

nonn,base

Author

Hieronymus Fischer, Jan 23 2013

Status

approved