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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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).

Original entry on oeis.org

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

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Author

Hieronymus Fischer, Mar 24 2012; additional comments and formulas Jan 23 2013

Keywords

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.

Examples

			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).

Links

Crossrefs

Cf. A006995, A206923, A206924, A206925, A206927, A070939.
Cf. A217100, A217101.

Programs

  • Smalltalk
    A206926
"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).
^s [by Hieronymus Fischer]

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)).

A217101 Minimal number such that the number of contiguous palindromic bit patterns in its binary representation is n, or 0, if there is no such number.

Original entry on oeis.org

1, 2, 3, 4, 0, 7, 8, 18, 17, 15, 16, 42, 33, 68, 31, 32, 133, 65, 267, 130, 63, 64, 260, 129, 341, 258, 447, 127, 128, 682, 257, 1040, 514, 895, 1029, 255, 256, 1919, 513, 2056, 1026, 1791, 2053, 2052, 511, 512, 5376, 1025, 5461, 2050, 3583, 4101, 4100, 8203, 1023, 1024, 8200
Offset: 1

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Author

Hieronymus Fischer, Jan 23 2013

Keywords

Comments

The set of numbers which have n contiguous palindromic bit patterns (in their binary representation) is not empty, provided n<>5. Proof: For even n we have A206925(A206927(n/2)) = 2*(n/2) = n. For n=1,3,7,9 we get A206925(k)=n if we set k=1,3,8,17. For odd n>10 we define b(n) := 14*2^((n-9)/2)+A206927((n-9)/2). The b(n) have the binary expansion 11110, 111100, 1111001, 11110010, 111100101, 1111001011, 11110010110, 111100101100, 1111001011001, 11110010110010, 111100101100101, ..., for n=11, 13, 15, 17, ... . Evidently, b(n) is constructed by the concatenation of 111 with repeated bit patterns of 100101 (=37) truncated to 4+(n-9)/2 digits. As a result, the number of contiguous palindromic bit patterns of b(n) is A206925(111_2) + 3 + A206925(A206927((n-9)/2)) = 6 + 3 + n - 9 = n. This proves that there is always a number with n contiguous palindromic bit patterns.
a(5)=0, and this is the only zero term. Proof: The inequality A206925(n) >= 2*floor(log_2(n)) (cf. A206925) implies A206925(n) > 5 for n >= 8. By direct search we find A206925(n)<>5 for n=1..7. Thus, there is no k with A206925(k)=5, which implies a(5)=0.

Examples

			a(3) = 3, since 3=11_2 has 3 contiguous palindromic bit patterns, and this is the least such number.
a(6) = 7. Since 7=111_2 has 6 contiguous palindromic bit patterns, and this is the least such number.
a(8) = 18. Since 18=10010_2 has 8 contiguous palindromic bit patterns (1, 0, 0, 1, 0, 00, 010 and 1001), and this is the least such number.
a(9) = 17. Since 17=10001_2 has 9 contiguous palindromic bit patterns (1, 0, 0, 0, 1, 00, 00, 000, and 10001), and this is the least such number.

Links

Crossrefs

Cf. A006995, A206923, A206924, A206925, A206926, A070939, A217100.

Formula

a(n) = min(k | A206925(k) = n), for n<>5.
A206925(a(n)) = n, n<>5.
a(n) <= A217100(n), equality holds for n = 1, 2, 3 and 5, only.
a(A000217(n)) = 2^n - 1.
a(A000217(n)+1) = 2^n.
a(A000217(n)+3) = 2^(n+1)+1, n>2.
a(A000217(n)+5) = 2^(n+2)+2, n>4.
a(A000217(n)+6) = 2^(n+3) - 2^n - 1, n>5.
a(A000217(n)+7) = 2^(n+3)+5, n>6.
a(A000217(n)+8) = 2^(n+3)+4, n>7.
a(A000217(n)+9) = 2^(n+4)+11, n>8.
a(A000217(n)+10) = 2^(n+4) - 2^n - 1, n>9.
a(A000217(n)+11) = 21*2^n, n>10.
a(A000217(n)+12) = 2^(n+4)+8, n>11.
a(A000217(n)+13) = 2^(n+5)+18, n>12.

A217098 Greatest binary palindrome (cf. A006995) with n binary digits such that the number of contiguous palindromic bit patterns is minimal.

Original entry on oeis.org

1, 3, 5, 9, 27, 51, 107, 165, 403, 843, 1675, 2661, 5709, 13515, 27083, 39513, 108235, 208083, 432843, 682341, 1664211, 3461835, 6922955, 10918245, 23434061, 55390923, 110785227, 161912409, 443134667, 852178131, 1772532427, 2795133285, 6817395923, 14180201163, 28360356555
Offset: 1

Views

Author

Hieronymus Fischer, Jan 23 2013

Keywords

Comments

Subsequence of A217099.
a(n) is the greatest binary palindrome with n binary digits which meets the minimal possible number of palindromic substrings for that number of digits.

Examples

			a(1) = 1, since 1 is the largest binary palindrome with 1 palindromic substring (=1) which is the minimum for binary palindromes with 1 place.
a(3) = 5, since 5=101_2 is the largest binary palindrome with 4 palindromic substrings which is the minimum for binary palindromes with 3 places.
a(6) = 51, since 51=110011_2 is the largest binary palindrome with 11 palindromic substrings which is the minimum for binary palindromes with 6 places.

Links

Crossrefs

Cf. A006995, A206923, A206924, A206925, A206926, A070939, A217097, A217099, A217100, A217101.

Formula

a(n) = max(p | p is binary palindrome with n binary digits and A206925(p) = min(A206925(q) | q is binary palindrome with n binary digits)).
a(n) = A006995(j), where j := j(n) = max(k > A206915(2^(n-1)) | A206924(k) = min(A206925(A006995(i)) | i > A206915(2^(n-1)))).
a(n) = max(p | p is binary palindrome with n binary digits and A206925(p) = 2*(n-1) + floor((n-3)/2)).
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