|
%I #14 Sep 23 2021 01:27:02
%S 1,2,3,4,0,7,8,18,17,15,16,42,33,68,31,32,133,65,267,130,63,64,260,
%T 129,341,258,447,127,128,682,257,1040,514,895,1029,255,256,1919,513,
%U 2056,1026,1791,2053,2052,511,512,5376,1025,5461,2050,3583,4101,4100,8203,1023,1024,8200
%N 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.
%C 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.
%C 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.
%H Hieronymus Fischer, <a href="/A217101/b217101.txt">Table of n, a(n) for n = 1..300</a>
%F a(n) = min(k | A206925(k) = n), for n<>5.
%F A206925(a(n)) = n, n<>5.
%F a(n) <= A217100(n), equality holds for n = 1, 2, 3 and 5, only.
%F a(A000217(n)) = 2^n - 1.
%F a(A000217(n)+1) = 2^n.
%F a(A000217(n)+3) = 2^(n+1)+1, n>2.
%F a(A000217(n)+5) = 2^(n+2)+2, n>4.
%F a(A000217(n)+6) = 2^(n+3) - 2^n - 1, n>5.
%F a(A000217(n)+7) = 2^(n+3)+5, n>6.
%F a(A000217(n)+8) = 2^(n+3)+4, n>7.
%F a(A000217(n)+9) = 2^(n+4)+11, n>8.
%F a(A000217(n)+10) = 2^(n+4) - 2^n - 1, n>9.
%F a(A000217(n)+11) = 21*2^n, n>10.
%F a(A000217(n)+12) = 2^(n+4)+8, n>11.
%F a(A000217(n)+13) = 2^(n+5)+18, n>12.
%e a(3) = 3, since 3=11_2 has 3 contiguous palindromic bit patterns, and this is the least such number.
%e a(6) = 7. Since 7=111_2 has 6 contiguous palindromic bit patterns, and this is the least such number.
%e 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.
%e 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.
%Y Cf. A006995, A206923, A206924, A206925, A206926, A070939, A217100.
%K nonn,base
%O 1,2
%A _Hieronymus Fischer_, Jan 23 2013
|
