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A206924
Number of contiguous palindromic bit patterns in the n-th binary palindrome.
10
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, 18, 18, 18, 18, 36, 31, 21, 19, 19, 19, 25, 21, 23, 23, 19, 21, 21, 21, 21, 25, 45, 39, 23, 25, 23, 23, 23, 21, 29, 29, 23, 21, 25, 25, 25, 27, 55, 48, 30, 26, 26
OFFSET
1,3
COMMENTS
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).
LINKS
FORMULA
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.
a(n) >= 2*floor(log_2(A006995(n))).
a(n) = A206925(A006995(n)).
a(n) <= ((floor(log_2(n)) + floor(log_2(n/3)) + 3) * (floor(log_2(n)) + floor(log_2(n/3))) + 2)/2.
a(n) >= 2*(floor(log_2(n)) + floor(log_2(n/3))), n>1. Equality holds for n=4 and n=6, only.
With m = 1+floor(log_2(A006995(n)), n>1:
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.
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.
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
a(n) >= |3*floor(log_2(n)) + 2*floor(log_2(n/3)) - 2|, n>1.
Asymptotic behavior:
a(n) = O(log(n)^2).
lim sup a(n)/log_2(n)^2 = 2, for n -> infinity.
lim inf a(n)/log_2(n) = 5, for n -> infinity.
lim inf (a(n) - 3*floor(log_2(n)) - 2*floor(log_2(n/3))) = -2, for n -> infinity.
lim inf a(n)/log_2(A006995(n)) = 5/2, for n -> infinity.
lim inf (2a(n) - 5*floor(log_2(A006995(n)))) = -3, for n -> infinity.
EXAMPLE
a(1) = a(2) = 1, since A006995(1) = 0 and A006995(2) = 1;
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;
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.
MATHEMATICA
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 *)
PROG
(Smalltalk)
"Calculates a(n)"
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Hieronymus Fischer, Mar 12 2012; additional formulas Jan 23 2013
STATUS
approved