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A206926
Numbers such that the number of contiguous palindromic bit patterns in their binary representation is minimal (for a given number of places).
10
2, 4, 5, 6, 9, 10, 11, 12, 13, 18, 19, 20, 22, 25, 26, 37, 38, 41, 44, 50, 52, 75, 77, 83, 89, 101, 105, 150, 154, 166, 178, 203, 211, 300, 308, 333, 357, 406, 422, 601, 617, 666, 715, 812, 845, 1202, 1235, 1332, 1430, 1625, 1690, 2405, 2470, 2665, 2860, 3250
OFFSET
1,1
COMMENTS
The only binary palindromes in the sequence are 5 and 9.
The sequence is infinite, since A206927 is an infinite subsequence.
a(n) is the least number > a(n-1) which have the same number of palindromic substrings than a(n-1). If such a number doesn't exist, a(n) is the least number with one additional digit which meets the minimal possible number of palindromic substrings for such increased number of digits.
The concatenation of the bit patterns of a(n) and the reversal of a(n) form a term of A217099. Same is true for the concatenation of the bit patterns of a(n) and the reversal of floor(a(n)/2).
For a given number of places m there are at least 2*(m-1) palindromic substrings in the binary representation (cf. A206925). According to the definition the sequence terms are those with minimal possible symmetry. In other words: numbers not in the sequence have a significantly 'higher grade of symmetry'.
The terms have characteristic bit patterns and can be subdivided into 6 different classes of minimal symmetry. There are the following basic bit patterns: '100101', '100110', '101001', '101100', '110010' and '110100' representing the numbers 37, 38, 41, 44, 50 and 52. Numbers which are not a concatenation of one of these basic bit patterns do not meet the minimality condition. Evidently, 37, 44 and 50 (=Set_1) are equivalent patterns, since they can be derived from each other by rotation of digits (bit rotation). Same is true for 38, 41 and 52 (=Set_2). These two sets reflect reverse (mirror inverted) patterns. Each of those numbers from these sets can be viewed as a substitute to represent minimal symmetry.
For a given number b>3 the number of contiguous palindromic bit patterns in its binary representation is minimal if and only if there exists a number c from Set_1 or Set_2 such that the bit pattern of b is contained in concatenated c bit patterns, or, what is equivalent, if and only if b is contained in the concatenated bit patterns of 37 or 41.
If b is a number with more than 4 binary digits such that the number of contiguous palindromic bit patterns in its binary representation is minimal, then b is contained in the concatenated bit patterns either of 37 or 41, but not in both.
LINKS
FORMULA
a(n) = min(k > a(n-1) | A206925(k) = A206925(a(n-1))), if this minimum exists, else a(n) = min(k >= 2*2^floor(log(a(n-1))) | A206925(k) = min(A206925(j) | j >= 2*2^floor(log(a(n-1)))).
A206925(a(n)) = 2*floor(log_2(a(n))).
A070939(a(n)) = 4 + floor((n-4)/6), for n>4.
A206925(a(n)) = 6 + 2*floor((n-4)/6), for n>4.
Iteration formulas for k>0:
a(6(k+1)+4) = 2a(6k+4) + floor(37*2^(k+5)/63) mod 2.
a(6(k+1)+5) = 2a(6k+5) + floor(41*2^(k+1)/63) mod 2.
a(6(k+1)+6) = 2a(6k+6) + floor(41*2^(k+5)/63) mod 2.
a(6(k+1)+7) = 2a(6k+7) + floor(37*2^(k+2)/63) mod 2.
a(6(k+1)+8) = 2a(6k+8) + floor(37*2^(k+4)/63) mod 2.
a(6(k+1)+9) = 2a(6k+9) + floor(41*2^(k+4)/63) mod 2.
Calculation formulas for k>0:
a(6k+4) = floor((37*2^(k+4)/63) mod 2^(k+4).
a(6k+5) = floor((41*2^(k+6)/63) mod 2^(k+4).
a(6k+6) = floor((41*2^(k+4)/63) mod 2^(k+4).
a(6k+7) = floor((37*2^(k+7)/63) mod 2^(k+4).
a(6k+8) = floor((37*2^(k+9)/63) mod 2^(k+4).
a(6k+9) = floor((41*2^(k+9)/63) mod 2^(k+4).
With q(i) = 1 - 2*(floor((i+5)/6) - floor((i+4)/6) + floor((i+2)/6) + floor(i/6)),
this means q(i) = -1, 1, 1, -1, -1, 1, for i = 1..6,
p(i) = - 4 + 9*floor((i+5)/6) - 4*floor((i+4)/6) + 4*floor((i+3)/6) - 3*floor((i+2)/6)) + 2*floor((i+1)/6)),
this means p(i) = 5, 1, 5, 2, 4, 4, for i = 1..6,
k := k(n) = floor((n-4)/6),
j := j(n) = 1 + (n-4) mod 6,
we get the following formulas:
a(n+6) = 2*a(n) + floor((39+2*q(j))*2^(k+p(j))/63) mod 2, for n>9.
