A006995 -id:A006995 - cp's OEIS frontend

A178225 Characteristic function of A006995 (binary palindromes).

1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0

Offset: 0

Jeremy Gardiner, May 23 2010

a(n)=1 if n is in A006995, a(n)=0 otherwise.
For n<43, identical to parity of A175096.
Comment by Franklin T. Adams-Watters: (Start)
Any permutation of the runs of n gives another such permutation when reversed. This pairs up all non-palindromic permutations of the runs of n. Thus the parity of A175096(n) is the parity of the number of palindromic run-permutations of n. For small n, this is 1 when n is a binary palindrome, and 0 otherwise.
The first exception is 43, binary 101011, which has a nontrivial palindromic run-permutation 45, binary 101101. Another kind of exception occurs first for n = 365, binary 101101101, which is a palindrome, but has another palindromic run-permutation 427, binary 110101011. (End)
Given an index n such that a(n)=1, then the following A164126(A206915(n))-1 terms will be 0. n'=A164126(A206915(n)) is the next term with a(n')=1. Therefore, if we subtract 1 from each term of A164126, we get the sequence of run lengths of 0's. - Hieronymus Fischer, Feb 19 2012.
Given an index n such that a(n)=0, then p=A206913(n) is the greatest index pA206914(n) is the least index q>n such that a(q)=1, which implies a(k)=0 for all k with n<=kHieronymus Fischer, Feb 19 2012.
Binary palindromes are distributed symmetrically with respect to threefold multiples of powers of 2. This becomes obvious by the generating function g(x) below. Example for the resulting factors of x^(3*2^5)=x^96: the factors are x^q and x^(-q) for q=3,11,23,31. Thus, the palindromes are 96+3, 96-3, 96+11, 96-11, 96+23, 96-23, 96+31, 96-31. The respective number of palindromes with this property is 2^(floor(m/2)), where m is the exponent of the corresponding power of 2. - Hieronymus Fischer, Apr 04 2012

			a(3)=1, since 3 is binary palindromic;
a(4)=0, since 4 is not palindromic.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for characteristic functions

Cf. A006995, A175096, A178226

Cf. A136522. See A206915 for the partial sums.

a178225 n = fromEnum $ n == a030101 n  -- Reinhard Zumkeller, Oct 21 2011

A178225[n_]:=Boole[PalindromeQ[IntegerDigits[n,2]]];
Array[A178225,100,0] (* Paolo Xausa, Oct 15 2023 *)

a(n) = my(b=binary(n)); b == Vecrev(b); \\ Michel Marcus, Feb 13 2019

a187225 = lambda n: int(bin(n)[2:] == bin(n)[:1:-1]) # David Radcliffe, May 05 2023

a(A006995(n)) = 1; a(A154809(n)) = 0. - Reinhard Zumkeller, Oct 21 2011
a(n) = if A030101(n) = n then 1, otherwise 0. - Reinhard Zumkeller, Jan 17 2012
a(n) = 1 - (A206916(n) - A206915(n)). - Hieronymus Fischer, Feb 18 2012
G.f.: g(x) = 1 + x + x^3 + Sum{j>=1} x^(3*2^j)*(f_j(x)+f_j(1/x)), where the f_j(x) are defined as follows:
  f_1(x)=x, and for j > 1,
  f_j(x) = x^3*Product_{k=1..floor((j-1)/2)} (1+x^b(j,k)), where b(j,k) = 2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k > 1, and b(j,1) = (2+(-1)^j)*2^(floor((j-1)/2)+1). The first explicit terms of this g.f. are
  g(x) = 1 + x + x^3 + (f_1(x) + f_1(1/x))*x^6 + (f_2(x) + f_2(1/x))*x^12 + (f_3(x)+f_3(1/x))*x^24 + (f_4(x) + f_4(1/x))*x^48 + (f_5(x) + f_5(1/x))*x^96 + ... = 1 + x + x^3 + (x+1/x)*x^6 + (x^3+1/x^3)*x^12 + (x^3*(1+x^4) + (1+1/x^4)/x^3)*x^24 + (x^3*(1+x^12) + (1+1/x^12)/x^3)*x^48 + (x^3*(1+x^8)(1+x^20) + (1+1/x^20)(1+1/x^8)/x^3)*x^96 + ...  - Hieronymus Fischer, Apr 02 2012

