A206925 -id:A206925 - cp's OEIS frontend

A215244 a(n) is the number of ways of writing the binary expansion of n as a product (or concatenation) of palindromes.

1, 1, 1, 2, 2, 2, 2, 4, 4, 3, 3, 3, 4, 3, 4, 8, 8, 5, 4, 5, 5, 5, 4, 6, 8, 5, 5, 6, 8, 6, 8, 16, 16, 9, 7, 9, 8, 6, 6, 10, 10, 6, 8, 8, 7, 7, 7, 12, 16, 9, 7, 10, 8, 8, 8, 11, 16, 10, 10, 11, 16, 12, 16, 32, 32, 17, 13, 17, 13, 11, 11, 18, 15, 11, 10, 10, 12, 9, 11, 20, 20, 11, 10, 11, 13, 13, 10, 16, 14, 9, 10, 12, 13, 11, 13, 24, 32, 17, 13, 18, 14, 11, 12, 19, 16, 10, 13

Offset: 0

N. J. A. Sloane, Aug 07 2012

"Product" is used here is the usual sense of combinatorics on words.
The sequence may be arranged into a triangle in which row k contains the 2^k terms [a(2^k), ..., a(2^(k+1)-1)]: (1), (1), (1,2), (2,2,2,4), ... The row sums of this triangle appear to be A063782. - R. J. Mathar, Aug 09 2012. I have a proof that A063782 does indeed give the row sums. - N. J. A. Sloane, Aug 10 2012
a(A220654(n)) = n and a(m) < n for m < A220654(n). - Reinhard Zumkeller, Dec 17 2012

			a(4)=2 since 4 = 100, and 100 can be written as either 1.0.0 or 1.00.
a(5)=2 from 1.0.1, 101.
a(10)=3 from 1.0.1.0, 1.010, 101.1
Written as a triangle, the sequence is:
1,
1,
1, 2,
2, 2, 2, 4,
4, 3, 3, 3, 4, 3, 4, 8,
8, 5, 4, 5, 5, 5, 4, 6, 8, 5, 5, 6, 8, 6, 8, 16,
16, 9, 7, 9, 8, 6, 6, 10, 10, 6, 8, 8, 7, 7, 7, 12, 16, 9, 7, 10, 8, 8, 8, 11, 16, 10, 10, 11, 16, 12, 16, 32,
...

Cf. A050430, A215245, A215246, A215250, A215253, A215254, A063782.

Cf. A206925, A030308.

import Data.Map (Map, singleton, (!), insert)
import Data.List (inits, tails)
newtype Bin = Bin [Int] deriving (Eq, Show, Read)
instance Ord Bin where
   Bin us <= Bin vs | length us == length vs = us <= vs
                    | otherwise              = length us <= length vs
a215244 n = a215244_list !! n
a215244_list = 1 : f [1] (singleton (Bin [0]) 1) where
   f bs m | last bs == 1 = y : f (succ bs) (insert (Bin bs) y m)
          | otherwise    = f (succ bs) (insert (Bin bs) y m) where
     y = fromEnum (pal bs) +
         sum (zipWith (\us vs -> if pal us then m ! Bin vs else 0)
                      (init $ drop 1 $ inits bs) (drop 1 $ tails bs))
     pal ds = reverse ds == ds
     succ [] = [0]; succ (0:ds) = 1 : ds; succ (1:ds) = 0 : succ ds
-- Reinhard Zumkeller, Dec 17 2012

isPal := proc(L)
    local d ;
    for d from 1 to nops(L)/2 do
        if op(d,L) <> op(-d,L) then
            return false;
        end if;
    end do:
    return true;
end proc:
A215244L := proc(L)
    local a,c;
    a := 0 ;
    if isPal(L) then
        a := 1;
    end if;
    for c from 1 to nops(L)-1 do
        if isPal( [op(1..c,L)] ) then
            a := a+procname([op(c+1..nops(L),L)]) ;
        end if;
    end do:
    return a;
end proc:
A215244 := proc(n)
    if n = 1 then
        1;
    else
        convert(n,base,2) ;
        A215244L(%) ;
    end if;
end proc: # R. J. Mathar, Aug 07 2012
# Caution: the last procedure applies A215244L to the reverse of the binary expansion of n, which is perfectly OK but might cause a problem if the procedure was used in a different problem. - N. J. A. Sloane, Aug 11 2012

