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A206924 Number of contiguous palindromic bit patterns in the n-th binary palindrome. 9
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 (list; graph; refs; listen; history; text; internal format)
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

Hieronymus Fischer, Table of n, a(n) for n = 1..10000

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)

A206924

"Calculates a(n)"

^self A006995 A206925

CROSSREFS

Cf. A006995, A070939, A206923, A206925, A206926, A217099.

Sequence in context: A023830 A063649 A053158 * A185443 A275258 A230593

Adjacent sequences:  A206921 A206922 A206923 * A206925 A206926 A206927

KEYWORD

nonn,base

AUTHOR

Hieronymus Fischer, Mar 12 2012; additional formulas Jan 23 2013

STATUS

approved

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Last modified April 18 10:44 EDT 2019. Contains 322209 sequences. (Running on oeis4.)