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A206924 Number of contiguous palindromic bit patterns in the n-th binary palindrome.

%I A206924 #24 Mar 05 2013 15:34:11
%S A206924 1,1,3,4,6,6,10,9,9,9,15,13,11,11,21,18,14,16,14,14,14,16,28,24,16,16,
%T A206924 18,18,18,18,36,31,21,19,19,19,25,21,23,23,19,21,21,21,21,25,45,39,23,
%U A206924 25,23,23,23,21,29,29,23,21,25,25,25,27,55,48,30,26,26
%N A206924 Number of contiguous palindromic bit patterns in the n-th binary palindrome.
%C A206924 For a given number of places m a binary palindrome has at least 2*(m-1) + floor((m-3)/2) palindromic substrings. To a certain extent, this number indicates the minimal possible grade of symmetry (cf. A210926 and A217099).
%H A206924 Hieronymus Fischer, <a href="/A206924/b206924.txt">Table of n, a(n) for n = 1..10000</a>
%F A206924 a(n) <= m*(m+1)/2, where m = 1+floor(log_2(A006995(n)), equality holds if n+1 is a power of 2 or n+1 is 3-times a power of 2.
%F A206924 a(n) >= 2*floor(log_2(A006995(n))).
%F A206924 a(n) = A206925(A006995(n)).
%F A206924 a(n) <= ((floor(log_2(n)) + floor(log_2(n/3)) + 3) * (floor(log_2(n)) + floor(log_2(n/3))) + 2)/2.
%F A206924 a(n) >= 2*(floor(log_2(n)) + floor(log_2(n/3))), n>1. Equality holds for n=4 and n=6, only.
%F A206924 With m = 1+floor(log_2(A006995(n)), n>1:
%F A206924 a(n) >= 2(m-1) + floor((m-3)/2). Equality holds infinitely often for those n>3 for which A006995(n) is a term of A217099.
%F A206924 a(n) >= (5m - 8)/2. Equality holds infinitely often for those n>3 for which A006995(n) is a term of A217099 with an even number of digits.
%F A206924 a(n) >= 3*floor(log_2(n)) + 2*floor(log_2(n/3)) - 2. Equality holds infinitely often for those n>3 for which A006995(n) is a term of A217099
%F A206924 a(n) >= |3*floor(log_2(n)) + 2*floor(log_2(n/3)) - 2|, n>1.
%F A206924 Asymptotic behavior:
%F A206924 a(n) = O(log(n)^2).
%F A206924 lim sup a(n)/log_2(n)^2 = 2, for n -> infinity.
%F A206924 lim inf a(n)/log_2(n) = 5, for n -> infinity.
%F A206924 lim inf (a(n) - 3*floor(log_2(n)) - 2*floor(log_2(n/3))) = -2, for n -> infinity.
%F A206924 lim inf a(n)/log_2(A006995(n)) = 5/2, for n -> infinity.
%F A206924 lim inf (2a(n) - 5*floor(log_2(A006995(n)))) = -3, for n -> infinity.
%e A206924 a(1) = a(2) = 1, since A006995(1) = 0 and A006995(2) = 1;
%e A206924 a(3) = 3, since A006995(3)=3=11_2 and so there are the following 3 palindromic bit patterns the left 1, the right 1 and 11;
%e A206924 a(10) = 9, since A006995(10) = 27 = 11011_2 and so there are the following 9 palindromic bit patterns: 1, 1, 0, 1, 1, 11, 11, 101, 11011.
%t A206924 palQ[w_] := w == Reverse@w; subs[w_] := Flatten[Table[Take[w, {j, i}], {i, Length@w}, {j,i}], 1]; seq={}; k=0; While[Length@seq < 100, u = IntegerDigits[k++,2]; If[palQ@u, AppendTo[seq, Length@Select[subs@u, palQ]]]]; seq (* _Giovanni Resta_, Feb 13 2013 *)
%o A206924 (Smalltalk)
%o A206924 A206924
%o A206924 "Calculates a(n)"
%o A206924 ^self A006995 A206925
%Y A206924 Cf. A006995, A070939, A206923, A206925, A206926, A217099.
%K A206924 nonn,base
%O A206924 1,3
%A A206924 _Hieronymus Fischer_, Mar 12 2012; additional formulas Jan 23 2013