cp's OEIS Frontend

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.

Showing 1-9 of 9 results.

A206925 Number of contiguous palindromic bit patterns in the binary representation of n.

Original entry on oeis.org

1, 2, 3, 4, 4, 4, 6, 7, 6, 6, 6, 6, 6, 7, 10, 11, 9, 8, 8, 8, 9, 8, 9, 9, 8, 8, 9, 9, 9, 11, 15, 16, 13, 11, 11, 11, 10, 10, 11, 11, 10, 12, 11, 10, 11, 11, 13, 13, 11, 10, 11, 10, 11, 11, 12, 12, 11, 11, 12, 13, 13, 16, 21, 22, 18, 15, 15, 14, 13, 13, 14, 14
Offset: 1

Views

Author

Hieronymus Fischer, Mar 12 2012

Keywords

Comments

The number of contiguous palindromic bit patterns in the binary representation of n is a measure for the grade of symmetry in an abstract arrangement of two kinds of elements (where the number of elements is the number of binary digits, of course).
The minimum value for a(n) is 2*floor(log_2(n)) and will be taken infinitely often (see A206926 and A206927). This means: For a given number of places m there are at least 2*(m-1) palindromic substrings in the binary representation. This lower bound indicates to a certain extent the minimal possible symmetry.

Examples

			a(1) = 1, since 1 = 1_2 is the only palindromic bit pattern;
a(4) = 4, since 4 = 100_2 and there are the following palindromic bit patterns: 1, 0, 0, 00;
a(5) = 4, since 5 = 101_2 and there are the following palindromic bit patterns: 1, 0, 1, 101;
a(8) = 7, since 8 = 1000_2 and there are the following palindromic bit patterns: 1, 0, 0, 0, 00, 00, 000.
		

Crossrefs

Programs

  • Haskell
    import Data.Map (fromList, (!), insert)
    import Data.List (inits, tails)
    a206925 n = a206925_list !! (n-1)
    a206925_list = 1 : f [0, 1] (fromList [(Bin [0], 1), (Bin [1], 1)]) where
       f bs'@(b:bs) m = y : f (succ bs') (insert (Bin bs') y m) where
         y = m ! (Bin bs) +
             length (filter (\ds -> ds == reverse ds) $ tail $ inits bs')
         succ [] = [1]; succ (0:ds) = 1 : ds; succ (1:ds) = 0 : succ ds
    -- Reinhard Zumkeller, Dec 17 2012
    
  • PARI
    a(n)=n=binary(n);sum(k=0,#n-1,sum(i=1,#n-k,prod(j=0, k\2,n[i+j]==n[i+k-j]))) \\ Charles R Greathouse IV, Mar 21 2012
    
  • Python
    def A206925(n):
        s = bin(n)[2:]
        k = len(s)
        return sum(1 for i in range(k) for j in range(i+1,k+1) if s[i:j] == s[j-1:i-1-k:-1]) # Chai Wah Wu, Jan 31 2023
  • Smalltalk
    A206925
    "Answers the number of symmetric bit patterns of n as a binary."
    | m p q n numSym |
    n := self.
    n < 2 ifTrue: [^1].
    m := n integerFloorLog: 2.
    p := n printStringRadix: 2.
    numSym := 0.
    1 to: m + 1
      do:
       [:k |
       1 to: k
        do:
         [:j |
         q := p copyFrom: j to: k.
         q = q reverse ifTrue: [numSym := numSym + 1]]].
    ^numSym // Hieronymus Fischer, Feb 16 2013
    

Formula

a(n) <= (m+1)*(m+2)/2, where m = floor(log_2(n)); equality holds if n + 1 is a power of 2.
a(n) >= 2*floor(log_2(n)).
This estimation cannot be improved in general, since equality holds for A206926(n): a(A206926(n)) = 2*floor(log_2(A206926(n))).
Asymptotic behavior:
a(n) = O(log(n)^2).
lim sup a(n)/log_2(n)^2 = 1/2, for n --> infinity.
lim inf a(n)/log_2(n) = 2, for n --> infinity.