A206915 The index (in A006995) of the greatest binary palindrome <= n; also the 'lower inverse' of A006995.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 16, 16, 16, 16, 16, 16, 16

Offset: 0

Hieronymus Fischer, Feb 15 2012

The greatest m such that A006995(m)<= n;
The number of binary palindromes <= n;
n is palindromic iff a(n)=A206916(n);
a(n) is the number of the binary palindrome A206913(n);
if n is a binary palindrome, then A006995(a(n))=n, so a(n) is 'inverse' with respect to A006995.
Partial sums of the binary palindromic characteristic function A178225.

			a(1)=2 since 2 is the index number of the greatest binary palindrome <= 1;
a(5)=4 since there are only 4 binary palindromes (namely 0,1,3 and 5) which are less than or equal to 5;
a(10)=6 since A006995(6)=9<=10, but A006995(7)=15>10, and so that, 6 is the index number of greatest binary palindrome <= 10;

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A006995, A206914, A206916, A206920.

A178225[n_]:=Boole[PalindromeQ[IntegerDigits[n,2]]];
Accumulate[Array[A178225,100,0]] (* Paolo Xausa, Oct 15 2023 *)

def A206915(n):
    l = n.bit_length()
    k = l+1>>1
    return (n>>l-k)-(int(bin(n)[k+1:1:-1] or '0',2)>(n&(1<Chai Wah Wu, Jul 24 2024

a(n) = max(m | A006995(m) <= n);
a(A006995(n)) = n;
A006995(a(n)) <= n, equality holds true iff n is a binary palindrome;
Let p = A206913(n), m = floor(log_2(p)) and p>2, then:
a(n) = (((5-(-1)^m)/2) + sum_{k=1..floor(m/2)} (floor(p/2^k) mod 2)/2^k)) * 2^floor(m/2).
a(n) = (1/2)*((6-(-1)^m)*2^floor(m/2) - 1 - sum_{k=1..floor(m/2)} (-1)^floor(p/2^k) * 2^(floor(m/2)-k))).
a(n) = (5-(-1)^m) * 2^floor(m/2)/2 - 3*sum_{k=2..floor(m/2)} (floor(p/2^k) * 2^floor(m/2)/2^k) + (floor(p/2) * 2^floor(m/2)/2 - 2*floor((p/2) * 2^floor(m/2)) * floor((m-1)/m+1/2).
Partial sums S(n) = sum_{k=0..n} a(k):
S(n) = (n+1)*a(n) - A206920(a(n)).
G.f.: g(x) = (1+x+x^3+sum_{j>=1} x^(3*2^j)*(f_j(x)+f_j(1/x)))/(1-x), where the f_j(x) are defined as follows:
  f_1(x) = x, and for j>1,
  f_j(x) = x^3*product_{k=1..floor((j-1)/2)} (1+x^b(j,k)), where b(j,k)=2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k>1, and b(j,1)=(2+(-1)^j)*2^(floor((j-1)/2)+1).

