A206924 Number of contiguous palindromic bit patterns in the n-th binary palindrome.
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
Examples
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.
Links
- Hieronymus Fischer, Table of n, a(n) for n = 1..10000
Programs
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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 *)
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Smalltalk
A206924 "Calculates a(n)" ^self A006995 A206925
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) <= ((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.
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