a(n+6) = 2*a(n) + b(k(n),j(n)), for n>9,
where b(k,j) is the 6x6-matrix:
(1 0 1 0 0 0)
(1 1 1 1 1 1)
(0 0 0 0 1 1)
(0 0 1 1 0 0)
(1 1 0 0 1 0)
(1 0 1 0 0 0).
a(n) = floor((39+2*q(j(n)))*2^(k(n)+p(j(n))+5)/63) mod 2^(k(n)+4), for n>4.
a(n) = (floor((39+2*q(j))*2^(6+p(j))/63) mod 32) * 2^(k-1) + (floor((39+2*q(j))*2^(6+p(j))/63) mod 64) * 2^(k mod 6 -1)*(2^(6*floor(k/6)) - 1)/63 + sum_{i=1..(k mod 6 - 1)} 2^(k mod 6 - 1 - i)*(floor((39+2*q(j))*2^(p(j)+i)/63) mod 2), for n>9.
a(n) = floor((39+2*q(j(n)))*2^(p(j(n))+floor((n+26)/6))/63) mod 2^floor((n+20)/6)), for n>4.
With: b(i) = floor((39+2*q(i))*2^(6+p(i))/63) mod 32, this means b(i) = 18, 19, 20, 22, 25, 26, for i = 1..6,
c(i) = (floor((39+2*q(i))*2^(6+p(i))/63) mod 64. This means c(i) = 50, 19, 52, 22, 25, 26, for i = 1..6:
a(n) = b(j)* 2^(k-1) + c(j)*2^(k mod 6 -1)*(2^(6*floor(k/6)) - 1)/63 + sum_{i=1..(k mod 6 - 1)} 2^(k mod 6 - 1 - i)*(floor(c(j)*2^i/63) mod 2), for n>9.
a(n) = floor(16*c(j)*2^floor((n+2)/6)/63) mod (8*2^floor((n+2)/6))), for n>4.
Asymptotic behavior:
a(n) = O(2^(n/6)).
lim inf a(n)/2^floor((n+2)/6) = 8*37/63 = 4.698412….
lim sup a(n)/2^floor((n+2)/6) = 8*52/63 = 6.603174….
lim inf a(n)/2^(n/6) = sqrt(2)*4*52/63 = 4.66914953….
lim sup a(n)/2^(n/6) = 2^(1/3)*8*37/63 = 5.91962906….
G.f. g(x) = x*(2 + 4x + 5x^2 + 6x^3 + 9x^4 + 10x^5 + 7x^6 + 3x^8 + 6x^9 + x^10 + x^13)/(1-2x^6) + (x^16*(1+x^2)(1+x^27) + x^22*(1-x^6)/(1-x) + x^32*(1-x^12)/((1-x^2)(1-x)) + x^47*(1+x^3)/(1-x))/(1-x^36).
Also: g(x) = x*(2 + 5x*(1-x^40) + 4x^2*(1+x^2+x^3-x^6-x^36-x^38-x^42)+ 2x^3*(1-x^3+x^6-x^7+x^39+x^43) - 6x^7*(1+x^4+x^38-x^40) - x^12*(1-x+x^7-x^8-x^10+x^15+x^16+x^18-x^20-x^23-x^24) - 3x^37*(1-x^6))/((1-x)(1+x^2)(1-x^9)(1+x^9)(1+x^18)).
EXAMPLE
a(2)=4=100_2 has 4 [=A206925(4)] contiguous palindromic bit patterns, this is the minimum value for binary numbers with 3 places. The other numbers with 3 places which meet that minimum value of 4 are 5 and 6.
a(7)=11=1011_2 has 6 [=A206925(11)] contiguous palindromic bit patterns, this is the minimum value for binary numbers with 4 places. The other numbers with 4 places which meet that minimum value of 6 are 9, 10, 12 and 13.
Examples to demonstrate the concatenation rule:
a(4) = 6 = 110_2 = (110010_2 truncated to 3 digits) = (50 truncated to 3 binary digits).
a(35) = 308 = 100110100_2 = (concatenation(100110, 100110) truncated to 9 digits) = (concatenation(38, 38) truncated to 9 binary digits).
a(94) = 307915 = 1001011001011001011_2 = (concatenation(101001, 101001, 101001, 101001) truncated to 19 digits) = (concatenation(41, 41, 41, 41) truncated to 19 binary digits).
PROG
(Smalltalk)
"Calculates a(n)"
| k j p q pArray qArray B n s |
n := self.
pArray := #(5 1 5 2 4 4).
qArray := #(-1 1 1 -1 -1 1).
B := #(2 4 5 6 9 10 11 12 13 18 19 20 22 25 26).
n < 10 ifTrue: [^s := B at: n].
k := (n - 4) // 6.
j := (n - 4) \\ 6 + 1.
p := pArray at: j.
q := qArray at: j.
s := (2 * q + 39) * (2 raisedToInteger: k + p + 5) // 63 \\ (2 raisedToInteger: k + 4).
KEYWORD
nonn,base
AUTHOR
Hieronymus Fischer, Mar 24 2012; additional comments and formulas Jan 23 2013
STATUS
approved