A044051 a(n) = (s(n)+1)/2, where s=A006995 (base-2 palindromes).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 11, 14, 16, 17, 23, 26, 32, 33, 37, 43, 47, 50, 54, 60, 64, 65, 77, 83, 95, 98, 110, 116, 128, 129, 137, 149, 157, 163, 171, 183, 191, 194, 202, 214, 222, 228, 236, 248, 256, 257, 281, 293, 317, 323, 347, 359, 383

Offset: 1

Melia Aldridge, ma38(AT)spruce.evansville.edu

A141707(a(n)) = 1. - Reinhard Zumkeller, Apr 20 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A006995, A141707.

a044051 = (`div` 2) . (+ 1) . a006995 . (+ 1)
-- Reinhard Zumkeller, Apr 20 2015

Let n >= 3, m=floor(log_2(n)), p=floor((3*2^(m-1)-1)/n); then a(n) = 2^(2*m-2-p) + 1 + p*(1-(-1)^n)*2^(m-1) + (1/2)*Sum_{k=1..m-1-p} (floor((n - (3-p)*2^(m-1))/2^(m-1-k)) mod 2)*(2^k + 2^(2*m-1-p-k)). - Hieronymus Fischer, Feb 18 2012

A206920 Sum of the first n binary palindromes; a(n) = Sum_{k=1..n} A006995(k).

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 40, 57, 78, 105, 136, 169, 214, 265, 328, 393, 466, 551, 644, 743, 850, 969, 1096, 1225, 1378, 1543, 1732, 1927, 2146, 2377, 2632, 2889, 3162, 3459, 3772, 4097, 4438, 4803, 5184, 5571, 5974, 6401, 6844, 7299, 7770, 8265, 8776, 9289, 9850

Offset: 1

Hieronymus Fischer, Feb 18 2012

Different from A206919.

			a(1) = 0, since A006995(1) = 0;
a(4) = 9, since the sum of the first 4 binary palindromes is 9 = 0 + 1 + 3 + 5.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A006995, A206918, A206919.

a206920 n = a206920_list !! (n-1)
a206920_list = scanl1 (+) a006995_list
-- Reinhard Zumkeller, Feb 27 2012

Accumulate@ Map[FromDigits[#, 2] &, Select[Array[IntegerDigits[#, 2] &, 600, 0], PalindromeQ]] (* Michael De Vlieger, Feb 20 2018 *)

Let n > 3, p = A006995(n), m = floor(log_2(p)), then
a(n) = (8/7)*((3/4)*(4 - (-1)^m)/(3 + (-1)^m)*2^(3*floor(m/2)) - 1) + (floor(p/2^floor(m/2)) mod 2)*p + 2^m + 1 + Sum_{k = 1..(floor(m/2) - 1)} (floor(p/2^k) mod 2)*(2^k + 2^(m-k) + 2^(m-floor(m/2) + 1)*(4^(floor(m/2) - k - 1) - 1) + (2 - (-1)^m)*2^floor(m/2) + 2^(floor(m/2) - k)*(p - floor((p mod (2^(m-k+1)))/2^k)*2^k)). - [Corrected; missing factor to the sum term (2 -(-1)^m) pasted by the author, Sep 07 2018]
From Hieronymus Fischer, Sep 07 2018: (Start)
Slightly simplified and better readable formula:
a(n) = A_m + B_m + 2^m + 1 + Sum_{k = 1..(m2-1)} C_k*(D_k + E_k + F_k + G_k),
where m2 = floor(m/2),
A_m = (8/7)*((3/4)*(4-(-1)^m)/(3+(-1)^m)*2^(3*m2)-1),
B_m = p*(floor(p/2^m2) mod 2),
C_k = floor(p/2^k) mod 2,
D_k = 2^k + 2^(m-k),
E_k = 2^(m-m2+1)*(4^(m2-k-1)-1),
F_k = (2 - (-1)^m)*2^m2,
G_k = 2^(m2-k) * (p - p mod (2^(m-k+1)) + p mod 2^k). (End)
G.f.: g(x) = (x^2 + 3x^3 + Sum_{j >= 1} (3*2^j*(1 - x^floor((j+1)/2))/(1-x)*x^((1/2) - floor((j+1)/2)) + f_j(x) - f_j(1/x))*x^(2*2^floor(j/2) + 3*2^floor((j-1)/2) - (1/2)))/(1-x), where the f_j(x) are the same as defined for the g.f. of A006995.