Extensions

Comments and formulas added by Hieronymus Fischer, Jan 23 2013

A206924 Number of contiguous palindromic bit patterns in the n-th binary palindrome.

Original entry on oeis.org

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

Views

Author

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

Keywords

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).

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.
		

Crossrefs

Programs

  • 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 *)
  • 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) = 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.

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

Views

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).
		

Crossrefs

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)).

A217099 Binary palindromes (cf. A006995) such that the number of contiguous palindromic bit patterns is minimal (for a given number of places).

Original entry on oeis.org

0, 1, 3, 5, 9, 17, 21, 27, 45, 51, 73, 93, 99, 107, 153, 165, 297, 313, 325, 403, 717, 843, 1241, 1421, 1619, 1675, 2409, 2661, 4841, 4953, 5349, 5709, 13011, 13515, 21349, 22861, 26067, 27083, 38505, 39513, 76905, 78937, 85349, 108235, 183117, 208083, 307817, 366413, 415955, 432843, 632409
Offset: 1

Views

Author

Hieronymus Fischer, Jan 23 2013

Keywords

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.
a(n) is the least binary palindrome > a(n-1) which have the same number of palindromic substrings than a(n-1). If such a palindrome doesn't exist, a(n) is the least binary palindrome with one additional digit which meets the minimal possible number of palindromic substrings for such increased number of digits.
b_left(n) := floor(a(n)/2^log_2(a(n))) is a term of A206926, if n > 3. More precise, the bit pattern of b_left(n) is contained in the concatenation of the bit patterns of 37 or of 41, provided n > 16.
b_right(n) := a(n) mod (2^(1+log_2(a(n))) is a term of A206926, if n > 6. More precise, the bit pattern of b_right(n) is contained in the concatenation of the bit patterns of 37 or of 41, provided n > 16.
Provided n > 16: The bit pattern of b_left(n) is contained in the continued concatenation of the bit pattern of 37 (or 41, respectively) if and only if the bit pattern of b_ right(n) is contained in the continued concatenation of the bit pattern of 41 (or 37, respectively).

Examples

			a(1) = 0, since 0 is a binary palindrome with 1 palindromic substring (=0) which is the minimum for binary palindromes with 1 place.
a(2) = 1, since 1 is a binary palindrome with 1 palindromic substring (=1) which is the minimum for binary palindromes with 1 place.
a(8) = 27, since 27=11011_2 is a binary palindrome with 9 palindromic substrings which is the minimum for binary palindromes with 5 places.
a(9) = 45, since 45=101101_2 is a binary palindrome with 11 palindromic substrings which is the minimum for binary palindromes with 6 places.
		

Crossrefs

Programs

  • Smalltalk
    "Calculates a(n) - not optimized.
    If the complete array 'answer' is answered instead of a separate term, the next 2 (if d is even) or 4 (if d is odd) terms are calculated simultaneously"
    | n min d B k j p q answer |
    answer := OrderedCollection new.
    n := self.
    B := #(0 1 3 5 9 17 21 27 45 51 73 93 99 107 153 165).
    n <= 16 ifTrue: [^s := B at: n].
    k := (n - 5) // 6 - 1.
    j := (n - 5) \\ 6 + 1.
    d := 2 * k + 7 + (j // 5).
    min := (d - 1) * 2 + ((d - 3) // 2).
    0 to: 5
      do:
       [:i |
       p := (6 * k + 4 + i) A206926.
       s := p * (2 raisedToInteger: d // 2).
       q := p // (2 - (j // 5)) reverse: 2.
       s A206925 = min ifTrue: [answer add: (s + q)]].
    ^answer at: j - (j // 5 * 4) [by Hieronymus Fischer]