palQ[L_] := SameQ[L, Reverse[L]];
b[L_] := b[L] = Module[{a = palQ[L] // Boole, c}, For[c = 1, c < Length[L], c++, If[palQ[L[[;;c]]], a = a + b[L[[c+1;;]]]]]; a];
a[n_] := If[n == 1, 1, b[IntegerDigits[n, 2]]];
a /@ Range[0, 106] (* Jean-François Alcover, Oct 27 2019 *)

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

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

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).

			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.

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

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

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 *)

A206924
"Calculates a(n)"
^self A006995 A206925

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

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

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.

			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).

Hieronymus Fischer, Table of n, a(n) for n = 1..2000

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

Cf. A217100, A217101.

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]

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)).

A206923 Number of bisections of the n-th binary palindrome bit pattern until the result is not palindromic.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 3, 1, 2, 1, 4, 1, 2, 1, 4, 1, 2, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Hieronymus Fischer, Mar 12 2012

Let k=1, p(1)=A006995(n) and m(1)=number of bits in p(1); if p(k) is a binary palindrome > 1 then iterate k=k+1, m(k)=floor((m(k-1)+1)/2), p(k)=leftmost m(k) bits of p(k-1); else set a(n)=k endif.

			a(1)=a(2)=1, since A006995(1)=0 and A006995(2)=1;
a(5)=3, since A006995(5)=7=111_2 and so the iteration is 11==>11==>1;
a(9)=2, since A006995(9)=21=10101_2 and so the iteration is 10101==>101;
a(13)=2, since A006995(13)=45=101101_2 and so the iteration is 101101==>101;
a(15)=4, since A006995(15)=63=111111_2 and so the iteration is 111111==>111==>11==>1;
a(37)=3, since A006995(37)=341=101010101_2 and so the iteration is 101010101==>10101==>101;

A006995, A206915, A178225, A154809, A206924, A206925.

/* quasi-C program fragment, omitting formal details, n>1 */
p=n;
p1=n+1;
k=0;
while (A178225(p)==1) && (p != p1)
{
  p1=p;
  k++;
  m=int(log(p)/log(2));
  p=int(p/2^int((m+1)/2));
}
return k;

Recursion: define f(x)=floor(A006995(x)/2^floor(floor(log_2(A006995(x))+1)/2)), for x=1,2,3,...
  Case 1: a(n)=1+a(A206915(f(n))), if f(n) is a binary palindrome;
  Case 2: a(n)=1, else.
Formally: a(n)=if (A178225(f(n))==1) then a(A206915(f(n)))+1 else 1.

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

Hieronymus Fischer, Jan 23 2013

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).

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..1000

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

"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]

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<=lWith 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

Hieronymus Fischer, Mar 24 2012

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')

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,1,-2).

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

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)).

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

Hieronymus Fischer, Jan 23 2013

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).

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..1000

Cf. A006995, A206923, A206924, A206925, A206926, A070939, A217101.

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

Hieronymus Fischer, Jan 23 2013

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.

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..300

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

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

Hieronymus Fischer, Feb 10 2013

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.

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..500

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

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

Hieronymus Fischer, Jan 23 2013

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.

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..500

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

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)).

Showing 1-10 of 10 results.

cp's OEIS Frontend

A215244 a(n) is the number of ways of writing the binary expansion of n as a product (or concatenation) of palindromes.

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

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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).

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A206923 Number of bisections of the n-th binary palindrome bit pattern until the result is not palindromic.

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

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

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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.

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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.

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