A206916 Index of the least binary palindrome >=n; also the "upper inverse" of A006995.

Original entry on oeis.org

1, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 17, 17, 17, 17, 17

Offset: 0

Hieronymus Fischer, Feb 17 2012

The least m such that A006995(m)>=n;
n is palindromic iff a(n)=A206915(n);
a(n) is the number of the binary palindrome A206914(n);
if n is a binary palindrome, then A006995(a(n))=n, so a(n) is 'inverse' with respect to A006995

			a(2)=3 since 3 is the index number of the least binary palindrome >= 2;
a(5)=4 since 4 is the index number of the least binary palindrome >= 5;
a(10)=7 since A006995(7)=15>=10, but A006995(6)=9<10, and so that, 7 is the index number of least binary palindrome >= 10;

Cf. A006995, A206920.

def A206916(n):
    l = n.bit_length()
    k = l+1>>1
    return (n>>l-k)+(int(bin(n)[k+1:1:-1] or '0',2)<(n&(1<Chai Wah Wu, Jul 24 2024

a(n)=min(m|A006995(m)>=n);
a(A006995(n))=n;
A006995(a(n))>=n, equality holds true iff n is a binary palindrome;
Let p=A206914(n), m=floor(log_2(p)) and p>2, then:
a(n)=(((5-(-1)^m)/2) + sum_{k=1..floor(m/2)} (floor(p/2^k) mod 2)/2^k))*2^floor(m/2);
a(n)=(1/2)*((6-(-1)^m)*2^floor(m/2)-1-sum_       {k=1..floor(m/2)} (-1)^floor(p/2^k)*2^(floor(m/2)-k)));
a(n)=(5-(-1)^m)*2^floor(m/2)/2-3*sum_{k=2..floor(m/2)} floor(p/2^k)*2^floor(m/2)/2^k)+(floor(p/2)*2^floor(m/2)/2-2*floor((p/2)*2^floor(m/2))*floor((m-1)/m+1/2).
Partial sums S(n) = sum_{k=0..n} a(k):
S(n) = 1+n*a(n)-A206920(a(n)-1), valid for n>0.
G.f.: g(x)=(x+x^2+x^3+sum_{j=1..infinity} x^(3*2^j)*(f_j(x)+f_j(1/x)))/(x(1-x)), where the f_j(x) are defined as follows:
  f_1(x)=x, and for j>1,
  f_j(x)=x^3*product_{k=1..floor((j-1)/2)} (1+x^b(j,k)), where b(j,k)=2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k>1, and b(j,1)=(2+(-1)^j)*2^(floor((j-1)/2)+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.

A164126 First differences of A006995.

Original entry on oeis.org

1, 2, 2, 2, 2, 6, 2, 4, 6, 4, 2, 12, 6, 12, 2, 8, 12, 8, 6, 8, 12, 8, 2, 24, 12, 24, 6, 24, 12, 24, 2, 16, 24, 16, 12, 16, 24, 16, 6, 16, 24, 16, 12, 16, 24, 16, 2, 48, 24, 48, 12, 48, 24, 48, 6, 48, 24, 48, 12, 48, 24, 48, 2, 32, 48, 32, 24, 32, 48, 32, 12, 32, 48, 32, 24, 32, 48, 32, 6