Formula

a(n) = min(p > a(n-1) | p is binary palindrome and A206925(p) = A206925(a(n-1))), if this minimum exists, else a(n) = min(p > 2*2^floor(log(a(n-1))) | p is binary palindrome and A206925(p) = min(A206925(q) | q is binary palindrome and q > 2*2^floor(log(a(n-1))))).
a(n) = A006995(j), where j := j(n) = min(k > A206915(a(n-1)) | A206924(k) = A206925(a(n-1)), if this minimum exists, else j(n) = min(k > A206915(2*2^floor(log(a(n-1)))) | A206924(k) = min(a206925(A006995(i)) | i > A206915(2*2^floor(log(a(n-1)))))).
With k := k(n) = floor((n - 5)/6) - 1, j := j(n) = (n - 5) mod 6 + 1, d = 2k+7+floor(j/5),
c = 2*(d-1) + floor((d-3)/2), f(i) = A206926(6k + 4 + i)*2^floor(d/2) + Reversal(floor((A206926(6k + 4 + i))/(2 - floor(j/5)))), for i=0..5, we have
a(n) = b(j - 4*floor(j/5)), where b(m) = f(min(m-1<=i<=5 | A206925(f(i)) = c and f(i) <> b(l) for 1<=l
With m = 1+floor(log_2(a(n)), n > 3:
A206924(k) = 2(m-1) + floor((m-3)/2), where k is that uniquely determined number for which A006995(k) = a(n).
A206924(A206915(a(n))) = 2(m-1) + floor((m-3)/2).
A206924(A206915(a(n))) = 3*floor(log_2(A206915(a(n)))) + 2*floor(log_2(A206915(a(n))/3)) - 2, n > 3.

A206927 Minimal numbers of binary length n+1 such that the number of contiguous palindromic bit patterns in the binary representation is minimal.

Original entry on oeis.org

2, 4, 9, 18, 37, 75, 150, 300, 601, 1202, 2405, 4811, 9622, 19244, 38489, 76978, 153957, 307915, 615830, 1231660, 2463321, 4926642, 9853285, 19706571, 39413142, 78826284, 157652569, 315305138, 630610277, 1261220555
Offset: 1

Author

Hieronymus Fischer, Mar 24 2012

Keywords

Comments

Subsequence of A206926.
From left to right, the binary representation of a(n) consists of a concatenation of the bit pattern 100101 (=37). If the number of places is not a multiple of 6, the least significant places are truncated. This leads to a simple linear recurrence.
Example: a(19)=615830=10010110010110_2=concatenate('100101','100101','10')

Examples

			a(3)=9=1001_2 has 6 [=A206925(9)] contiguous palindromic bit patterns. This is the minimum value for binary numbers with 4 places and 9 is the least number with this property.
a(9)=601=1001011001_2 has 18 [=A206925(601)] contiguous palindromic bit patterns. This is the minimum value for binary numbers with 10 places and 601 is the least number with this property.
		

Formula

a(n) = 37*2^(1+n mod 6)*(2^(6*floor(n/6))-1)/63 + floor(37*2^(n mod 6)/2^5).
a(n) = floor((37*2^(n+1)/63)) mod 2^(n+1).
A206925(a(n)) = 2*floor(log_2(a(n))).
a(n+1) = 2a(n) + floor(37*2^(n+2)/63) mod 2.
G.f.: x*( 2+x^2+x^4+x^5-2*x^6 ) / ( (x-1)*(2*x-1)*(1+x)*(x^2-x+1)*(1+x+x^2) ). - R. J. Mathar, Apr 02 2012
Also, g.f. x*(2+x^2+x^4+x^5-2*x^6)/((1-2*x)*(1-x^6)).

Extensions

Further formulas added by Hieronymus Fischer, Jan 13 2013

A217100 Greatest 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, 6, 0, 13, 14, 26, 29, 52, 58, 105, 116, 211, 233, 422, 467, 845, 934, 1690, 1869, 3380, 3738, 6761, 7476, 13523, 14953, 27046, 29907, 54093, 59814, 108186, 119629, 216372, 239258, 432745, 478516, 865491, 957033, 1730982, 1914067, 3461965, 3828134
Offset: 1