Offset: 1

Vladimir Joseph Stephan Orlovsky, Aug 10 2009

Contribution from Hieronymus Fischer, Feb 18 2012: (Start)
From the formula section it follows that a(2^m - 1 + 2^(m-1) - k) = a(2^m - 1 + k) for 0 <= k <= 2^(m-1), as well as a(2^m - 1 + 2^(m-1) - k) = 2 for k=0, 2^(m-1) and a(2^m - 1 + 2^(m-1) - k) = 6 for k=2^(m-2), hence, starting from positions n=2^m-1, the following 2^(m-1) terms form symmetric tuples limited on the left and on the right by a '2' and always having a '6' as the center element.
Example: for n = 15 = 2^4 - 1, we have the (2^3+1)-tuple (2,8,12,8,6,8,12,8,2).
Further on, since a(2^m - 1 + 2^(m-1) + k) = a(2^(m+1) - 1 - k) for 0 <= k <= 2^(m-1) an analogous statement holds true for starting positions n = 2^m + 2^(m-1) - 1.
Example: for n = 23 = 2^4 + 2^3 - 1, we find the (2^3+1)-tuple (2,24,12,24,6,24,12,24,2).
If we group the sequence terms according to the value of m=floor(log_2(n)), writing those terms together in separate lines and opening each new line for n >= 2^m + 2^(m-1), then a kind of a 'logarithmic shaped' cone end will be formed, where both the symmetry and the calculation rules become obvious. The first 63 terms are depicted below:
                          1
                          2
                          2
                        2  2
                        6  2
                      4 6  4  2
                     12 6 12  2
                8 12  8 6  8 12  8 2
               24 12 24 6 24 12 24 2
   16 24 16 12 16 24 16 6 16 24 16 12 16 24 16 2
   48 24 48 12 48 24 48 6 48 24 48 12 48 24 48 2
.
(End)
Decremented by 1, also the sequence of run lengths of 0's in A178225. - Hieronymus Fischer, Feb 19 2012

			a(1) = A006995(2) - A006995(1) = 1 - 0 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A006995, A164124, A029971, A016041, A164125.

See A178225.

f[n_]:=FromDigits[RealDigits[n,2][[1]]]==FromDigits[Reverse[RealDigits[n, 2][[1]]]]; a=1;lst={};Do[If[f[n],AppendTo[lst,n-a];a=n],{n,1,8!,1}]; lst

def A164126(n):
    if n == 1: return 1
    m = (a:=1<<(l:=n.bit_length()-2))|(n&a-1)
    k = (m<Chai Wah Wu, Jun 11 2024

a(n) = A006995(n+1) - A006995(n).
Contribution from Hieronymus Fischer, Feb 17 2012: (Start)
a(4*2^m - 1) = a(6*2^m - 1) = 2;
a(5*2^m - 1) = a(7*2^m - 1) = 6 (for m > 0);
Let m = floor(log_2(n)), then
  Case 1: a(n) = 2, if n+1 = 2^(m+1) or n+1 = 3*2^(m-1);
  Case 2: a(n) = 2^(m-1), if n = 0(mod 2) and n < 3*2^(m-1);
  Case 3: a(n) = 3*2^(m-1), if n = 0(mod 2) and n >= 3*2^(m-1);
  Case 4: a(n) = 3*2^(m-1)/gcd(n+1-2^m, 2^m), otherwise.
Cases 2-4 above can be combined as
  Case 2': a(n) = (2 - (-1)^(n-(n-1)*floor(2*n/(3*2^m))))*2^(m-1)/gcd(n+1-2^m, 2^m).
Recursion formula:
Let m = floor(log_2(n)); then
  Case 1: a(n) = 2*a(n-2^(m-1)), if 2^m <= n < 2^m + 2^(m-2) - 1;
  Case 2: a(n) = 6, if n = 2^m + 2^(m-2) - 1;
  Case 3: a(n) = a(n-2^(m-2)), if 2^m + 2^(m-2) <= n < 2^m + 2^(m-1) - 1;
  Case 4: a(n) = 2, if n = 2^m + 2^(m-1) - 1;
  Case 5: a(n) = (2 + (-1)^n)*a(n-2^(m-1)), otherwise (which means 2^m + 2^(m-1) <= n < 2^(m+1)).
(End)

a(1) changed to 1 and keyword:base added by R. J. Mathar, Aug 26 2009

A261679 Number of ordered pairs (u,v) of binary palindromes (see A006995) with u+v=n.