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 and finite, provided n<>5. Proof: compare A217101 for the proof of existence. The finiteness follows from A206925, since each number p > 2^floor((n+1)/2) has at least A206925(p) >= 2*floor(log_2(p)) > 2*floor((n+1)/2) = n + n mod 2 palindromic substrings. Thus, there is a boundary b <= 2^floor((n+1)/2) such that all numbers > b have more than n palindromic substrings. It follows, that the set of numbers with n palindromic substrings is finite.
a(5)=0, and this is the only zero term. Proof: cf. A217101.
The binary expansion of a(n) has 1 + floor(n/2) digits (n<>5).
a(2n) is the maximal number with n+1 binary digits such that the number of contiguous palindromic bit patterns in the binary representation is minimal (cf. A206926).

Examples

			a(3)=3, since 3=11_2 has 3 contiguous palindromic bit patterns, and this is the greatest such number.
a(6)=13. Since 13=1101_2 has 6 contiguous palindromic bit patterns, and this is the greatest such number.
a(8)=26. Since 26=11010_2 has 8 contiguous palindromic bit patterns (1, 1, 0, 1, 0, 11, 101 and 010), and this is the greatest such number.
a(9)=27. Since 17=11011_2 has 9 contiguous palindromic bit patterns (1, 1, 0, 1, 1, 11, 11, 101 and 11011), and this is the greatest such number.
		

Formula

a(n) = max(k | A206925(k) = n), for n<>5.
A206925(a(n)) = n, n<>5.
a(n) => A217101(n), equality holds for n = 1, 2, 3 and 5, only.
Iteration formula:
a(n+2) = 2*a(n) + floor(52*2^floor((n+4-(-1)^n)/2)/63) mod 2, n>5.
a(n+2) = 2*a(n) + floor(52*2^((2*n+7-3*(-1)^n)/4)/63) mod 2, n>5.
Direct calculation formula:
a(n) = floor(52*2^floor((n+1+(-1)^n)/2)/63) mod 2^floor((n+1+(-1)^n)/2)) + (1-(-1)^n)*2^floor((n-2)/2)), for n>5.
a(n) = floor(52*2^((2*n + 1 + 3*(-1)^n)/4)/63) + (1-(-1)^n)*2^((2*n - 5 + (-1)^n)/4), for n>5.
a(2k) = A206926(6k-9), k>2.
G.f.: x*(1 +2*x +x^2 +2*x^3 -6*x^4 +x^5 +14*x^6 +x^8 +x^11 -x^12 -x^13 -2*x^15 +7*x^16 -14*x^18)/((1-2*x^2)*(1-x^3)*(1+x^3)*(1+x^6)); also:
x*(1 -x +3*x^2 +x*(3+14*x^5)/(1-2*x^2) +x^5*(1+x^3)*(1+x^6+x^8)/((1-2*x^2)*(1-x^12))).

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

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.
		

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.

A217097 Least binary palindrome (cf. A006995) with n binary digits such that the number of contiguous palindromic bit patterns is minimal.

Original entry on oeis.org

0, 3, 5, 9, 17, 45, 73, 153, 297, 717, 1241, 2409, 4841, 13011, 21349, 38505, 76905, 183117, 307817, 632409, 1231465, 2929485, 5060185, 9853545, 19708521, 53261523, 87349605, 157653609, 315300457, 749917005, 1261214313, 2590611033, 5044869737, 11998647117, 20724946521
Offset: 1

Author

Hieronymus Fischer, Feb 10 2013

Keywords

Comments

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

Examples

			a(1) = 0, since 0 is the least binary palindrome with 1 palindromic substring (=0) which is the minimum for binary palindromes with 1 place.
a(3) = 5, since 5=101_2 is the least binary palindrome with 4 palindromic substrings which is the minimum for binary palindromes with 3 places.
a(6) = 45, since 45=101101_2 is the least binary palindrome with 11 palindromic substrings which is the minimum for binary palindromes with 6 places.
		

Crossrefs

Cf. A006995, A206923, A206924, A206925, A206926, A070939, A217098, 217099, 217100, 217101.

Formula

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

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

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.
		

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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