Original entry on oeis.org

1, 2, 1, 2, 2, 2, 3, 2, 4, 2, 5, 0, 4, 0, 3, 2, 4, 2, 5, 0, 4, 2, 6, 0, 6, 0, 4, 2, 4, 0, 5, 2, 6, 2, 7, 0, 8, 0, 6, 0, 4, 0, 5, 0, 2, 2, 4, 0, 8, 0, 4, 2, 6, 0, 7, 0, 2, 0, 4, 0, 6, 0, 3, 2, 4, 2, 9, 0, 6, 0, 4, 0, 8, 2, 4, 0, 4, 0, 8, 0, 6, 0, 6, 0, 4, 2, 4, 0, 4

Offset: 0

N. J. A. Sloane, Sep 04 2015

			8=1+7=3+5=5+3=7+1, so a(8)=4.

N. J. A. Sloane, Table of n, a(n) for n = 0..9999

Cf. A006995, A261680.

For zeros see A241491, A261678.

G.f. = (Sum_{p in A006995} x^p)^2.

A261680 Number of ordered quadruples (u,v,w,x) of binary palindromes (see A006995) with u+v+w+x=n.

Original entry on oeis.org

1, 4, 6, 8, 13, 16, 22, 28, 34, 44, 50, 60, 59, 72, 70, 80, 92, 88, 114, 96, 125, 104, 152, 120, 172, 144, 188, 152, 215, 144, 242, 160, 272, 172, 302, 180, 329, 216, 352, 240, 388, 228, 430, 228, 442, 212, 476, 192, 506, 228, 496, 248, 540, 252, 582, 276, 592

Offset: 0

N. J. A. Sloane, Sep 04 2015

Conjecture: a(n)>0: every number is the sum of four binary palindromes. (Compare A261422, A261675.)

N. J. A. Sloane, Table of n, a(n) for n = 0..9999
Aayush Rajasekaran, Jeffrey Shallit, and Tim Smith, Sums of Palindromes: an Approach via Nested-Word Automata, preprint arXiv:1706.10206 [cs.FL], June 30 2017.

Cf. A006995, A261422, A261675, A261679.

G.f. = (Sum_{p in A006995} x^p)^4.

A305468 Positive integers that can be expressed as the quotient of two binary palindromic numbers (that is, terms of A006995).

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 27, 31, 33, 39, 43, 45, 51, 53, 55, 57, 61, 63, 65, 71, 73, 77, 79, 83, 85, 91, 93, 95, 99, 107, 109, 117, 119, 121, 127, 129, 133, 143, 149, 151, 153, 157, 159, 163, 165, 171, 173, 179, 181, 187, 189, 191, 195, 203, 205

Offset: 1

Jeffrey Shallit, Jun 02 2018

			79 is in the sequence because 888987 and 11253 are both binary palindromes, and 79 = 888987/11253.  These are in fact the smallest such numbers for 79.

Joseph Meleshko, Table of n, a(n) for n = 1..1271
James Haoyu Bai, Joseph Meleshko, Samin Riasat, and Jeffrey Shallit, Quotients of Palindromic and Antipalindromic Numbers, arXiv:2202.13694 [math.NT], 2022.
James Haoyu Bai, Joseph Meleshko, Samin Riasat, and Jeffrey Shallit, Quotients of Palindromic and Antipalindromic Numbers, INTEGERS 22 (2022), #A96.

Cf. A006995, A305469, A305470.

Showing 1-10 of 233 results. Next

cp's OEIS Frontend

A178225 Characteristic function of A006995 (binary palindromes).

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A206915 The index (in A006995) of the greatest binary palindrome <= n; also the 'lower inverse' of A006995.

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A044051 a(n) = (s(n)+1)/2, where s=A006995 (base-2 palindromes).

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A206920 Sum of the first n binary palindromes; a(n) = Sum_{k=1..n} A006995(k).

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A206916 Index of the least binary palindrome >=n; also the "upper inverse" of A006995.

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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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A164126 First differences of A006995.

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